Chapter 7: System of Particles and Rotational Motion

Introduction to Rigid Body Motion

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Motion Types of Rigid Bodies

Key Concepts

• Pure translation: All particles have same velocity v and acceleration α; α₁ = α₂ = α₃.
• Combination of translation and rotation: Velocities and accelerations differ; α₁ ≠ α₂ ≠ α₃.
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration α₁ = α₂ = α₃ = 2 m/s².

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles α₁ ≠ α₂ ≠ α₃ indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Key Concepts

• Pure translation: All particles have same velocity v and acceleration α; α₁ = α₂ = α₃.
• Combination of translation and rotation: Velocities and accelerations differ; α₁ ≠ α₂ ≠ α₃.
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration α₁ = α₂ = α₃ = 2 m/s².

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles α₁ ≠ α₂ ≠ α₃ indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Key Concepts

• Pure translation: All particles have same velocity v and acceleration α; α₁ = α₂ = α₃.
• Combination of translation and rotation: Velocities and accelerations differ; α₁ ≠ α₂ ≠ α₃.
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration α₁ = α₂ = α₃ = 2 m/s².

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles α₁ ≠ α₂ ≠ α₃ indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Key Concepts

• Pure translation: All particles have same velocity v and acceleration α; α₁ = α₂ = α₃.
• Combination of translation and rotation: Velocities and accelerations differ; α₁ ≠ α₂ ≠ α₃.
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration α₁ = α₂ = α₃ = 2 m/s².

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles α₁ ≠ α₂ ≠ α₃ indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle θ=30°.
• Find the component of gravitational force down the slope: P₁ = mg sin θ = 5 × 9.8 × sin 30° = 24.5 N.

Ex 2

• Point P₁ at radius r₁=0.2 m rotates about z-axis with angular velocity ω=10, rad/s.
• Calculate linear speed: v = rω = 0.2 × 10 = 2 m/s.

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Key Concepts

• Rigid body: Body with fixed shape; distances between particles remain constant.
• Pure translational motion: All particles move with the same velocity at any instant.
• Example: Block sliding down an inclined plane with forces P₁ (gravity component) and P₂ (normal force).
• Rotational motion: Body rotates about a fixed axis; particles move in circles perpendicular to the axis.
• Radius of rotation: Perpendicular distance of a point from the axis.
• Axis of rotation may be fixed or moving (e.g., spinning top with precession).
• Motion types: Pure translation, pure rotation (pivoted), or combination of both.

Solved Examples

Ex 1

• Block of mass 5 kg slides down an inclined plane at angle \(\theta=30^\circ\).
• Find the component of gravitational force down the slope: \(P_1 = mg \sin \theta = 5 \times 9.8 \times \sin 30^\circ = 24.5\,\text{N}\).

Ex 2

• Point \(P_1\) at radius \(r_1=0.2\,m\) rotates about z-axis with angular velocity \(\omega=10\,\text{rad/s}\).
• Calculate linear speed: \(v = r \omega = 0.2 \times 10 = 2\,\text{m/s}\).

Practice Set

Q1

In pure translational motion, the velocity of all particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

A rigid body rotates about a fixed axis. The path of each particle is:

• A) Straight line
• B) Ellipse
• C) Circle
• D) Parabola

Q3

A spinning top's axis sweeps out a cone. This motion is called:

• A) Precession
• B) Nutation
• C) Translation
• D) Rotation

Answer Key

• Q1: C - All particles have the same velocity in pure translation.
• Q2: C - Particles move in circles around the axis in rotation.
• Q3: A - The axis movement around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.
• Translation = same velocity for all particles.
• Rotation = particles move in circles about axis.
• Precession = axis sweeps a cone (spinning top).

Motion Types of Rigid Bodies

Key Concepts

• Pure translation: All particles have same velocity v and acceleration α; α₁ = α₂ = α₃.
• Combination of translation and rotation: Velocities and accelerations differ; α₁ ≠ α₂ ≠ α₃.
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration α₁ = α₂ = α₃ = 2 m/s².

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles α₁ ≠ α₂ ≠ α₃ indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Key Concepts

• Pure translation: All particles have same velocity v and acceleration α; α₁ = α₂ = α₃.
• Combination of translation and rotation: Velocities and accelerations differ; α₁ ≠ α₂ ≠ α₃.
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration α₁ = α₂ = α₃ = 2 m/s².

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles α₁ ≠ α₂ ≠ α₃ indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Key Concepts

• Pure translation: All particles have same velocity v and acceleration α; α₁ = α₂ = α₃.
• Combination of translation and rotation: Velocities and accelerations differ; α₁ ≠ α₂ ≠ α₃.
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration α₁ = α₂ = α₃ = 2 m/s².

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles α₁ ≠ α₂ ≠ α₃ indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Key Concepts

• Pure translation: All particles have same velocity \(v\) and acceleration \(\alpha\); \(\alpha_1 = \alpha_2 = \alpha_3\).
• Combination of translation and rotation: Velocities and accelerations differ; \(\alpha_1 \neq \alpha_2 \neq \alpha_3\).
• Pivoted/fixed rigid body motion is pure rotation.

Solved Examples

Ex 1

• Rigid body moves with pure translation: velocity of all points = 5 m/s.
• Acceleration \(\alpha_1 = \alpha_2 = \alpha_3 = 2\,\text{m/s}^2\).

Ex 2

• Rigid body moves with combined motion; velocity at point O = 3 m/s, at point P = 5 m/s.
• Angles \(\alpha_1 \neq \alpha_2 \neq \alpha_3\) indicate rotation plus translation.

Practice Set

Q1

In pure translational motion, acceleration of particles is:

• A) Different
• B) Zero
• C) Same
• D) Varies with position

Q2

In combined translation and rotation, velocities of points O and P are:

• A) Equal
• B) Different
• C) Zero
• D) Constant

Q3

Which motion occurs if a rigid body is pivoted?

• A) Pure translation
• B) Combination of translation and rotation
• C) Pure rotation
• D) No motion

Answer Key

• Q1: C - All particles have same acceleration in pure translation.
• Q2: B - Velocities differ in combined motion.
• Q3: C - Pivoted body undergoes pure rotation.

Memory Aids

• Pure translation: equal velocity & acceleration.
• Combined motion: different velocities & angles.
• Pivoted body = pure rotation.

Centre of Mass

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance d:
• Position of CM from m₁: x_CM = m₂/(m₁ + m₂) d
• General formula with coordinates x₁, x₂: x_CM = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)
• For n particles with masses m_i and position vectors vector r_i:
• vector R_CM = (∑ m_i vector r_i)/(∑ m_i) = (1/M) ∑ m_i vector r_i, where M = ∑ m_i
• Coordinates of CM in 3D:
• x_CM = (∑ m_i x_i)/M, y_CM = (∑ m_i y_i)/M, z_CM = (∑ m_i z_i)/M
• For continuous mass distribution (rigid body):
• x_CM = (1/M) ∫ x , dm, y_CM = (1/M) ∫ y , dm, z_CM = (1/M) ∫ z , dm

Solved Examples

Ex 1

• Two particles: m₁=2 kg, m₂=3 kg, distance d=5 m.
• Calculate x_CM = 3/(2+3) × 5 = 3 m from m₁.

Ex 2

• Three particles at x₁=0, x₂=4, x₃=6 m with masses 1, 2, 3 kg.
• Calculate x_CM = (1×0 + 2×4 + 3×6)/(1+2+3) = 26/6 = 4.33 m.

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) (∑ vector r_i)/n
• B) (∑ m_i vector r_i)/(∑ m_i)
• C) ∑ m_i vector r_i
• D) (∑ m_i)/(∑ vector r_i)

Q3

For a continuous mass distribution, the coordinate x_CM is given by:

• A) ∫ x , dm
• B) (1/M) ∫ x , dm
• C) (1/m) ∫ x , dm
• D) ∫ x , dx

Answer Key

• Q1: B - x_CM = 3/5 × 5 = 3 m from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass M.

Memory Aids

• CM = weighted average position of mass.
• Two particles: x_CM = m₂/(m₁ + m₂) d.
• n particles: vector R_CM = (∑ m_i vector r_i)/(∑ m_i).
• Continuous mass: use integrals (1/M) ∫ x , dm.

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance d:
• Position of CM from m₁: x_CM = m₂/(m₁ + m₂) d
• General formula with coordinates x₁, x₂: x_CM = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)
• For n particles with masses m_i and position vectors vector r_i:
• vector R_CM = (∑ m_i vector r_i)/(∑ m_i) = (1/M) ∑ m_i vector r_i, where M = ∑ m_i
• Coordinates of CM in 3D:
• x_CM = (∑ m_i x_i)/M, y_CM = (∑ m_i y_i)/M, z_CM = (∑ m_i z_i)/M
• For continuous mass distribution (rigid body):
• x_CM = (1/M) ∫ x , dm, y_CM = (1/M) ∫ y , dm, z_CM = (1/M) ∫ z , dm

Solved Examples

Ex 1

• Two particles: m₁=2 kg, m₂=3 kg, distance d=5 m.
• Calculate x_CM = 3/(2+3) × 5 = 3 m from m₁.

Ex 2

• Three particles at x₁=0, x₂=4, x₃=6 m with masses 1, 2, 3 kg.
• Calculate x_CM = (1×0 + 2×4 + 3×6)/(1+2+3) = 26/6 = 4.33 m.

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) (∑ vector r_i)/n
• B) (∑ m_i vector r_i)/(∑ m_i)
• C) ∑ m_i vector r_i
• D) (∑ m_i)/(∑ vector r_i)

Q3

For a continuous mass distribution, the coordinate x_CM is given by:

• A) ∫ x , dm
• B) (1/M) ∫ x , dm
• C) (1/m) ∫ x , dm
• D) ∫ x , dx

Answer Key

• Q1: B - x_CM = 3/5 × 5 = 3 m from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass M.

Memory Aids

• CM = weighted average position of mass.
• Two particles: x_CM = m₂/(m₁ + m₂) d.
• n particles: vector R_CM = (∑ m_i vector r_i)/(∑ m_i).
• Continuous mass: use integrals (1/M) ∫ x , dm.

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance d:
• Position of CM from m₁: x_CM = m₂/(m₁ + m₂) d
• General formula with coordinates x₁, x₂: x_CM = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)
• For n particles with masses m_i and position vectors vector r_i:
• vector R_CM = (∑ m_i vector r_i)/(∑ m_i) = (1/M) ∑ m_i vector r_i, where M = ∑ m_i
• Coordinates of CM in 3D:
• x_CM = (∑ m_i x_i)/M, y_CM = (∑ m_i y_i)/M, z_CM = (∑ m_i z_i)/M
• For continuous mass distribution (rigid body):
• x_CM = (1/M) ∫ x , dm, y_CM = (1/M) ∫ y , dm, z_CM = (1/M) ∫ z , dm

Solved Examples

Ex 1

• Two particles: m₁=2 kg, m₂=3 kg, distance d=5 m.
• Calculate x_CM = 3/(2+3) × 5 = 3 m from m₁.

Ex 2

• Three particles at x₁=0, x₂=4, x₃=6 m with masses 1, 2, 3 kg.
• Calculate x_CM = (1×0 + 2×4 + 3×6)/(1+2+3) = 26/6 = 4.33 m.

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) (∑ vector r_i)/n
• B) (∑ m_i vector r_i)/(∑ m_i)
• C) ∑ m_i vector r_i
• D) (∑ m_i)/(∑ vector r_i)

Q3

For a continuous mass distribution, the coordinate x_CM is given by:

• A) ∫ x , dm
• B) (1/M) ∫ x , dm
• C) (1/m) ∫ x , dm
• D) ∫ x , dx

Answer Key

• Q1: B - x_CM = 3/5 × 5 = 3 m from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass M.

Memory Aids

• CM = weighted average position of mass.
• Two particles: x_CM = m₂/(m₁ + m₂) d.
• n particles: vector R_CM = (∑ m_i vector r_i)/(∑ m_i).
• Continuous mass: use integrals (1/M) ∫ x , dm.

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance d:
• Position of CM from m₁: x_CM = m₂/(m₁ + m₂) d
• General formula with coordinates x₁, x₂: x_CM = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)
• For n particles with masses m_i and position vectors vector r_i:
• vector R_CM = (∑ m_i vector r_i)/(∑ m_i) = (1/M) ∑ m_i vector r_i, where M = ∑ m_i
• Coordinates of CM in 3D:
• x_CM = (∑ m_i x_i)/M, y_CM = (∑ m_i y_i)/M, z_CM = (∑ m_i z_i)/M
• For continuous mass distribution (rigid body):
• x_CM = (1/M) ∫ x , dm, y_CM = (1/M) ∫ y , dm, z_CM = (1/M) ∫ z , dm

Solved Examples

Ex 1

• Two particles: m₁=2 kg, m₂=3 kg, distance d=5 m.
• Calculate x_CM = 3/(2+3) × 5 = 3 m from m₁.

Ex 2

• Three particles at x₁=0, x₂=4, x₃=6 m with masses 1, 2, 3 kg.
• Calculate x_CM = (1×0 + 2×4 + 3×6)/(1+2+3) = 26/6 = 4.33 m.

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) (∑ vector r_i)/n
• B) (∑ m_i vector r_i)/(∑ m_i)
• C) ∑ m_i vector r_i
• D) (∑ m_i)/(∑ vector r_i)

Q3

For a continuous mass distribution, the coordinate x_CM is given by:

• A) ∫ x , dm
• B) (1/M) ∫ x , dm
• C) (1/m) ∫ x , dm
• D) ∫ x , dx

Answer Key

• Q1: B - x_CM = 3/5 × 5 = 3 m from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass M.

Memory Aids

• CM = weighted average position of mass.
• Two particles: x_CM = m₂/(m₁ + m₂) d.
• n particles: vector R_CM = (∑ m_i vector r_i)/(∑ m_i).
• Continuous mass: use integrals (1/M) ∫ x , dm.

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance d:
• Position of CM from m₁: x_CM = m₂/(m₁ + m₂) d
• General formula with coordinates x₁, x₂: x_CM = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)
• For n particles with masses m_i and position vectors vector r_i:
• vector R_CM = (∑ m_i vector r_i)/(∑ m_i) = (1/M) ∑ m_i vector r_i, where M = ∑ m_i
• Coordinates of CM in 3D:
• x_CM = (∑ m_i x_i)/M, y_CM = (∑ m_i y_i)/M, z_CM = (∑ m_i z_i)/M
• For continuous mass distribution (rigid body):
• x_CM = (1/M) ∫ x , dm, y_CM = (1/M) ∫ y , dm, z_CM = (1/M) ∫ z , dm

Solved Examples

Ex 1

• Two particles: m₁=2 kg, m₂=3 kg, distance d=5 m.
• Calculate x_CM = 3/(2+3) × 5 = 3 m from m₁.

Ex 2

• Three particles at x₁=0, x₂=4, x₃=6 m with masses 1, 2, 3 kg.
• Calculate x_CM = (1×0 + 2×4 + 3×6)/(1+2+3) = 26/6 = 4.33 m.

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) (∑ vector r_i)/n
• B) (∑ m_i vector r_i)/(∑ m_i)
• C) ∑ m_i vector r_i
• D) (∑ m_i)/(∑ vector r_i)

Q3

For a continuous mass distribution, the coordinate x_CM is given by:

• A) ∫ x , dm
• B) (1/M) ∫ x , dm
• C) (1/m) ∫ x , dm
• D) ∫ x , dx

Answer Key

• Q1: B - x_CM = 3/5 × 5 = 3 m from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass M.

Memory Aids

• CM = weighted average position of mass.
• Two particles: x_CM = m₂/(m₁ + m₂) d.
• n particles: vector R_CM = (∑ m_i vector r_i)/(∑ m_i).
• Continuous mass: use integrals (1/M) ∫ x , dm.

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance d:
• Position of CM from m₁: x_CM = m₂/(m₁ + m₂) d
• General formula with coordinates x₁, x₂: x_CM = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)
• For n particles with masses m_i and position vectors vector r_i:
• vector R_CM = (∑ m_i vector r_i)/(∑ m_i) = (1/M) ∑ m_i vector r_i, where M = ∑ m_i
• Coordinates of CM in 3D:
• x_CM = (∑ m_i x_i)/M, y_CM = (∑ m_i y_i)/M, z_CM = (∑ m_i z_i)/M
• For continuous mass distribution (rigid body):
• x_CM = (1/M) ∫ x , dm, y_CM = (1/M) ∫ y , dm, z_CM = (1/M) ∫ z , dm

Solved Examples

Ex 1

• Two particles: m₁=2 kg, m₂=3 kg, distance d=5 m.
• Calculate x_CM = 3/(2+3) × 5 = 3 m from m₁.

Ex 2

• Three particles at x₁=0, x₂=4, x₃=6 m with masses 1, 2, 3 kg.
• Calculate x_CM = (1×0 + 2×4 + 3×6)/(1+2+3) = 26/6 = 4.33 m.

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) (∑ vector r_i)/n
• B) (∑ m_i vector r_i)/(∑ m_i)
• C) ∑ m_i vector r_i
• D) (∑ m_i)/(∑ vector r_i)

Q3

For a continuous mass distribution, the coordinate x_CM is given by:

• A) ∫ x , dm
• B) (1/M) ∫ x , dm
• C) (1/m) ∫ x , dm
• D) ∫ x , dx

Answer Key

• Q1: B - x_CM = 3/5 × 5 = 3 m from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass M.

Memory Aids

• CM = weighted average position of mass.
• Two particles: x_CM = m₂/(m₁ + m₂) d.
• n particles: vector R_CM = (∑ m_i vector r_i)/(∑ m_i).
• Continuous mass: use integrals (1/M) ∫ x , dm.

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance d:
• Position of CM from m₁: x_CM = m₂/(m₁ + m₂) d
• General formula with coordinates x₁, x₂: x_CM = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)
• For n particles with masses m_i and position vectors vector r_i:
• vector R_CM = (∑ m_i vector r_i)/(∑ m_i) = (1/M) ∑ m_i vector r_i, where M = ∑ m_i
• Coordinates of CM in 3D:
• x_CM = (∑ m_i x_i)/M, y_CM = (∑ m_i y_i)/M, z_CM = (∑ m_i z_i)/M
• For continuous mass distribution (rigid body):
• x_CM = (1/M) ∫ x , dm, y_CM = (1/M) ∫ y , dm, z_CM = (1/M) ∫ z , dm

Solved Examples

Ex 1

• Two particles: m₁=2 kg, m₂=3 kg, distance d=5 m.
• Calculate x_CM = 3/(2+3) × 5 = 3 m from m₁.

Ex 2

• Three particles at x₁=0, x₂=4, x₃=6 m with masses 1, 2, 3 kg.
• Calculate x_CM = (1×0 + 2×4 + 3×6)/(1+2+3) = 26/6 = 4.33 m.

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) (∑ vector r_i)/n
• B) (∑ m_i vector r_i)/(∑ m_i)
• C) ∑ m_i vector r_i
• D) (∑ m_i)/(∑ vector r_i)

Q3

For a continuous mass distribution, the coordinate x_CM is given by:

• A) ∫ x , dm
• B) (1/M) ∫ x , dm
• C) (1/m) ∫ x , dm
• D) ∫ x , dx

Answer Key

• Q1: B - x_CM = 3/5 × 5 = 3 m from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass M.

Memory Aids

• CM = weighted average position of mass.
• Two particles: x_CM = m₂/(m₁ + m₂) d.
• n particles: vector R_CM = (∑ m_i vector r_i)/(∑ m_i).
• Continuous mass: use integrals (1/M) ∫ x , dm.

Key Concepts

• Centre of mass (CM): Point where total mass is considered concentrated and external forces act.
• For two particles (m₁, m₂) separated by distance \(d\):
• Position of CM from m₁: \(x_{CM} = \frac{m_2}{m_1 + m_2} d\)
• General formula with coordinates \(x_1, x_2\): \(x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\)
• For n particles with masses \(m_i\) and position vectors \(\vec{r}_i\):
• \(\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i} = \frac{1}{M} \sum m_i \vec{r}_i\), where \(M = \sum m_i\)
• Coordinates of CM in 3D:
• \(x_{CM} = \frac{\sum m_i x_i}{M}, y_{CM} = \frac{\sum m_i y_i}{M}, z_{CM} = \frac{\sum m_i z_i}{M}\)
• For continuous mass distribution (rigid body):
• \(x_{CM} = \frac{1}{M} \int x \, dm, y_{CM} = \frac{1}{M} \int y \, dm, z_{CM} = \frac{1}{M} \int z \, dm\)

Solved Examples

Ex 1

• Two particles: \(m_1=2\,kg\), \(m_2=3\,kg\), distance \(d=5\,m\).
• Calculate \(x_{CM} = \frac{3}{2+3} \times 5 = 3\,m\) from \(m_1\).

Ex 2

• Three particles at \(x_1=0\), \(x_2=4\), \(x_3=6\) m with masses 1, 2, 3 kg.
• Calculate \(x_{CM} = \frac{1\times0 + 2\times4 + 3\times6}{1+2+3} = \frac{26}{6} = 4.33\,m\).

Practice Set

Q1

For two particles with masses 2 kg and 3 kg separated by 5 m, the CM from the 2 kg mass is:

• A) 2 m
• B) 3 m
• C) 4 m
• D) 5 m

Q2

In a system of particles, the position vector of CM is:

• A) \(\frac{\sum \vec{r}_i}{n}\)
• B) \(\frac{\sum m_i \vec{r}_i}{\sum m_i}\)
• C) \(\sum m_i \vec{r}_i\)
• D) \(\frac{\sum m_i}{\sum \vec{r}_i}\)

Q3

For a continuous mass distribution, the coordinate \(x_{CM}\) is given by:

• A) \(\int x \, dm\)
• B) \(\frac{1}{M} \int x \, dm\)
• C) \(\frac{1}{m} \int x \, dm\)
• D) \(\int x \, dx\)

Answer Key

• Q1: B - \(x_{CM} = \frac{3}{5} \times 5 = 3\,m\) from 2 kg mass.
• Q2: B - Weighted average position by mass.
• Q3: B - Integral divided by total mass \(M\).

Memory Aids

• CM = weighted average position of mass.
• Two particles: \(x_{CM} = \frac{m_2}{m_1 + m_2} d\).
• n particles: \(\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}\).
• Continuous mass: use integrals \(\frac{1}{M} \int x \, dm\).

Motion of Centre of Mass

Definition and Formula

• Consider a system of n particles with masses m₁, m₂, ..., m_n.
• Position vectors of particles are vector r₁, vector r₂, ..., vector r_n.
• Position vector of centre of mass (CM) is vector R.
• Formula: Mvector R = m₁vector r₁ + m₂vector r₂ + ... + m_nvector r_n, where M = ∑ m_i.
• Differentiating w.r.t time gives velocity relation: Mvector V_CM = m₁vector v₁ + m₂vector v₂ + ... + m_nvector v_n.
• Acceleration relation: Mvector A_CM = m₁vector a₁ + m₂vector a₂ + ... + m_nvector a_n.
• Newton's 2nd law for CM: Mvector A_CM = vector F_ext (only external forces affect CM motion).
• CM moves as if all mass is concentrated at CM and all external forces act there.

Solved Examples

• Ex 1: Find vector R for two particles m₁=2 kg at vector r₁=3î m and m₂=3 kg at vector r₂=5î m.
  • Use vector R = (m₁vector r₁ + m₂vector r₂)/(m₁ + m₂) = (2×3 + 3×5)/(2+3) = (6+15)/5 = 4.2 mî.
• Ex 2: Calculate velocity of CM if m₁=1 kg with velocity 2ĵ m/s and m₂=3 kg with velocity 4ĵ m/s.
  • vector V_CM = (m₁vector v₁ + m₂vector v₂)/(m₁ + m₂) = (1×2 + 3×4)/4 = 3.5 m/sĵ.

Practice Set

• Q1: The position vector of CM for two particles of masses 2 kg and 3 kg at positions 1 m and 4 m respectively is:
  • A) 2.5 m
  • B) 3.2 m
  • C) 3.8 m
  • D) 4.0 m
• Q2: If the total external force on a system is zero, the acceleration of the centre of mass is:
  • A) Zero
  • B) Constant
  • C) Increasing
  • D) Decreasing
• Q3: A projectile explodes at the highest point of its path. The centre of mass of fragments:
  • A) Falls vertically down
  • B) Continues along original parabolic path
  • C) Moves horizontally
  • D) Stops moving

Answer Key

• Q1: C) 3.8 m - Weighted average position using vector R = (m₁r₁ + m₂r₂)/(m₁ + m₂).
• Q2: A) Zero - No external force means no acceleration of CM.
• Q3: B) Continues along original parabolic path - CM follows same trajectory as if no explosion.

Memory Aids

• Mvector R = ∑ m_ivector r_i - CM position vector formula.
• Velocity of CM: vector V_CM = (∑ m_ivector v_i)/M.
• Acceleration of CM: Mvector A_CM = vector F_ext.
• CM moves as if all mass and external forces act at CM.

Definition and Formula

• Consider a system of (00n00) particles with masses (00m_1, m_2, \dots, m_n00).
• Position vectors of particles are (00\vec{r}_1, \vec{r}_2, \dots, \vec{r}_n00).
• Position vector of centre of mass (CM) is (00\vec{R}).
• Formula: (M\vec{R} = m_1\vec{r}_1 + m_2\vec{r}_2 + \dots + m_n\vec{r}_n), where (M = \sum m_i).
• Differentiating w.r.t time gives velocity relation: (M\vec{V}_{CM} = m_1\vec{v}_1 + m_2\vec{v}_2 + \dots + m_n\vec{v}_n).
• Acceleration relation: (M\vec{A}_{CM} = m_1\vec{a}_1 + m_2\vec{a}_2 + \dots + m_n\vec{a}_n).
• Newton's 2nd law for CM: (M\vec{A}_{CM} = \vec{F}_{ext}) (only external forces affect CM motion).
• CM moves as if all mass is concentrated at CM and all external forces act there.

Solved Examples

• Ex 1: Find (\vec{R}) for two particles (m_1=2\,kg) at (\vec{r}_1=3\hat{i}) m and (m_2=3\,kg) at (\vec{r}_2=5\hat{i}) m.
  • Use (\vec{R} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2} = \frac{2\times3 + 3\times5}{2+3} = \frac{6+15}{5} = 4.2\,m\hat{i}).
• Ex 2: Calculate velocity of CM if (m_1=1\,kg) with velocity (2\hat{i}\,m/s) and (m_2=3\,kg) with velocity (4\hat{i}\,m/s).
  • (\vec{V}_{CM} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2} = \frac{1\times2 + 3\times4}{4} = 3.5\,m/s\hat{i}).

Practice Set

• Q1: The position vector of CM for two particles of masses 2 kg and 3 kg at positions 1 m and 4 m respectively is:
  • A) 2.5 m
  • B) 3.2 m
  • C) 3.8 m
  • D) 4.0 m
• Q2: If the total external force on a system is zero, the acceleration of the centre of mass is:
  • A) Zero
  • B) Constant
  • C) Increasing
  • D) Decreasing
• Q3: A projectile explodes at the highest point of its path. The centre of mass of fragments:
  • A) Falls vertically down
  • B) Continues along original parabolic path
  • C) Moves horizontally
  • D) Stops moving

Answer Key

• Q1: C) 3.8 m - Weighted average position using (\vec{R} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}).
• Q2: A) Zero - No external force means no acceleration of CM.
• Q3: B) Continues along original parabolic path - CM follows same trajectory as if no explosion.

Memory Aids

• (M\vec{R} = \sum m_i\vec{r}_i) - CM position vector formula.
• Velocity of CM: (\vec{V}_{CM} = \frac{\sum m_i\vec{v}_i}{M}).
• Acceleration of CM: (M\vec{A}_{CM} = \vec{F}_{ext}).
• CM moves as if all mass and external forces act at CM.

Definition and Formula

• Consider a system of n particles with masses m₁, m₂, ..., m_n.
• Position vectors of particles are vector r₁, vector r₂, ..., vector r_n.
• Position vector of centre of mass (CM) is vector R.
• Formula: Mvector R = m₁vector r₁ + m₂vector r₂ + ... + m_nvector r_n, where M = ∑ m_i.
• Differentiating w.r.t time gives velocity relation: Mvector V_CM = m₁vector v₁ + m₂vector v₂ + ... + m_nvector v_n.
• Acceleration relation: Mvector A_CM = m₁vector a₁ + m₂vector a₂ + ... + m_nvector a_n.
• Newton's 2nd law for CM: Mvector A_CM = vector F_ext (only external forces affect CM motion).
• CM moves as if all mass is concentrated at CM and all external forces act there.

Solved Examples

• Ex 1: Find vector R for two particles m₁=2 kg at vector r₁=3î m and m₂=3 kg at vector r₂=5î m.
  • Use vector R = (m₁vector r₁ + m₂vector r₂)/(m₁ + m₂) = (2×3 + 3×5)/(2+3) = (6+15)/5 = 4.2 mî.
• Ex 2: Calculate velocity of CM if m₁=1 kg with velocity 2ĵ m/s and m₂=3 kg with velocity 4ĵ m/s.
  • vector V_CM = (m₁vector v₁ + m₂vector v₂)/(m₁ + m₂) = (1×2 + 3×4)/4 = 3.5 m/sĵ.

Practice Set

• Q1: The position vector of CM for two particles of masses 2 kg and 3 kg at positions 1 m and 4 m respectively is:
  • A) 2.5 m
  • B) 3.2 m
  • C) 3.8 m
  • D) 4.0 m
• Q2: If the total external force on a system is zero, the acceleration of the centre of mass is:
  • A) Zero
  • B) Constant
  • C) Increasing
  • D) Decreasing
• Q3: A projectile explodes at the highest point of its path. The centre of mass of fragments:
  • A) Falls vertically down
  • B) Continues along original parabolic path
  • C) Moves horizontally
  • D) Stops moving

Answer Key

• Q1: C) 3.8 m - Weighted average position using vector R = (m₁r₁ + m₂r₂)/(m₁ + m₂).
• Q2: A) Zero - No external force means no acceleration of CM.
• Q3: B) Continues along original parabolic path - CM follows same trajectory as if no explosion.

Memory Aids

• Mvector R = ∑ m_ivector r_i - CM position vector formula.
• Velocity of CM: vector V_CM = (∑ m_ivector v_i)/M.
• Acceleration of CM: Mvector A_CM = vector F_ext.
• CM moves as if all mass and external forces act at CM.

Definition and Formula

• Consider a system of \(n\) particles with masses \(m_1, m_2, \dots, m_n\).
• Position vectors of particles are \(\vec{r}_1, \vec{r}_2, \dots, \vec{r}_n\).
• Position vector of centre of mass (CM) is \(\vec{R}\).
• Formula: \(M\vec{R} = m_1\vec{r}_1 + m_2\vec{r}_2 + \dots + m_n\vec{r}_n\), where \(M = \sum m_i\).
• Differentiating w.r.t time gives velocity relation: \(M\vec{V}_{CM} = m_1\vec{v}_1 + m_2\vec{v}_2 + \dots + m_n\vec{v}_n\).
• Acceleration relation: \(M\vec{A}_{CM} = m_1\vec{a}_1 + m_2\vec{a}_2 + \dots + m_n\vec{a}_n\).
• Newton's 2nd law for CM: \(M\vec{A}_{CM} = \vec{F}_{ext}\) (only external forces affect CM motion).
• CM moves as if all mass is concentrated at CM and all external forces act there.

Solved Examples

• Ex 1: Find \(\vec{R}\) for two particles \(m_1=2\,kg\) at \(\vec{r}_1=3\hat{i}\) m and \(m_2=3\,kg\) at \(\vec{r}_2=5\hat{i}\) m.
  • Use \(\vec{R} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2} = \frac{2\times3 + 3\times5}{2+3} = \frac{6+15}{5} = 4.2\,m\hat{i}\).
• Ex 2: Calculate velocity of CM if \(m_1=1\,kg\) with velocity \(2\hat{i}\,m/s\) and \(m_2=3\,kg\) with velocity \(4\hat{i}\,m/s\).
  • \(\vec{V}_{CM} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2} = \frac{1\times2 + 3\times4}{4} = 3.5\,m/s\hat{i}\).

Practice Set

• Q1: The position vector of CM for two particles of masses 2 kg and 3 kg at positions 1 m and 4 m respectively is:
  • A) 2.5 m
  • B) 3.2 m
  • C) 3.8 m
  • D) 4.0 m
• Q2: If the total external force on a system is zero, the acceleration of the centre of mass is:
  • A) Zero
  • B) Constant
  • C) Increasing
  • D) Decreasing
• Q3: A projectile explodes at the highest point of its path. The centre of mass of fragments:
  • A) Falls vertically down
  • B) Continues along original parabolic path
  • C) Moves horizontally
  • D) Stops moving

Answer Key

• Q1: C) 3.8 m - Weighted average position using \(\vec{R} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}\).
• Q2: A) Zero - No external force means no acceleration of CM.
• Q3: B) Continues along original parabolic path - CM follows same trajectory as if no explosion.

Memory Aids

• \(M\vec{R} = \sum m_i\vec{r}_i\) - CM position vector formula.
• Velocity of CM: \(\vec{V}_{CM} = \frac{\sum m_i\vec{v}_i}{M}\).
• Acceleration of CM: \(M\vec{A}_{CM} = \vec{F}_{ext}\).
• CM moves as if all mass and external forces act at CM.

Linear Momentum of a System of Particles

Key Concepts

• Linear momentum vector p of a system of n particles is the vector sum of individual momenta:
  vector p = vector p₁ + vector p₂ + vector p₃ + ... + vector p_n
• Relation with center of mass velocity vector V_CM:
  M vector V_CM = m₁ vector v₁ + m₂ vector v₂ + m₃ vector v₃ + ... + m_n vector v_n
• Newton's 2nd law for system:
  dvector p/dt = M dvector V_CM/dt = M vector A_CM = vector F_ext
• Conservation of linear momentum:
  If external force vector F_ext = 0, then
  dvector p/dt = 0 ⇒ vector p = constant
• If total external force is zero, center of mass moves with constant velocity.
• Example: Alpha decay of radium nucleus (Ra) into radon nucleus (Rn) and alpha particle (He) shows conservation of momentum with center of mass in uniform motion or at rest.

Solved Examples

• Ex 1: Calculate total momentum of two particles with masses m₁ = 2 kg, m₂ = 3 kg moving at velocities vector v₁ = 4î m/s and vector v₂ = -2î m/s.
  Solution:
  vector p = m₁ vector v₁ + m₂ vector v₂ = 2 × 4 + 3 × (-2) = 8 - 6 = 2 kg·m/s
• Ex 2: A system with total mass M = 5 kg has center of mass velocity vector V_CM = 3ĵ m/s. Find total momentum.
  Solution:
  vector p = M vector V_CM = 5 × 3 = 15ĵ ,kg·m/s

Practice Set

• Q1: Two particles of masses 1 kg and 2 kg move with velocities 3 m/s and 4 m/s respectively in the same direction. Total momentum is:
  A) 5 kg·m/s
  B) 11 kg·m/s
  C) 14 kg·m/s
  D) 7 kg·m/s
• Q2: If no external force acts on a system, the total momentum:
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero
• Q3: A radium nucleus at rest splits into radon nucleus and alpha particle. If alpha particle moves with velocity 2 × 10⁷ m/s and mass ratio m_Rn : m_He = 4:1, velocity of radon nucleus is:
  A) 0.5 × 10⁷ m/s
  B) 2 × 10⁷ m/s
  C) 8 × 10⁷ m/s
  D) 1 × 10⁷ m/s

Answer Key

• Q1: B - Total momentum = 1 × 3 + 2 × 4 = 3 + 8 = 11 kg·m/s
• Q2: C - No external force means momentum is conserved (constant)
• Q3: A - Using conservation: m_Rn v_Rn = m_He v_He ⇒ v_Rn = m_He/m_Rn v_He = 1/4 × 2 × 10⁷ = 0.5 × 10⁷ m/s

Memory Aids

• vector p = ∑ m_i vector v_i = M vector V_CM
• dvector p/dt = vector F_ext
• Zero external force → momentum constant
• Center of mass velocity constant if vector F_ext = 0
• Alpha decay: particles recoil back-to-back conserving momentum

Key Concepts

• Linear momentum vector p of a system of n particles is the vector sum of individual momenta:
  vector p = vector p₁ + vector p₂ + vector p₃ + ... + vector p_n
• Relation with center of mass velocity vector V_CM:
  M vector V_CM = m₁ vector v₁ + m₂ vector v₂ + m₃ vector v₃ + ... + m_n vector v_n
• Newton's 2nd law for system:
  dvector p/dt = M dvector V_CM/dt = M vector A_CM = vector F_ext
• Conservation of linear momentum:
  If external force vector F_ext = 0, then
  dvector p/dt = 0 ⇒ vector p = constant
• If total external force is zero, center of mass moves with constant velocity.
• Example: Alpha decay of radium nucleus (Ra) into radon nucleus (Rn) and alpha particle (He) shows conservation of momentum with center of mass in uniform motion or at rest.

Solved Examples

• Ex 1: Calculate total momentum of two particles with masses m₁ = 2 kg, m₂ = 3 kg moving at velocities vector v₁ = 4î m/s and vector v₂ = -2î m/s.
  Solution:
  vector p = m₁ vector v₁ + m₂ vector v₂ = 2 × 4 + 3 × (-2) = 8 - 6 = 2 kg·m/s
• Ex 2: A system with total mass M = 5 kg has center of mass velocity vector V_CM = 3ĵ m/s. Find total momentum.
  Solution:
  vector p = M vector V_CM = 5 × 3 = 15ĵ ,kg·m/s

Practice Set

• Q1: Two particles of masses 1 kg and 2 kg move with velocities 3 m/s and 4 m/s respectively in the same direction. Total momentum is:
  A) 5 kg·m/s
  B) 11 kg·m/s
  C) 14 kg·m/s
  D) 7 kg·m/s
• Q2: If no external force acts on a system, the total momentum:
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero
• Q3: A radium nucleus at rest splits into radon nucleus and alpha particle. If alpha particle moves with velocity 2 × 10⁷ m/s and mass ratio m_Rn : m_He = 4:1, velocity of radon nucleus is:
  A) 0.5 × 10⁷ m/s
  B) 2 × 10⁷ m/s
  C) 8 × 10⁷ m/s
  D) 1 × 10⁷ m/s

Answer Key

• Q1: B - Total momentum = 1 × 3 + 2 × 4 = 3 + 8 = 11 kg·m/s
• Q2: C - No external force means momentum is conserved (constant)
• Q3: A - Using conservation: m_Rn v_Rn = m_He v_He ⇒ v_Rn = m_He/m_Rn v_He = 1/4 × 2 × 10⁷ = 0.5 × 10⁷ m/s

Memory Aids

• vector p = ∑ m_i vector v_i = M vector V_CM
• dvector p/dt = vector F_ext
• Zero external force → momentum constant
• Center of mass velocity constant if vector F_ext = 0
• Alpha decay: particles recoil back-to-back conserving momentum

Key Concepts

• Linear momentum p of a system of n particles is the vector sum of individual momenta:
  p = p₁ + p₂ + p₃ + ... + p_n
• Relation with center of mass velocity V_CM:
  M V_CM = m₁ v₁ + m₂ v₂ + m₃ v₃ + ... + m_n v_n
• Newton's 2nd law for system:
  dp/dt = M dV_CM/dt = M A_CM = F_ext
• Conservation of linear momentum:
  If external force F_ext = 0, then
  dp/dt = 0 ⇒ p = constant
• If total external force is zero, center of mass moves with constant velocity.
• Example: Alpha decay of radium nucleus (Ra) into radon nucleus (Rn) and alpha particle (He) shows conservation of momentum with center of mass in uniform motion or at rest.

Solved Examples

• Ex 1: Calculate total momentum of two particles with masses m₁ = 2 kg, m₂ = 3 kg moving at velocities v₁ = 4î m/s and v₂ = -2î m/s.
  Solution:
  p = m₁ v₁ + m₂ v₂ = 2 × 4 + 3 × (-2) = 8 - 6 = 2 kg·m/s
• Ex 2: A system with total mass M = 5 kg has center of mass velocity V_CM = 3ĵ m/s. Find total momentum.
  Solution:
  p = M V_CM = 5 × 3 = 15ĵ ,kg·m/s

Practice Set

• Q1: Two particles of masses 1 kg and 2 kg move with velocities 3 m/s and 4 m/s respectively in the same direction. Total momentum is:
  A) 5 kg·m/s
  B) 11 kg·m/s
  C) 14 kg·m/s
  D) 7 kg·m/s
• Q2: If no external force acts on a system, the total momentum:
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero
• Q3: A radium nucleus at rest splits into radon nucleus and alpha particle. If alpha particle moves with velocity 2 × 10⁷ m/s and mass ratio m_Rn : m_He = 4:1, velocity of radon nucleus is:
  A) 0.5 × 10⁷ m/s
  B) 2 × 10⁷ m/s
  C) 8 × 10⁷ m/s
  D) 1 × 10⁷ m/s

Answer Key

• Q1: B - Total momentum = 1 × 3 + 2 × 4 = 3 + 8 = 11 kg·m/s
• Q2: C - No external force means momentum is conserved (constant)
• Q3: A - Using conservation: m_Rn v_Rn = m_He v_He ⇒ v_Rn = m_He/m_Rn v_He = 1/4 × 2 × 10⁷ = 0.5 × 10⁷ m/s

Memory Aids

• p = ∑ m_i v_i = M V_CM
• dp/dt = F_ext
• Zero external force → momentum constant
• Center of mass velocity constant if F_ext = 0
• Alpha decay: particles recoil back-to-back conserving momentum

Key Concepts

• Linear momentum vector p of a system of n particles is the vector sum of individual momenta:
  vector p = vector p₁ + vector p₂ + vector p₃ + ... + vector p_n
• Relation with center of mass velocity vector V_CM:
  M vector V_CM = m₁ vector v₁ + m₂ vector v₂ + m₃ vector v₃ + ... + m_n vector v_n
• Newton's 2nd law for system:
  dvector p/dt = M dvector V_CM/dt = M vector A_CM = vector F_ext
• Conservation of linear momentum:
  If external force vector F_ext = 0, then
  dvector p/dt = 0 ⇒ vector p = constant
• If total external force is zero, center of mass moves with constant velocity.
• Example: Alpha decay of radium nucleus (Ra) into radon nucleus (Rn) and alpha particle (He) shows conservation of momentum with center of mass in uniform motion or at rest.

Solved Examples

• Ex 1: Calculate total momentum of two particles with masses m₁ = 2 kg, m₂ = 3 kg moving at velocities vector v₁ = 4î m/s and vector v₂ = -2î m/s.
  Solution:
  vector p = m₁ vector v₁ + m₂ vector v₂ = 2 × 4 + 3 × (-2) = 8 - 6 = 2 kg·m/s
• Ex 2: A system with total mass M = 5 kg has center of mass velocity vector V_CM = 3ĵ m/s. Find total momentum.
  Solution:
  vector p = M vector V_CM = 5 × 3 = 15ĵ ,kg·m/s

Practice Set

• Q1: Two particles of masses 1 kg and 2 kg move with velocities 3 m/s and 4 m/s respectively in the same direction. Total momentum is:
  A) 5 kg·m/s
  B) 11 kg·m/s
  C) 14 kg·m/s
  D) 7 kg·m/s
• Q2: If no external force acts on a system, the total momentum:
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero
• Q3: A radium nucleus at rest splits into radon nucleus and alpha particle. If alpha particle moves with velocity 2 × 10⁷ m/s and mass ratio m_Rn : m_He = 4:1, velocity of radon nucleus is:
  A) 0.5 × 10⁷ m/s
  B) 2 × 10⁷ m/s
  C) 8 × 10⁷ m/s
  D) 1 × 10⁷ m/s

Answer Key

• Q1: B - Total momentum = 1 × 3 + 2 × 4 = 3 + 8 = 11 kg·m/s
• Q2: C - No external force means momentum is conserved (constant)
• Q3: A - Using conservation: m_Rn v_Rn = m_He v_He ⇒ v_Rn = m_He/m_Rn v_He = 1/4 × 2 × 10⁷ = 0.5 × 10⁷ m/s

Memory Aids

• vector p = ∑ m_i vector v_i = M vector V_CM
• dvector p/dt = vector F_ext
• Zero external force → momentum constant
• Center of mass velocity constant if vector F_ext = 0
• Alpha decay: particles recoil back-to-back conserving momentum

Key Concepts

• Linear momentum (\( \vec{p} \)) of a system of (\( n \)) particles is the vector sum of individual momenta:
  (\( \vec{p} = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 + \ldots + \vec{p}_n \))
• Relation with center of mass velocity (\( \vec{V}_{CM} \)):
  (\( M \vec{V}_{CM} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3 + \ldots + m_n \vec{v}_n \))
• Newton's 2nd law for system:
  (\( \frac{d\vec{p}}{dt} = M \frac{d\vec{V}_{CM}}{dt} = M \vec{A}_{CM} = \vec{F}_{ext} \))
• Conservation of linear momentum:
  If external force (\( \vec{F}_{ext} = 0 \)), then
  (\( \frac{d\vec{p}}{dt} = 0 \Rightarrow \vec{p} = \text{constant} \))
• If total external force is zero, center of mass moves with constant velocity.
• Example: Alpha decay of radium nucleus (Ra) into radon nucleus (Rn) and alpha particle (He) shows conservation of momentum with center of mass in uniform motion or at rest.

Solved Examples

• Ex 1: Calculate total momentum of two particles with masses (\( m_1 = 2\,kg \)), (\( m_2 = 3\,kg \)) moving at velocities (\( \vec{v}_1 = 4\hat{i} \)) m/s and (\( \vec{v}_2 = -2\hat{i} \)) m/s.
  Solution:
  (\( \vec{p} = m_1 \vec{v}_1 + m_2 \vec{v}_2 = 2 \times 4 + 3 \times (-2) = 8 - 6 = 2\,kg\cdot m/s \))
• Ex 2: A system with total mass (\( M = 5\,kg \)) has center of mass velocity (\( \vec{V}_{CM} = 3\hat{j} \)) m/s. Find total momentum.
  Solution:
  (\( \vec{p} = M \vec{V}_{CM} = 5 \times 3 = 15\hat{j} \,kg\cdot m/s \))

Practice Set

• Q1: Two particles of masses 1 kg and 2 kg move with velocities 3 m/s and 4 m/s respectively in the same direction. Total momentum is:
  A) 5 kg·m/s
  B) 11 kg·m/s
  C) 14 kg·m/s
  D) 7 kg·m/s
• Q2: If no external force acts on a system, the total momentum:
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero
• Q3: A radium nucleus at rest splits into radon nucleus and alpha particle. If alpha particle moves with velocity (\( 2 \times 10^7 \)) m/s and mass ratio (\( m_{Rn} : m_{He} = 4:1 \)), velocity of radon nucleus is:
  A) (\( 0.5 \times 10^7 \)) m/s
  B) (\( 2 \times 10^7 \)) m/s
  C) (\( 8 \times 10^7 \)) m/s
  D) (\( 1 \times 10^7 \)) m/s

Answer Key

• Q1: B - Total momentum = (\( 1 \times 3 + 2 \times 4 = 3 + 8 = 11 \)) kg·m/s
• Q2: C - No external force means momentum is conserved (constant)
• Q3: A - Using conservation: (\( m_{Rn} v_{Rn} = m_{He} v_{He} \Rightarrow v_{Rn} = \frac{m_{He}}{m_{Rn}} v_{He} = \frac{1}{4} \times 2 \times 10^7 = 0.5 \times 10^7 \)) m/s

Memory Aids

• (\( \vec{p} = \sum m_i \vec{v}_i = M \vec{V}_{CM} \))
• (\( \frac{d\vec{p}}{dt} = \vec{F}_{ext} \))
• Zero external force → momentum constant
• Center of mass velocity constant if (\( \vec{F}_{ext} = 0 \))
• Alpha decay: particles recoil back-to-back conserving momentum

Key Concepts

• Linear momentum vector p of a system of n particles is the vector sum of individual momenta:
  vector p = vector p₁ + vector p₂ + vector p₃ + ... + vector p_n
• Relation with center of mass velocity vector V_CM:
  M vector V_CM = m₁ vector v₁ + m₂ vector v₂ + m₃ vector v₃ + ... + m_n vector v_n
• Newton's 2nd law for system:
  dvector p/dt = M dvector V_CM/dt = M vector A_CM = vector F_ext
• Conservation of linear momentum:
  If external force vector F_ext = 0, then
  dvector p/dt = 0 ⇒ vector p = constant
• If total external force is zero, center of mass moves with constant velocity.
• Example: Alpha decay of radium nucleus (Ra) into radon nucleus (Rn) and alpha particle (He) shows conservation of momentum with center of mass in uniform motion or at rest.

Solved Examples

• Ex 1: Calculate total momentum of two particles with masses m₁ = 2 kg, m₂ = 3 kg moving at velocities vector v₁ = 4î m/s and vector v₂ = -2î m/s.
  Solution:
  vector p = m₁ vector v₁ + m₂ vector v₂ = 2 × 4 + 3 × (-2) = 8 - 6 = 2 kg·m/s
• Ex 2: A system with total mass M = 5 kg has center of mass velocity vector V_CM = 3ĵ m/s. Find total momentum.
  Solution:
  vector p = M vector V_CM = 5 × 3 = 15ĵ ,kg·m/s

Practice Set

• Q1: Two particles of masses 1 kg and 2 kg move with velocities 3 m/s and 4 m/s respectively in the same direction. Total momentum is:
  A) 5 kg·m/s
  B) 11 kg·m/s
  C) 14 kg·m/s
  D) 7 kg·m/s
• Q2: If no external force acts on a system, the total momentum:
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero
• Q3: A radium nucleus at rest splits into radon nucleus and alpha particle. If alpha particle moves with velocity 2 × 10⁷ m/s and mass ratio m_Rn : m_He = 4:1, velocity of radon nucleus is:
  A) 0.5 × 10⁷ m/s
  B) 2 × 10⁷ m/s
  C) 8 × 10⁷ m/s
  D) 1 × 10⁷ m/s

Answer Key

• Q1: B - Total momentum = 1 × 3 + 2 × 4 = 3 + 8 = 11 kg·m/s
• Q2: C - No external force means momentum is conserved (constant)
• Q3: A - Using conservation: m_Rn v_Rn = m_He v_He ⇒ v_Rn = m_He/m_Rn v_He = 1/4 × 2 × 10⁷ = 0.5 × 10⁷ m/s

Memory Aids

• vector p = ∑ m_i vector v_i = M vector V_CM
• dvector p/dt = vector F_ext
• Zero external force → momentum constant
• Center of mass velocity constant if vector F_ext = 0
• Alpha decay: particles recoil back-to-back conserving momentum

Key Concepts

• Linear momentum \( \vec{p} \) of a system of \( n \) particles is the vector sum of individual momenta:
  \( \vec{p} = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 + \ldots + \vec{p}_n \)
• Relation with center of mass velocity \( \vec{V}_{CM} \):
  \( M \vec{V}_{CM} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3 + \ldots + m_n \vec{v}_n \)
• Newton's 2nd law for system:
  \( \frac{d\vec{p}}{dt} = M \frac{d\vec{V}_{CM}}{dt} = M \vec{A}_{CM} = \vec{F}_{ext} \)
• Conservation of linear momentum:
  If external force \( \vec{F}_{ext} = 0 \), then
  \( \frac{d\vec{p}}{dt} = 0 \Rightarrow \vec{p} = \text{constant} \)
• If total external force is zero, center of mass moves with constant velocity.
• Example: Alpha decay of radium nucleus (Ra) into radon nucleus (Rn) and alpha particle (He) shows conservation of momentum with center of mass in uniform motion or at rest.

Solved Examples

• Ex 1: Calculate total momentum of two particles with masses \( m_1 = 2\,kg \), \( m_2 = 3\,kg \) moving at velocities \( \vec{v}_1 = 4\hat{i} \) m/s and \( \vec{v}_2 = -2\hat{i} \) m/s.
  Solution:
  \( \vec{p} = m_1 \vec{v}_1 + m_2 \vec{v}_2 = 2 \times 4 + 3 \times (-2) = 8 - 6 = 2\,kg\cdot m/s \)
• Ex 2: A system with total mass \( M = 5\,kg \) has center of mass velocity \( \vec{V}_{CM} = 3\hat{j} \) m/s. Find total momentum.
  Solution:
  \( \vec{p} = M \vec{V}_{CM} = 5 \times 3 = 15\hat{j} \,kg\cdot m/s \)

Practice Set

• Q1: Two particles of masses 1 kg and 2 kg move with velocities 3 m/s and 4 m/s respectively in the same direction. Total momentum is:
  A) 5 kg·m/s
  B) 11 kg·m/s
  C) 14 kg·m/s
  D) 7 kg·m/s
• Q2: If no external force acts on a system, the total momentum:
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero
• Q3: A radium nucleus at rest splits into radon nucleus and alpha particle. If alpha particle moves with velocity \( 2 \times 10^7 \) m/s and mass ratio \( m_{Rn} : m_{He} = 4:1 \), velocity of radon nucleus is:
  A) \( 0.5 \times 10^7 \) m/s
  B) \( 2 \times 10^7 \) m/s
  C) \( 8 \times 10^7 \) m/s
  D) \( 1 \times 10^7 \) m/s

Answer Key

• Q1: B - Total momentum = \( 1 \times 3 + 2 \times 4 = 3 + 8 = 11 \) kg·m/s
• Q2: C - No external force means momentum is conserved (constant)
• Q3: A - Using conservation: \( m_{Rn} v_{Rn} = m_{He} v_{He} \Rightarrow v_{Rn} = \frac{m_{He}}{m_{Rn}} v_{He} = \frac{1}{4} \times 2 \times 10^7 = 0.5 \times 10^7 \) m/s

Memory Aids

• \( \vec{p} = \sum m_i \vec{v}_i = M \vec{V}_{CM} \)
• \( \frac{d\vec{p}}{dt} = \vec{F}_{ext} \)
• Zero external force → momentum constant
• Center of mass velocity constant if \( \vec{F}_{ext} = 0 \)
• Alpha decay: particles recoil back-to-back conserving momentum

Vector Product of Two Vectors

Definition and Formula

• Consider two vectors a and b with angle θ between them.
• Vector product (cross product) is a × b = |a||b| sin θ n̂.
• a × b is a vector perpendicular to both a and b.
• n̂ is the unit vector in the direction of a × b.

Direction of Cross Product

• Direction is perpendicular to the plane of a and b.
• Use right-hand thumb rule: curl fingers from a to b, thumb points to a × b.
• Unit vectors î, ĵ, k̂ satisfy:
  • î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic order)
  • î × î = ĵ × ĵ = k̂ × k̂ = 0
  • Anticyclic order: î × k̂ = -ĵ, ĵ × î = -k̂, k̂ × ĵ = -î

Calculation in Component Form

• For vectors a = a_x î + a_y ĵ + a_z k̂ and b = b_x î + b_y ĵ + b_z k̂,
• Cross product: [ a × b = (a_y b_z - a_z b_y) î - (a_x b_z - a_z b_x) ĵ + (a_x b_y - a_y b_x) k̂ ]

Solved Examples

Ex 1

• Given a = 2î + 3ĵ + k̂, b = î - 2ĵ + 4k̂.
• Calculate a × b using component formula.
• a × b = (3 × 4 - 1 × (-2))î - (2 × 4 - 1 × 1)ĵ + (2 × (-2) - 3 × 1)k̂ = (12 + 2)î - (8 - 1)ĵ + (-4 - 3)k̂ = 14î - 7ĵ - 7k̂ ).

Practice Set

• Q1: If a = 3î + 2ĵ and b = î + 4ĵ, what is a × b?
  • A. 10k̂
  • B. -10k̂
  • C. 5k̂
  • D. 0
• Q2: The angle between two vectors is 90°. If |a| = 5 and |b| = 3, what is the magnitude of a × b?
  • A. 15
  • B. 8
  • C. 0
  • D. 1
• Q3: For a = î + 2ĵ + 3k̂ and b = 4î - ĵ + 2k̂, find the ĵ component of a × b.
  • A. 10
  • B. -10
  • C. 5
  • D. -5

Answer Key

• Q1: A. 10k̂ - Cross product (3 × 4 - 2 × 1)k̂ = 10k̂.
• Q2: A. 15 - Magnitude = |a||b| sin 90° = 5 × 3 × 1 = 15.
• Q3: B. -10 - ĵ component = -(a_x b_z - a_z b_x) = -(1 × 2 - 3 × 4) = - (2 - 12) = -(-10) = 10 but sign is negative in formula, so answer is -10.

Memory Aids

• Right-hand rule: Fingers from a to b, thumb points a × b.
• Cyclic order: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ.
• Component formula: Use determinant pattern for quick calculation.

Definition and Formula

• Consider two vectors a and b with angle θ between them.
• Vector product (cross product) is a × b = |a||b| sin θ n̂.
• a × b is a vector perpendicular to both a and b.
• n̂ is the unit vector in the direction of a × b.

Direction of Cross Product

• Direction is perpendicular to the plane of a and b.
• Use right-hand thumb rule: curl fingers from a to b, thumb points to a × b.
• Unit vectors î, ĵ, k̂ satisfy:
  • î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic order)
  • î × î = ĵ × ĵ = k̂ × k̂ = 0
  • Anticyclic order: î × k̂ = -ĵ, ĵ × î = -k̂, k̂ × ĵ = -î

Calculation in Component Form

• For vectors a = a_x î + a_y ĵ + a_z k̂ and b = b_x î + b_y ĵ + b_z k̂,
• Cross product: [ a × b = (a_y b_z - a_z b_y) î - (a_x b_z - a_z b_x) ĵ + (a_x b_y - a_y b_x) k̂ ]

Solved Examples

Ex 1

• Given a = 2î + 3ĵ + k̂, b = î - 2ĵ + 4k̂.
• Calculate a × b using component formula.
• a × b = (3 × 4 - 1 × (-2))î - (2 × 4 - 1 × 1)ĵ + (2 × (-2) - 3 × 1)k̂ = (12 + 2)î - (8 - 1)ĵ + (-4 - 3)k̂ = 14î - 7ĵ - 7k̂ ).

Practice Set

• Q1: If a = 3î + 2ĵ and b = î + 4ĵ, what is a × b?
  • A. 10k̂
  • B. -10k̂
  • C. 5k̂
  • D. 0
• Q2: The angle between two vectors is 90°. If |a| = 5 and |b| = 3, what is the magnitude of a × b?
  • A. 15
  • B. 8
  • C. 0
  • D. 1
• Q3: For a = î + 2ĵ + 3k̂ and b = 4î - ĵ + 2k̂, find the ĵ component of a × b.
  • A. 10
  • B. -10
  • C. 5
  • D. -5

Answer Key

• Q1: A. 10k̂ - Cross product (3 × 4 - 2 × 1)k̂ = 10k̂.
• Q2: A. 15 - Magnitude = |a||b| sin 90° = 5 × 3 × 1 = 15.
• Q3: B. -10 - ĵ component = -(a_x b_z - a_z b_x) = -(1 × 2 - 3 × 4) = - (2 - 12) = -(-10) = 10 but sign is negative in formula, so answer is -10.

Memory Aids

• Right-hand rule: Fingers from a to b, thumb points a × b.
• Cyclic order: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ.
• Component formula: Use determinant pattern for quick calculation.

Definition and Formula

• Consider two vectors a and b with angle θ between them.
• Vector product (cross product) is a × b = |a||b| sin θ n̂.
• a × b is a vector perpendicular to both a and b.
• n̂ is the unit vector in the direction of a × b.

Direction of Cross Product

• Direction is perpendicular to the plane of a and b.
• Use right-hand thumb rule: curl fingers from a to b, thumb points to a × b.
• Unit vectors î, ĵ, k̂ satisfy:
  • î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic order)
  • î × î = ĵ × ĵ = k̂ × k̂ = 0
  • Anticyclic order: î × k̂ = -ĵ, ĵ × î = -k̂, k̂ × ĵ = -î

Calculation in Component Form

• For vectors a = a_x î + a_y ĵ + a_z k̂ and b = b_x î + b_y ĵ + b_z k̂,
• Cross product: [ a × b = (a_y b_z - a_z b_y) î - (a_x b_z - a_z b_x) ĵ + (a_x b_y - a_y b_x) k̂ ]

Solved Examples

Ex 1

• Given a = 2î + 3ĵ + k̂, b = î - 2ĵ + 4k̂.
• Calculate a × b using component formula.
• a × b = (3 × 4 - 1 × (-2))î - (2 × 4 - 1 × 1)ĵ + (2 × (-2) - 3 × 1)k̂ = (12 + 2)î - (8 - 1)ĵ + (-4 - 3)k̂ = 14î - 7ĵ - 7k̂ ).

Practice Set

• Q1: If a = 3î + 2ĵ and b = î + 4ĵ, what is a × b?
  • A. 10k̂
  • B. -10k̂
  • C. 5k̂
  • D. 0
• Q2: The angle between two vectors is 90°. If |a| = 5 and |b| = 3, what is the magnitude of a × b?
  • A. 15
  • B. 8
  • C. 0
  • D. 1
• Q3: For a = î + 2ĵ + 3k̂ and b = 4î - ĵ + 2k̂, find the ĵ component of a × b.
  • A. 10
  • B. -10
  • C. 5
  • D. -5

Answer Key

• Q1: A. 10k̂ - Cross product (3 × 4 - 2 × 1)k̂ = 10k̂.
• Q2: A. 15 - Magnitude = |a||b| sin 90° = 5 × 3 × 1 = 15.
• Q3: B. -10 - ĵ component = -(a_x b_z - a_z b_x) = -(1 × 2 - 3 × 4) = - (2 - 12) = -(-10) = 10 but sign is negative in formula, so answer is -10.

Memory Aids

• Right-hand rule: Fingers from a to b, thumb points a × b.
• Cyclic order: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ.
• Component formula: Use determinant pattern for quick calculation.

Definition and Formula

• Consider two vectors a and b with angle θ between them.
• Vector product (cross product) is a × b = |a||b| sin θ n̂.
• a × b is a vector perpendicular to both a and b.
• n̂ is the unit vector in the direction of a × b.

Direction of Cross Product

• Direction is perpendicular to the plane of a and b.
• Use right-hand thumb rule: curl fingers from a to b, thumb points to a × b.
• Unit vectors î, ĵ, k̂ satisfy:
  • î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic order)
  • î × î = ĵ × ĵ = k̂ × k̂ = 0
  • Anticyclic order: î × k̂ = -ĵ, ĵ × î = -k̂, k̂ × ĵ = -î

Calculation in Component Form

• For vectors a = a_x î + a_y ĵ + a_z k̂ and b = b_x î + b_y ĵ + b_z k̂,
• Cross product: [ a × b = (a_y b_z - a_z b_y) î - (a_x b_z - a_z b_x) ĵ + (a_x b_y - a_y b_x) k̂ ]

Solved Examples

Ex 1

• Given a = 2î + 3ĵ + k̂, b = î - 2ĵ + 4k̂.
• Calculate a × b using component formula.
• a × b = (3 × 4 - 1 × (-2))î - (2 × 4 - 1 × 1)ĵ + (2 × (-2) - 3 × 1)k̂ = (12 + 2)î - (8 - 1)ĵ + (-4 - 3)k̂ = 14î - 7ĵ - 7k̂ ).

Practice Set

• Q1: If a = 3î + 2ĵ and b = î + 4ĵ, what is a × b?
  • A. 10k̂
  • B. -10k̂
  • C. 5k̂
  • D. 0
• Q2: The angle between two vectors is 90°. If |a| = 5 and |b| = 3, what is the magnitude of a × b?
  • A. 15
  • B. 8
  • C. 0
  • D. 1
• Q3: For a = î + 2ĵ + 3k̂ and b = 4î - ĵ + 2k̂, find the ĵ component of a × b.
  • A. 10
  • B. -10
  • C. 5
  • D. -5

Answer Key

• Q1: A. 10k̂ - Cross product (3 × 4 - 2 × 1)k̂ = 10k̂.
• Q2: A. 15 - Magnitude = |a||b| sin 90° = 5 × 3 × 1 = 15.
• Q3: B. -10 - ĵ component = -(a_x b_z - a_z b_x) = -(1 × 2 - 3 × 4) = - (2 - 12) = -(-10) = 10 but sign is negative in formula, so answer is -10.

Memory Aids

• Right-hand rule: Fingers from a to b, thumb points a × b.
• Cyclic order: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ.
• Component formula: Use determinant pattern for quick calculation.

Definition and Formula

• Consider two vectors (\( \mathbf{a} \)) and (\( \mathbf{b} \)) with angle (\( \theta \)) between them.
• Vector product (cross product) is (\( \mathbf{a} \times \mathbf{b} = |a||b| \sin \theta \hat{n} \)).
• (\( \mathbf{a} \times \mathbf{b} \)) is a vector perpendicular to both (\( \mathbf{a} \)) and (\( \mathbf{b} \)).
• (\( \hat{n} \)) is the unit vector in the direction of (\( \mathbf{a} \times \mathbf{b} \)).

Direction of Cross Product

• Direction is perpendicular to the plane of (\( \mathbf{a} \)) and (\( \mathbf{b} \)).
• Use right-hand thumb rule: curl fingers from (\( \mathbf{a} \)) to (\( \mathbf{b} \)), thumb points to (\( \mathbf{a} \times \mathbf{b} \)).
• Unit vectors (\( \hat{i}, \hat{j}, \hat{k} \)) satisfy:
  • (\( \hat{i} \times \hat{j} = \hat{k} \)), (\( \hat{j} \times \hat{k} = \hat{i} \)), (\( \hat{k} \times \hat{i} = \hat{j} \)) (cyclic order)
  • (\( \hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0 \))
  • Anticyclic order: (\( \hat{i} \times \hat{k} = -\hat{j} \)), (\( \hat{j} \times \hat{i} = -\hat{k} \)), (\( \hat{k} \times \hat{j} = -\hat{i} \))

Calculation in Component Form

• For vectors (\( \mathbf{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \)) and (\( \mathbf{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k} \)),
• Cross product: [ (\( \mathbf{a} \times \mathbf{b} = (a_y b_z - a_z b_y) \hat{i} - (a_x b_z - a_z b_x) \hat{j} + (a_x b_y - a_y b_x) \hat{k} \)) ]

Solved Examples

Ex 1

• Given (\( \mathbf{a} = 2\hat{i} + 3\hat{j} + \hat{k} \)), (\( \mathbf{b} = \hat{i} - 2\hat{j} + 4\hat{k} \)).
• Calculate (\( \mathbf{a} \times \mathbf{b} \)) using component formula.
• (\( \mathbf{a} \times \mathbf{b} = (3 \times 4 - 1 \times (-2))\hat{i} - (2 \times 4 - 1 \times 1)\hat{j} + (2 \times (-2) - 3 \times 1)\hat{k} = (12 + 2)\hat{i} - (8 - 1)\hat{j} + (-4 - 3)\hat{k} = 14\hat{i} - 7\hat{j} - 7\hat{k} \)).

Practice Set

• Q1: If (\( \mathbf{a} = 3\hat{i} + 2\hat{j} \)) and (\( \mathbf{b} = \hat{i} + 4\hat{j} \)), what is (\( \mathbf{a} \times \mathbf{b} \))?
  • A. (\( 10\hat{k} \))
  • B. (\( -10\hat{k} \))
  • C. (\( 5\hat{k} \))
  • D. (\( 0 \))
• Q2: The angle between two vectors is 90°. If (\( |\mathbf{a}| = 5 \)) and (\( |\mathbf{b}| = 3 \)), what is the magnitude of (\( \mathbf{a} \times \mathbf{b} \))?
  • A. 15
  • B. 8
  • C. 0
  • D. 1
• Q3: For (\( \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k} \)) and (\( \mathbf{b} = 4\hat{i} - \hat{j} + 2\hat{k} \)), find the (\( \hat{j} \)) component of (\( \mathbf{a} \times \mathbf{b} \)).
  • A. 10
  • B. -10
  • C. 5
  • D. -5

Answer Key

• Q1: A. (\( 10\hat{k} \)) - Cross product (\( 3 \times 4 - 2 \times 1 \))k̂ = 10k̂.
• Q2: A. 15 - Magnitude (\( = |a||b| \sin 90^\circ = 5 \times 3 \times 1 = 15 \)).
• Q3: B. -10 - (\( \hat{j} \)) component (\( = -(a_x b_z - a_z b_x) = -(1 \times 2 - 3 \times 4) = - (2 - 12) = -(-10) = 10 \)) but sign is negative in formula, so answer is -10.

Memory Aids

• Right-hand rule: Fingers from (\( \mathbf{a} \)) to (\( \mathbf{b} \)), thumb points (\( \mathbf{a} \times \mathbf{b} \)).
• Cyclic order: (\( \hat{i} \times \hat{j} = \hat{k} \)), (\( \hat{j} \times \hat{k} = \hat{i} \)), (\( \hat{k} \times \hat{i} = \hat{j} \)).
• Component formula: Use determinant pattern for quick calculation.

Definition and Formula

• Consider two vectors a and b with angle θ between them.
• Vector product (cross product) is a × b = |a||b| sin θ n̂.
• a × b is a vector perpendicular to both a and b.
• n̂ is the unit vector in the direction of a × b.

Direction of Cross Product

• Direction is perpendicular to the plane of a and b.
• Use right-hand thumb rule: curl fingers from a to b, thumb points to a × b.
• Unit vectors î, ĵ, k̂ satisfy:
  • î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic order)
  • î × î = ĵ × ĵ = k̂ × k̂ = 0
  • Anticyclic order: î × k̂ = -ĵ, ĵ × î = -k̂, k̂ × ĵ = -î

Calculation in Component Form

• For vectors a = a_x î + a_y ĵ + a_z k̂ and b = b_x î + b_y ĵ + b_z k̂,
• Cross product: [ a × b = (a_y b_z - a_z b_y) î - (a_x b_z - a_z b_x) ĵ + (a_x b_y - a_y b_x) k̂ ]

Solved Examples

Ex 1

• Given a = 2î + 3ĵ + k̂, b = î - 2ĵ + 4k̂.
• Calculate a × b using component formula.
• a × b = (3 × 4 - 1 × (-2))î - (2 × 4 - 1 × 1)ĵ + (2 × (-2) - 3 × 1)k̂ = (12 + 2)î - (8 - 1)ĵ + (-4 - 3)k̂ = 14î - 7ĵ - 7k̂ ).

Practice Set

• Q1: If a = 3î + 2ĵ and b = î + 4ĵ, what is a × b?
  • A. 10k̂
  • B. -10k̂
  • C. 5k̂
  • D. 0
• Q2: The angle between two vectors is 90°. If |a| = 5 and |b| = 3, what is the magnitude of a × b?
  • A. 15
  • B. 8
  • C. 0
  • D. 1
• Q3: For a = î + 2ĵ + 3k̂ and b = 4î - ĵ + 2k̂, find the ĵ component of a × b.
  • A. 10
  • B. -10
  • C. 5
  • D. -5

Answer Key

• Q1: A. 10k̂ - Cross product (3 × 4 - 2 × 1)k̂ = 10k̂.
• Q2: A. 15 - Magnitude = |a||b| sin 90° = 5 × 3 × 1 = 15.
• Q3: B. -10 - ĵ component = -(a_x b_z - a_z b_x) = -(1 × 2 - 3 × 4) = - (2 - 12) = -(-10) = 10 but sign is negative in formula, so answer is -10.

Memory Aids

• Right-hand rule: Fingers from a to b, thumb points a × b.
• Cyclic order: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ.
• Component formula: Use determinant pattern for quick calculation.

Definition and Formula

• Consider two vectors \( \mathbf{a} \) and \( \mathbf{b} \) with angle \( \theta \) between them.
• Vector product (cross product) is \( \mathbf{a} \times \mathbf{b} = |a||b| \sin \theta \hat{n} \).
• \( \mathbf{a} \times \mathbf{b} \) is a vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).
• \( \hat{n} \) is the unit vector in the direction of \( \mathbf{a} \times \mathbf{b} \).

Direction of Cross Product

• Direction is perpendicular to the plane of \( \mathbf{a} \) and \( \mathbf{b} \).
• Use right-hand thumb rule: curl fingers from \( \mathbf{a} \) to \( \mathbf{b} \), thumb points to \( \mathbf{a} \times \mathbf{b} \).
• Unit vectors \( \hat{i}, \hat{j}, \hat{k} \) satisfy:
  • \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \) (cyclic order)
  • \( \hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0 \)
  • Anticyclic order: \( \hat{i} \times \hat{k} = -\hat{j} \), \( \hat{j} \times \hat{i} = -\hat{k} \), \( \hat{k} \times \hat{j} = -\hat{i} \)

Calculation in Component Form

• For vectors \( \mathbf{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \) and \( \mathbf{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k} \),
• Cross product: \[ \mathbf{a} \times \mathbf{b} = (a_y b_z - a_z b_y) \hat{i} - (a_x b_z - a_z b_x) \hat{j} + (a_x b_y - a_y b_x) \hat{k} \]

Solved Examples

Ex 1

• Given \( \mathbf{a} = 2\hat{i} + 3\hat{j} + \hat{k} \), \( \mathbf{b} = \hat{i} - 2\hat{j} + 4\hat{k} \).
• Calculate \( \mathbf{a} \times \mathbf{b} \) using component formula.
• \( \mathbf{a} \times \mathbf{b} = (3 \times 4 - 1 \times (-2))\hat{i} - (2 \times 4 - 1 \times 1)\hat{j} + (2 \times (-2) - 3 \times 1)\hat{k} = (12 + 2)\hat{i} - (8 - 1)\hat{j} + (-4 - 3)\hat{k} = 14\hat{i} - 7\hat{j} - 7\hat{k} \).

Practice Set

• Q1: If \( \mathbf{a} = 3\hat{i} + 2\hat{j} \) and \( \mathbf{b} = \hat{i} + 4\hat{j} \), what is \( \mathbf{a} \times \mathbf{b} \)?
  • A. \( 10\hat{k} \)
  • B. \( -10\hat{k} \)
  • C. \( 5\hat{k} \)
  • D. \( 0 \)
• Q2: The angle between two vectors is 90°. If \( |\mathbf{a}| = 5 \) and \( |\mathbf{b}| = 3 \), what is the magnitude of \( \mathbf{a} \times \mathbf{b} \)?
  • A. 15
  • B. 8
  • C. 0
  • D. 1
• Q3: For \( \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \mathbf{b} = 4\hat{i} - \hat{j} + 2\hat{k} \), find the \( \hat{j} \) component of \( \mathbf{a} \times \mathbf{b} \).
  • A. 10
  • B. -10
  • C. 5
  • D. -5

Answer Key

• Q1: A. \( 10\hat{k} \) - Cross product \( (3 \times 4 - 2 \times 1)\hat{k} = 10\hat{k} \).
• Q2: A. 15 - Magnitude \( = |a||b| \sin 90^\circ = 5 \times 3 \times 1 = 15 \).
• Q3: B. -10 - \( \hat{j} \) component \( = -(a_x b_z - a_z b_x) = -(1 \times 2 - 3 \times 4) = - (2 - 12) = -(-10) = 10 \) but sign is negative in formula, so answer is -10.

Memory Aids

• Right-hand rule: Fingers from \( \mathbf{a} \) to \( \mathbf{b} \), thumb points \( \mathbf{a} \times \mathbf{b} \).
• Cyclic order: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \).
• Component formula: Use determinant pattern for quick calculation.

Angular Velocity and its Relation with Linear Velocity

Key Concepts

• In pure rotational motion, all parts of a rigid body have the same angular velocity ω at any instant.
• For rotation about a fixed axis, ω lies along the axis of rotation.
• Direction of ω follows the right-hand rule: points in the direction a right-handed screw advances when rotated.
• Linear velocity v at a point is related to angular velocity by v = ω × r, where r is the position vector from the axis.
• Angular acceleration α = dω/dt is the rate of change of angular velocity.

Solved Examples

• Ex 1: A disk rotates with angular velocity ω = 10 text rad/s. Find linear velocity at radius r = 0.5 text m.
  • Use v = ωr = 10 × 0.5 = 5 text m/s.
• Ex 2: Angular velocity changes from 5 rad/s to 15 rad/s in 2 s. Find angular acceleration.
  • α = Δω/Δt = (15 - 5)/2 = 5 text rad/s².

Practice Set

• Q1: What is the direction of angular velocity vector for a clockwise rotating disk?
  • A) Along axis towards observer
  • B) Along axis away from observer
  • C) Perpendicular to axis
  • D) Opposite to axis
• Q2: A point on a rotating wheel at radius 0.3 m has linear velocity 6 m/s. Find angular velocity.
  • A) 10 rad/s
  • B) 15 rad/s
  • C) 20 rad/s
  • D) 25 rad/s
• Q3: Angular velocity changes from 8 rad/s to 20 rad/s in 3 s. What is angular acceleration?
  • A) 2 rad/s²
  • B) 4 rad/s²
  • C) 6 rad/s²
  • D) 8 rad/s²

Answer Key

• Q1: B) Along axis away from observer — Clockwise rotation vector points away by right-hand rule.
• Q2: A) 10 rad/s — ω = v/r = 6/0.3 = 20 rad/s (Corrected: 20 rad/s is option C, so correct answer is C)
• Q3: B) 4 rad/s² — α = (20 - 8)/3 = 4 rad/s².

Memory Aids

• Right-hand rule: Thumb points along ω.
• Linear velocity v = ωr.
• Angular acceleration α = dω/dt.

Key Concepts

• In pure rotational motion, all parts of a rigid body have the same angular velocity (\( \vec{\omega} \)) at any instant.
• For rotation about a fixed axis, (\( \vec{\omega} \)) lies along the axis of rotation.
• Direction of (\( \vec{\omega} \)) follows the right-hand rule: points in the direction a right-handed screw advances when rotated.
• Linear velocity (\( \vec{v} \)) at a point is related to angular velocity by (\( \vec{v} = \vec{\omega} \times \vec{r} \)), where (\( \vec{r} \)) is the position vector from the axis.
• Angular acceleration (\( \vec{\alpha} = \frac{d\vec{\omega}}{dt} \)) is the rate of change of angular velocity.

Solved Examples

• Ex 1: A disk rotates with angular velocity (\( \omega = 10 \text{ rad/s} \)). Find linear velocity at radius (\( r = 0.5 \text{ m} \)).
  • Use (\( v = \omega r = 10 \times 0.5 = 5 \text{ m/s} \)).
• Ex 2: Angular velocity changes from 5 rad/s to 15 rad/s in 2 s. Find angular acceleration.
  • (\( \alpha = \frac{\Delta \omega}{\Delta t} = \frac{15 - 5}{2} = 5 \text{ rad/s}^2 \)).

Practice Set

• Q1: What is the direction of angular velocity vector for a clockwise rotating disk?
  • A) Along axis towards observer
  • B) Along axis away from observer
  • C) Perpendicular to axis
  • D) Opposite to axis
• Q2: A point on a rotating wheel at radius 0.3 m has linear velocity 6 m/s. Find angular velocity.
  • A) 10 rad/s
  • B) 15 rad/s
  • C) 20 rad/s
  • D) 25 rad/s
• Q3: Angular velocity changes from 8 rad/s to 20 rad/s in 3 s. What is angular acceleration?
  • A) 2 rad/s²
  • B) 4 rad/s²
  • C) 6 rad/s²
  • D) 8 rad/s²

Answer Key

• Q1: B) Along axis away from observer — Clockwise rotation vector points away by right-hand rule.
• Q2: A) 10 rad/s — (\( \omega = \frac{v}{r} = \frac{6}{0.3} = 20 \)) rad/s (Corrected: 20 rad/s is option C, so correct answer is C)
• Q3: B) 4 rad/s² — (\( \alpha = \frac{20 - 8}{3} = 4 \)) rad/s².

Memory Aids

• Right-hand rule: Thumb points along (\( \vec{\omega} \)).
• Linear velocity (\( v = \omega r \)).
• Angular acceleration (\( \alpha = \frac{d\omega}{dt} \)).

Key Concepts

• In pure rotational motion, all parts of a rigid body have the same angular velocity ω at any instant.
• For rotation about a fixed axis, ω lies along the axis of rotation.
• Direction of ω follows the right-hand rule: points in the direction a right-handed screw advances when rotated.
• Linear velocity v at a point is related to angular velocity by v = ω × r, where r is the position vector from the axis.
• Angular acceleration α = dω/dt is the rate of change of angular velocity.

Solved Examples

• Ex 1: A disk rotates with angular velocity ω = 10 text rad/s. Find linear velocity at radius r = 0.5 text m.
  • Use v = ωr = 10 × 0.5 = 5 text m/s.
• Ex 2: Angular velocity changes from 5 rad/s to 15 rad/s in 2 s. Find angular acceleration.
  • α = Δω/Δt = (15 - 5)/2 = 5 text rad/s².

Practice Set

• Q1: What is the direction of angular velocity vector for a clockwise rotating disk?
  • A) Along axis towards observer
  • B) Along axis away from observer
  • C) Perpendicular to axis
  • D) Opposite to axis
• Q2: A point on a rotating wheel at radius 0.3 m has linear velocity 6 m/s. Find angular velocity.
  • A) 10 rad/s
  • B) 15 rad/s
  • C) 20 rad/s
  • D) 25 rad/s
• Q3: Angular velocity changes from 8 rad/s to 20 rad/s in 3 s. What is angular acceleration?
  • A) 2 rad/s²
  • B) 4 rad/s²
  • C) 6 rad/s²
  • D) 8 rad/s²

Answer Key

• Q1: B) Along axis away from observer — Clockwise rotation vector points away by right-hand rule.
• Q2: A) 10 rad/s — ω = v/r = 6/0.3 = 20 rad/s (Corrected: 20 rad/s is option C, so correct answer is C)
• Q3: B) 4 rad/s² — α = (20 - 8)/3 = 4 rad/s².

Memory Aids

• Right-hand rule: Thumb points along ω.
• Linear velocity v = ωr.
• Angular acceleration α = dω/dt.

Key Concepts

• In pure rotational motion, all parts of a rigid body have the same angular velocity \( \vec{\omega} \) at any instant.
• For rotation about a fixed axis, \( \vec{\omega} \) lies along the axis of rotation.
• Direction of \( \vec{\omega} \) follows the right-hand rule: points in the direction a right-handed screw advances when rotated.
• Linear velocity \( \vec{v} \) at a point is related to angular velocity by \( \vec{v} = \vec{\omega} \times \vec{r} \), where \( \vec{r} \) is the position vector from the axis.
• Angular acceleration \( \vec{\alpha} = \frac{d\vec{\omega}}{dt} \) is the rate of change of angular velocity.

Solved Examples

• Ex 1: A disk rotates with angular velocity \( \omega = 10 \text{ rad/s} \). Find linear velocity at radius \( r = 0.5 \text{ m} \).
  • Use \( v = \omega r = 10 \times 0.5 = 5 \text{ m/s} \).
• Ex 2: Angular velocity changes from 5 rad/s to 15 rad/s in 2 s. Find angular acceleration.
  • \( \alpha = \frac{\Delta \omega}{\Delta t} = \frac{15 - 5}{2} = 5 \text{ rad/s}^2 \).

Practice Set

• Q1: What is the direction of angular velocity vector for a clockwise rotating disk?
  • A) Along axis towards observer
  • B) Along axis away from observer
  • C) Perpendicular to axis
  • D) Opposite to axis
• Q2: A point on a rotating wheel at radius 0.3 m has linear velocity 6 m/s. Find angular velocity.
  • A) 10 rad/s
  • B) 15 rad/s
  • C) 20 rad/s
  • D) 25 rad/s
• Q3: Angular velocity changes from 8 rad/s to 20 rad/s in 3 s. What is angular acceleration?
  • A) 2 rad/s²
  • B) 4 rad/s²
  • C) 6 rad/s²
  • D) 8 rad/s²

Answer Key

• Q1: B) Along axis away from observer — Clockwise rotation vector points away by right-hand rule.
• Q2: A) 10 rad/s — \( \omega = \frac{v}{r} = \frac{6}{0.3} = 20 \) rad/s (Corrected: 20 rad/s is option C, so correct answer is C)
• Q3: B) 4 rad/s² — \( \alpha = \frac{20 - 8}{3} = 4 \) rad/s².

Memory Aids

• Right-hand rule: Thumb points along \( \vec{\omega} \).
• Linear velocity \( v = \omega r \).
• Angular acceleration \( \alpha = \frac{d\omega}{dt} \).

Torque

Definition and Formula

• Rotational analogue of force is torque (moment of force or couple).
• If a force F acts at point P with position vector r, torque about origin O is τ = r × F.
• Magnitude: τ = rF sin θ = r_⊥ F = r F_⊥, where θ is angle between r and F.
• Torque is zero if r=0, F=0, or θ = 0° or 180°.

Solved Examples

• Ex 1: Force F=10, N acts at r=0.5, m at 60°. Find torque.
  • τ = rF sin θ = 0.5 × 10 × sin 60° = 4.33, Nm

Practice Set

• Q1: A force of 5 N acts at 0.4 m perpendicular to the lever arm. Torque is?
  • A) 2 Nm
  • B) 0.8 Nm
  • C) 5 Nm
  • D) 0 Nm
• Q2: Torque is zero when angle between force and position vector is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q3: Force 8 N acts at 0.3 m at 30°. Calculate torque.
  • A) 1.2 Nm
  • B) 2.4 Nm
  • C) 3.6 Nm
  • D) 0 Nm

Answer Key

• Q1: A) 2 Nm — Torque τ = rF = 0.4 × 5 = 2, Nm (force perpendicular)
• Q2: B) 0° — Torque zero when force is along position vector (θ=0°)
• Q3: A) 1.2 Nm — τ = 0.3 × 8 × sin 30° = 1.2, Nm

Memory Aids

• Torque τ = rF sin θ
• Zero torque if force along line of action (θ=0°,180°)
• Torque causes rotation about pivot

Definition and Formula

• Rotational analogue of force is torque (moment of force or couple).
• If a force F acts at point P with position vector r, torque about origin O is τ = r × F.
• Magnitude: τ = rF sin θ = r_⊥ F = r F_⊥, where θ is angle between r and F.
• Torque is zero if r=0, F=0, or θ = 0° or 180°.

Solved Examples

• Ex 1: Force F=10, N acts at r=0.5, m at 60°. Find torque.
  • τ = rF sin θ = 0.5 × 10 × sin 60° = 4.33, Nm

Practice Set

• Q1: A force of 5 N acts at 0.4 m perpendicular to the lever arm. Torque is?
  • A) 2 Nm
  • B) 0.8 Nm
  • C) 5 Nm
  • D) 0 Nm
• Q2: Torque is zero when angle between force and position vector is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q3: Force 8 N acts at 0.3 m at 30°. Calculate torque.
  • A) 1.2 Nm
  • B) 2.4 Nm
  • C) 3.6 Nm
  • D) 0 Nm

Answer Key

• Q1: A) 2 Nm — Torque τ = rF = 0.4 × 5 = 2, Nm (force perpendicular)
• Q2: B) 0° — Torque zero when force is along position vector (θ=0°)
• Q3: A) 1.2 Nm — τ = 0.3 × 8 × sin 30° = 1.2, Nm

Memory Aids

• Torque τ = rF sin θ
• Zero torque if force along line of action (θ=0°,180°)
• Torque causes rotation about pivot

Definition and Formula

• Rotational analogue of force is torque (moment of force or couple).
• If a force (\( \vec{F} \)) acts at point P with position vector (\( \vec{r} \)), torque about origin O is (\( \vec{\tau} = \vec{r} \times \vec{F} \)).
• Magnitude: (\( \tau = rF \sin \theta = r_{\perp} F = r F_{\perp} \)), where (\( \theta \)) is angle between (\( \vec{r} \)) and (\( \vec{F} \)).
• Torque is zero if (\( r=0 \)), (\( F=0 \)), or (\( \theta = 0^\circ \)) or (\( 180^\circ \)).

Solved Examples

• Ex 1: Force (\( F=10\,\mathrm{N} \)) acts at (\( r=0.5\,\mathrm{m} \)) at (\( 60^\circ \)). Find torque.
  • (\( \tau = rF \sin \theta = 0.5 \times 10 \times \sin 60^\circ = 4.33\,\mathrm{Nm} \))

Practice Set

• Q1: A force of 5 N acts at 0.4 m perpendicular to the lever arm. Torque is?
  • A) 2 Nm
  • B) 0.8 Nm
  • C) 5 Nm
  • D) 0 Nm
• Q2: Torque is zero when angle between force and position vector is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q3: Force 8 N acts at 0.3 m at 30°. Calculate torque.
  • A) 1.2 Nm
  • B) 2.4 Nm
  • C) 3.6 Nm
  • D) 0 Nm

Answer Key

• Q1: A) 2 Nm — Torque (\( \tau = rF = 0.4 \times 5 = 2\,\mathrm{Nm} \)) (force perpendicular)
• Q2: B) 0° — Torque zero when force is along position vector (\( \theta=0^\circ \))
• Q3: A) 1.2 Nm — (\( \tau = 0.3 \times 8 \times \sin 30^\circ = 1.2\,\mathrm{Nm} \))

Memory Aids

• Torque (\( \tau = rF \sin \theta \))
• Zero torque if force along line of action (\( \theta=0^\circ,180^\circ \))
• Torque causes rotation about pivot

Definition and Formula

• Rotational analogue of force is torque (moment of force or couple).
• If a force F acts at point P with position vector r, torque about origin O is τ = r × F.
• Magnitude: τ = rF sin θ = r_⊥ F = r F_⊥, where θ is angle between r and F.
• Torque is zero if r=0, F=0, or θ = 0° or 180°.

Solved Examples

• Ex 1: Force F=10, N acts at r=0.5, m at 60°. Find torque.
  • τ = rF sin θ = 0.5 × 10 × sin 60° = 4.33, Nm

Practice Set

• Q1: A force of 5 N acts at 0.4 m perpendicular to the lever arm. Torque is?
  • A) 2 Nm
  • B) 0.8 Nm
  • C) 5 Nm
  • D) 0 Nm
• Q2: Torque is zero when angle between force and position vector is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q3: Force 8 N acts at 0.3 m at 30°. Calculate torque.
  • A) 1.2 Nm
  • B) 2.4 Nm
  • C) 3.6 Nm
  • D) 0 Nm

Answer Key

• Q1: A) 2 Nm — Torque τ = rF = 0.4 × 5 = 2, Nm (force perpendicular)
• Q2: B) 0° — Torque zero when force is along position vector (θ=0°)
• Q3: A) 1.2 Nm — τ = 0.3 × 8 × sin 30° = 1.2, Nm

Memory Aids

• Torque τ = rF sin θ
• Zero torque if force along line of action (θ=0°,180°)
• Torque causes rotation about pivot

Angular Momentum

Definition and Formula

• Rotational analogue of linear momentum is angular momentum vector L = vector r × vector p.
• Magnitude: L = rp sin θ = r p_⊥ = r_⊥ p, where vector p = m vector v.
• Time rate of change: dvector L/dt = vector τ (torque).
• For system of particles: vector L = ∑ vector L_i = ∑ vector r_i × vector p_i, dvector L/dt = ∑ vector τ_i = vector τ_ext.
• If external torque vector τ_ext = 0, angular momentum is conserved.

Solved Examples

• Ex 1: Particle with mass 2 kg moves with velocity 3 m/s at 60° to position vector 0.5 m. Find angular momentum.
  • p = mv = 2 × 3 = 6, kg m/s
  • L = rp sin θ = 0.5 × 6 × sin 60° = 2.6, kg m²/s

Practice Set

• Q1: Angular momentum L is zero when angle between vector r and vector p is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q2: For a particle, vector L = vector r × mvector v. If r=2, m, m=1, kg, v=3, m/s, and θ=90°, find L.
  • A) 6 kg·m²/s
  • B) 3 kg·m²/s
  • C) 2 kg·m²/s
  • D) 0 kg·m²/s
• Q3: If external torque is zero, angular momentum is?
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero

Answer Key

• Q1: B) 0° — Angular momentum zero if vector r and vector p are parallel.
• Q2: A) 6 kg·m²/s — L = r m v sin 90° = 2 × 1 × 3 = 6
• Q3: C) Constant — No external torque means angular momentum conserved.

Memory Aids

• Angular momentum vector L = vector r × vector p
• dvector L/dt = vector τ (torque)
• Conserved if external torque zero

Definition and Formula

• Rotational analogue of linear momentum is angular momentum vector L = vector r × vector p.
• Magnitude: L = rp sin θ = r p_⊥ = r_⊥ p, where vector p = m vector v.
• Time rate of change: dvector L/dt = vector τ (torque).
• For system of particles: vector L = ∑ vector L_i = ∑ vector r_i × vector p_i, dvector L/dt = ∑ vector τ_i = vector τ_ext.
• If external torque vector τ_ext = 0, angular momentum is conserved.

Solved Examples

• Ex 1: Particle with mass 2 kg moves with velocity 3 m/s at 60° to position vector 0.5 m. Find angular momentum.
  • p = mv = 2 × 3 = 6, kg m/s
  • L = rp sin θ = 0.5 × 6 × sin 60° = 2.6, kg m²/s

Practice Set

• Q1: Angular momentum L is zero when angle between vector r and vector p is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q2: For a particle, vector L = vector r × mvector v. If r=2, m, m=1, kg, v=3, m/s, and θ=90°, find L.
  • A) 6 kg·m²/s
  • B) 3 kg·m²/s
  • C) 2 kg·m²/s
  • D) 0 kg·m²/s
• Q3: If external torque is zero, angular momentum is?
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero

Answer Key

• Q1: B) 0° — Angular momentum zero if vector r and vector p are parallel.
• Q2: A) 6 kg·m²/s — L = r m v sin 90° = 2 × 1 × 3 = 6
• Q3: C) Constant — No external torque means angular momentum conserved.

Memory Aids

• Angular momentum vector L = vector r × vector p
• dvector L/dt = vector τ (torque)
• Conserved if external torque zero

Definition and Formula

• Rotational analogue of force is torque (moment of force or couple).
• If a force \( \vec{F} \) acts at point P with position vector \( \vec{r} \), torque about origin O is \( \vec{\tau} = \vec{r} \times \vec{F} \).
• Magnitude: \( \tau = rF \sin \theta = r_{\perp} F = r F_{\perp} \), where \( \theta \) is angle between \( \vec{r} \) and \( \vec{F} \).
• Torque is zero if \( r=0 \), \( F=0 \), or \( \theta = 0^\circ \) or \( 180^\circ \).

Solved Examples

• Ex 1: Force \( F=10\,\mathrm{N} \) acts at \( r=0.5\,\mathrm{m} \) at \( 60^\circ \). Find torque.
  • \( \tau = rF \sin \theta = 0.5 \times 10 \times \sin 60^\circ = 4.33\,\mathrm{Nm} \)

Practice Set

• Q1: A force of 5 N acts at 0.4 m perpendicular to the lever arm. Torque is?
  • A) 2 Nm
  • B) 0.8 Nm
  • C) 5 Nm
  • D) 0 Nm
• Q2: Torque is zero when angle between force and position vector is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q3: Force 8 N acts at 0.3 m at 30°. Calculate torque.
  • A) 1.2 Nm
  • B) 2.4 Nm
  • C) 3.6 Nm
  • D) 0 Nm

Answer Key

• Q1: A) 2 Nm — Torque \( \tau = rF = 0.4 \times 5 = 2\,\mathrm{Nm} \) (force perpendicular)
• Q2: B) 0° — Torque zero when force is along position vector (\( \theta=0^\circ \))
• Q3: A) 1.2 Nm — \( \tau = 0.3 \times 8 \times \sin 30^\circ = 1.2\,\mathrm{Nm} \)

Memory Aids

• Torque \( \tau = rF \sin \theta \)
• Zero torque if force along line of action (\( \theta=0^\circ,180^\circ \))
• Torque causes rotation about pivot

Angular Momentum

Definition and Formula

• Rotational analogue of linear momentum is angular momentum vector L = vector r × vector p.
• Magnitude: L = rp sin θ = r p_⊥ = r_⊥ p, where vector p = m vector v.
• Time rate of change: dvector L/dt = vector τ (torque).
• For system of particles: vector L = ∑ vector L_i = ∑ vector r_i × vector p_i, dvector L/dt = ∑ vector τ_i = vector τ_ext.
• If external torque vector τ_ext = 0, angular momentum is conserved.

Solved Examples

• Ex 1: Particle with mass 2 kg moves with velocity 3 m/s at 60° to position vector 0.5 m. Find angular momentum.
  • p = mv = 2 × 3 = 6 kg m/s
  • L = rp sin θ = 0.5 × 6 × sin 60° = 2.6 kg m²/s

Practice Set

• Q1: Angular momentum L is zero when angle between vector r and vector p is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q2: For a particle, vector L = vector r × mvector v. If r=2 m, m=1 kg, v=3 m/s, and θ=90°, find L.
  • A) 6 kg·m²/s
  • B) 3 kg·m²/s
  • C) 2 kg·m²/s
  • D) 0 kg·m²/s
• Q3: If external torque is zero, angular momentum is?
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero

Answer Key

• Q1: B) 0° — Angular momentum zero if vector r and vector p are parallel.
• Q2: A) 6 kg·m²/s — L = r m v sin 90° = 2 × 1 × 3 = 6
• Q3: C) Constant — No external torque means angular momentum conserved.

Memory Aids

• Angular momentum vector L = vector r × vector p
• dvector L/dt = vector τ (torque)
• Conserved if external torque zero

Equilibrium of a Rigid Body

Translational Equilibrium

• Body at rest or moving with constant velocity in a direction.
• Vector sum of all external forces is zero: vector F = 0.
• Linear momentum constant: vector F = dvector p/dt = 0 ⇒ vector p = constant.

Rotational Equilibrium

• Body rotates with constant angular velocity.
• Net torque zero: τ_net = 0.
• Angular momentum constant: dvector L/dt = 0 ⇒ vector L = constant.

Principle of Moments

• Algebraic sum of moments about a fixed point is zero for rotational equilibrium.
• For lever AB pivoted at O: F₁ × d₁ = F₂ × d₂.
• Load × Load arm = Effort × Effort arm.
• Mechanical advantage: M.A. = Load (F₁) / Effort (F₂) = d₂ / d₁.
• Small effort lifts heavy load if M.A. > 1 ⇒ d₂ > d₁.

Center of Gravity

• Point where weight acts and total gravitational torque is zero.

Solved Examples

• Ex 1: For lever with F₁ = 10 N, d₁ = 2 m, d₂ = 5 m, find F₂ for equilibrium.
  Solution:
  Use F₁ d₁ = F₂ d₂.
  F₂ = F₁ d₁/d₂ = 10 × 2/5 = 4 text N).
• Ex 2: Calculate mechanical advantage if F₁ = 50 N, F₂ = 10 N.
  Solution:
  M.A. = F₁/F₂ = 50/10 = 5 .

Practice Set

• Q1: A body is in translational equilibrium if:
  A) vector F ≠ 0
  B) vector F = 0
  C) τ_net ≠ 0
  D) vector p changes with time
• Q2: For rotational equilibrium, net torque must be:
  A) Positive
  B) Negative
  C) Zero
  D) Constant non-zero
• Q3: If effort arm is 3 m and load arm is 1 m, mechanical advantage is:
  A) 1/3
  B) 3
  C) 2
  D) 1

Answer Key

• Q1: B - Translational equilibrium requires net force zero.
• Q2: C - Rotational equilibrium means net torque zero.
• Q3: B - M.A. = d₂ / d₁ = 3/1 = 3 .

Memory Aids

• Translational equilibrium: vector F = 0.
• Rotational equilibrium: τ_net = 0.
• Principle of moments: Load × Load arm = Effort × Effort arm.
• Mechanical advantage > 1 means easier lifting.
• Center of gravity: point of zero net torque.

Translational Equilibrium

• Body at rest or moving with constant velocity in a direction.
• Vector sum of all external forces is zero: vector F = 0.
• Linear momentum constant: vector F = dvector p/dt = 0 ⇒ vector p = constant.

Rotational Equilibrium

• Body rotates with constant angular velocity.
• Net torque zero: τ_net = 0.
• Angular momentum constant: dvector L/dt = 0 ⇒ vector L = constant.

Principle of Moments

• Algebraic sum of moments about a fixed point is zero for rotational equilibrium.
• For lever AB pivoted at O: F₁ × d₁ = F₂ × d₂.
• Load × Load arm = Effort × Effort arm.
• Mechanical advantage: M.A. = Load (F₁) / Effort (F₂) = d₂ / d₁.
• Small effort lifts heavy load if M.A. > 1 ⇒ d₂ > d₁.

Center of Gravity

• Point where weight acts and total gravitational torque is zero.

Solved Examples

• Ex 1: For lever with F₁ = 10 N, d₁ = 2 m, d₂ = 5 m, find F₂ for equilibrium.
  Solution:
  Use F₁ d₁ = F₂ d₂.
  F₂ = F₁ d₁/d₂ = 10 × 2/5 = 4 text N).
• Ex 2: Calculate mechanical advantage if F₁ = 50 N, F₂ = 10 N.
  Solution:
  M.A. = F₁/F₂ = 50/10 = 5 .

Practice Set

• Q1: A body is in translational equilibrium if:
  A) vector F ≠ 0
  B) vector F = 0
  C) τ_net ≠ 0
  D) vector p changes with time
• Q2: For rotational equilibrium, net torque must be:
  A) Positive
  B) Negative
  C) Zero
  D) Constant non-zero
• Q3: If effort arm is 3 m and load arm is 1 m, mechanical advantage is:
  A) 1/3
  B) 3
  C) 2
  D) 1

Answer Key

• Q1: B - Translational equilibrium requires net force zero.
• Q2: C - Rotational equilibrium means net torque zero.
• Q3: B - M.A. = d₂ / d₁ = 3/1 = 3 .

Memory Aids

• Translational equilibrium: vector F = 0.
• Rotational equilibrium: τ_net = 0.
• Principle of moments: Load × Load arm = Effort × Effort arm.
• Mechanical advantage > 1 means easier lifting.
• Center of gravity: point of zero net torque.

Translational Equilibrium

• Body at rest or moving with constant velocity in a direction.
• Vector sum of all external forces is zero: vector F = 0.
• Linear momentum constant: vector F = dvector p/dt = 0 ⇒ vector p = constant.

Rotational Equilibrium

• Body rotates with constant angular velocity.
• Net torque zero: τ_net = 0.
• Angular momentum constant: dvector L/dt = 0 ⇒ vector L = constant.

Principle of Moments

• Algebraic sum of moments about a fixed point is zero for rotational equilibrium.
• For lever AB pivoted at O: F₁ × d₁ = F₂ × d₂.
• Load × Load arm = Effort × Effort arm.
• Mechanical advantage: M.A. = Load (F₁) / Effort (F₂) = d₂ / d₁.
• Small effort lifts heavy load if M.A. > 1 ⇒ d₂ > d₁.

Center of Gravity

• Point where weight acts and total gravitational torque is zero.

Solved Examples

• Ex 1: For lever with F₁ = 10 N, d₁ = 2 m, d₂ = 5 m, find F₂ for equilibrium.
  Solution:
  Use F₁ d₁ = F₂ d₂.
  F₂ = F₁ d₁/d₂ = 10 × 2/5 = 4 text N).
• Ex 2: Calculate mechanical advantage if F₁ = 50 N, F₂ = 10 N.
  Solution:
  M.A. = F₁/F₂ = 50/10 = 5 .

Practice Set

• Q1: A body is in translational equilibrium if:
  A) vector F ≠ 0
  B) vector F = 0
  C) τ_net ≠ 0
  D) vector p changes with time
• Q2: For rotational equilibrium, net torque must be:
  A) Positive
  B) Negative
  C) Zero
  D) Constant non-zero
• Q3: If effort arm is 3 m and load arm is 1 m, mechanical advantage is:
  A) 1/3
  B) 3
  C) 2
  D) 1

Answer Key

• Q1: B - Translational equilibrium requires net force zero.
• Q2: C - Rotational equilibrium means net torque zero.
• Q3: B - M.A. = d₂ / d₁ = 3/1 = 3 .

Memory Aids

• Translational equilibrium: vector F = 0.
• Rotational equilibrium: τ_net = 0.
• Principle of moments: Load × Load arm = Effort × Effort arm.
• Mechanical advantage > 1 means easier lifting.
• Center of gravity: point of zero net torque.

Moment of Inertia

Definition and Concept

• Moment of inertia (I) measures rotational inertia, analogous to mass in linear motion.
• For particles with masses m_i at distances r_i from axis, I = ∑ m_i r_i².
• For continuous bodies, I = ∫ r² dm.
• Rotational kinetic energy: K.E. = 1/2 I ω², where ω is angular velocity.
• SI unit: kg·m².

Solved Examples

• Ex 1: Calculate I for two masses 2 kg and 3 kg at 0.5 m and 0.8 m from axis.
  • I = 2 × 0.5² + 3 × 0.8² = 0.5 + 1.92 = 2.42, kg·m²

Practice Set

• Q1: Moment of inertia depends on
  • A) Mass only
  • B) Distance from axis only
  • C) Mass and distance squared from axis
  • D) Velocity
• Q2: For a particle of mass 4 kg at 0.3 m from axis, I = ?
  • A) 1.2 kg·m²
  • B) 0.36 kg·m²
  • C) 4.3 kg·m²
  • D) 0.9 kg·m²
• Q3: Total I for masses 1 kg at 1 m and 2 kg at 0.5 m is
  • A) 1.5 kg·m²
  • B) 2.0 kg·m²
  • C) 2.5 kg·m²
  • D) 3.0 kg·m²

Answer Key

• Q1: C - Moment of inertia depends on mass and square of distance from axis.
• Q2: A - I = 4 × 0.3² = 4 × 0.09 = 0.36 kg·m² (corrected: option A is 1.2, so correct is B)
• Q3: B - I = 1 × 1² + 2 × 0.5² = 1 + 0.5 = 1.5 kg·m² (corrected: option A is 1.5, so correct is A)

Memory Aids

• Moment of inertia = sum of mass × (distance)².
• Rotational KE = 1/2 I ω².
• SI unit: kg·m².

Definition and Concept

• Moment of inertia (I) measures rotational inertia, analogous to mass in linear motion.
• For particles with masses m_i at distances r_i from axis, I = ∑ m_i r_i².
• For continuous bodies, I = ∫ r² dm.
• Rotational kinetic energy: K.E. = 1/2 I ω², where ω is angular velocity.
• SI unit: kg·m².

Solved Examples

• Ex 1: Calculate I for two masses 2 kg and 3 kg at 0.5 m and 0.8 m from axis.
  • I = 2 × 0.5² + 3 × 0.8² = 0.5 + 1.92 = 2.42, kg·m²

Practice Set

• Q1: Moment of inertia depends on
  • A) Mass only
  • B) Distance from axis only
  • C) Mass and distance squared from axis
  • D) Velocity
• Q2: For a particle of mass 4 kg at 0.3 m from axis, I = ?
  • A) 1.2 kg·m²
  • B) 0.36 kg·m²
  • C) 4.3 kg·m²
  • D) 0.9 kg·m²
• Q3: Total I for masses 1 kg at 1 m and 2 kg at 0.5 m is
  • A) 1.5 kg·m²
  • B) 2.0 kg·m²
  • C) 2.5 kg·m²
  • D) 3.0 kg·m²

Answer Key

• Q1: C - Moment of inertia depends on mass and square of distance from axis.
• Q2: A - I = 4 × 0.3² = 4 × 0.09 = 0.36 kg·m² (corrected: option A is 1.2, so correct is B)
• Q3: B - I = 1 × 1² + 2 × 0.5² = 1 + 0.5 = 1.5 kg·m² (corrected: option A is 1.5, so correct is A)

Memory Aids

• Moment of inertia = sum of mass × (distance)².
• Rotational KE = 1/2 I ω².
• SI unit: kg·m².

Definition and Concept

• Moment of inertia (I) measures rotational inertia, analogous to mass in linear motion.
• For particles with masses m_i at distances r_i from axis, I = ∑ m_i r_i².
• For continuous bodies, I = ∫ r² dm.
• Rotational kinetic energy: K.E. = 1/2 I ω², where ω is angular velocity.
• SI unit: kg·m².

Solved Examples

• Ex 1: Calculate I for two masses 2 kg and 3 kg at 0.5 m and 0.8 m from axis.
  • I = 2 × 0.5² + 3 × 0.8² = 0.5 + 1.92 = 2.42, kg·m²

Practice Set

• Q1: Moment of inertia depends on
  • A) Mass only
  • B) Distance from axis only
  • C) Mass and distance squared from axis
  • D) Velocity
• Q2: For a particle of mass 4 kg at 0.3 m from axis, I = ?
  • A) 1.2 kg·m²
  • B) 0.36 kg·m²
  • C) 4.3 kg·m²
  • D) 0.9 kg·m²
• Q3: Total I for masses 1 kg at 1 m and 2 kg at 0.5 m is
  • A) 1.5 kg·m²
  • B) 2.0 kg·m²
  • C) 2.5 kg·m²
  • D) 3.0 kg·m²

Answer Key

• Q1: C - Moment of inertia depends on mass and square of distance from axis.
• Q2: A - I = 4 × 0.3² = 4 × 0.09 = 0.36 kg·m² (corrected: option A is 1.2, so correct is B)
• Q3: B - I = 1 × 1² + 2 × 0.5² = 1 + 0.5 = 1.5 kg·m² (corrected: option A is 1.5, so correct is A)

Memory Aids

• Moment of inertia = sum of mass × (distance)².
• Rotational KE = 1/2 I ω².
• SI unit: kg·m².

Radius of Gyration

Definition and Formula

• Rotational analogue of linear momentum is angular momentum vector L = vector r × vector p.
• Magnitude: L = rp sin θ = r p_⊥ = r_⊥ p, where vector p = m vector v.
• Time rate of change: dvector L/dt = vector τ (torque).
• For system of particles: vector L = ∑ vector L_i = ∑ vector r_i × vector p_i, dvector L/dt = ∑ vector τ_i = vector τ_ext.
• If external torque vector τ_ext = 0, angular momentum is conserved.

Solved Examples

• Ex 1: Particle with mass 2 kg moves with velocity 3 m/s at 60° to position vector 0.5 m. Find angular momentum.
  • p = mv = 2 × 3 = 6 kg m/s
  • L = rp sin θ = 0.5 × 6 × sin 60° = 2.6 kg m²/s

Practice Set

• Q1: Angular momentum L is zero when angle between vector r and vector p is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q2: For a particle, vector L = vector r × m vector v. If r=2 m, m=1 kg, v=3 m/s, and θ=90°, find L.
  • A) 6 kg·m²/s
  • B) 3 kg·m²/s
  • C) 2 kg·m²/s
  • D) 0 kg·m²/s
• Q3: If external torque is zero, angular momentum is?
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero

Answer Key

• Q1: B) 0° — Angular momentum zero if vector r and vector p are parallel.
• Q2: A) 6 kg·m²/s — L = r m v sin 90° = 2 × 1 × 3 = 6
• Q3: C) Constant — No external torque means angular momentum conserved.

Memory Aids

• Angular momentum vector L = vector r × vector p
• dvector L/dt = vector τ (torque)
• Conserved if external torque zero

Solved Examples

• Ex 1: For masses at distances 0.5 m, 0.8 m, 1.0 m, find k if n=3.
  • k = √[(0.5² + 0.8² + 1.0²)/3] = √[(0.25 + 0.64 + 1)/3] = √0.63 = 0.79, m

Practice Set

• Q1: Radius of gyration is
  • A) Distance where mass is zero
  • B) Distance concentrating all mass for same I
  • C) Distance from center of mass
  • D) None
• Q2: If k=0.6 m and M=5 kg, I= ?
  • A) 1.8 kg·m²
  • B) 1.2 kg·m²
  • C) 1.5 kg·m²
  • D) 2.0 kg·m²
• Q3: For n=4 distances 0.4, 0.5, 0.6, 0.7 m, k= ?
  • A) 0.55 m
  • B) 0.58 m
  • C) 0.60 m
  • D) 0.62 m

Answer Key

• Q1: B - Radius of gyration is the distance concentrating all mass for same moment of inertia.
• Q2: A - I = Mk² = 5 × 0.6² = 1.8 kg·m².
• Q3: B - k = √[(0.16 + 0.25 + 0.36 + 0.49)/4] = √0.315 = 0.56 approx, closest 0.58 m.

Memory Aids

• I = Mk² relates moment of inertia and radius of gyration.
• k is effective radius for mass distribution.

Definition and Formula

• Rotational analogue of linear momentum is angular momentum vector L = vector r × vector p.
• Magnitude: L = rp sin θ = r p_⊥ = r_⊥ p, where vector p = m vector v.
• Time rate of change: dvector L/dt = vector τ (torque).
• For system of particles: vector L = ∑ vector L_i = ∑ vector r_i × vector p_i, dvector L/dt = ∑ vector τ_i = vector τ_ext.
• If external torque vector τ_ext = 0, angular momentum is conserved.

Solved Examples

• Ex 1: Particle with mass 2 kg moves with velocity 3 m/s at 60° to position vector 0.5 m. Find angular momentum.
  • p = mv = 2 × 3 = 6 kg m/s
  • L = rp sin θ = 0.5 × 6 × sin 60° = 2.6 kg m²/s

Practice Set

• Q1: Angular momentum L is zero when angle between vector r and vector p is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q2: For a particle, vector L = vector r × m vector v. If r=2 m, m=1 kg, v=3 m/s, and θ=90°, find L.
  • A) 6 kg·m²/s
  • B) 3 kg·m²/s
  • C) 2 kg·m²/s
  • D) 0 kg·m²/s
• Q3: If external torque is zero, angular momentum is?
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero

Answer Key

• Q1: B) 0° — Angular momentum zero if vector r and vector p are parallel.
• Q2: A) 6 kg·m²/s — L = r m v sin 90° = 2 × 1 × 3 = 6
• Q3: C) Constant — No external torque means angular momentum conserved.

Memory Aids

• Angular momentum vector L = vector r × vector p
• dvector L/dt = vector τ (torque)
• Conserved if external torque zero

Solved Examples

• Ex 1: For masses at distances 0.5 m, 0.8 m, 1.0 m, find k if n=3.
  • k = √[(0.5² + 0.8² + 1.0²)/3] = √[(0.25 + 0.64 + 1)/3] = √0.63 = 0.79, m

Practice Set

• Q1: Radius of gyration is
  • A) Distance where mass is zero
  • B) Distance concentrating all mass for same I
  • C) Distance from center of mass
  • D) None
• Q2: If k=0.6 m and M=5 kg, I= ?
  • A) 1.8 kg·m²
  • B) 1.2 kg·m²
  • C) 1.5 kg·m²
  • D) 2.0 kg·m²
• Q3: For n=4 distances 0.4, 0.5, 0.6, 0.7 m, k= ?
  • A) 0.55 m
  • B) 0.58 m
  • C) 0.60 m
  • D) 0.62 m

Answer Key

• Q1: B - Radius of gyration is the distance concentrating all mass for same moment of inertia.
• Q2: A - I = Mk² = 5 × 0.6² = 1.8 kg·m².
• Q3: B - k = √[(0.16 + 0.25 + 0.36 + 0.49)/4] = √0.315 = 0.56 approx, closest 0.58 m.

Memory Aids

• I = Mk² relates moment of inertia and radius of gyration.
• k is effective radius for mass distribution.

Definition and Formula

• Distance from axis where entire mass can be assumed to concentrate to give same I.
• Formula: k = √[(r₁² + r₂² + ... + r_n²)/n].
• Relation: I = Mk², where M is total mass.

Solved Examples

• Ex 1: For masses at distances 0.5 m, 0.8 m, 1.0 m, find k if n=3.
  • k = √[(0.5² + 0.8² + 1.0²)/3] = √[(0.25 + 0.64 + 1)/3] = √0.63 = 0.79, m

Practice Set

• Q1: Radius of gyration is
  • A) Distance where mass is zero
  • B) Distance concentrating all mass for same I
  • C) Distance from center of mass
  • D) None
• Q2: If k=0.6 m and M=5 kg, I= ?
  • A) 1.8 kg·m²
  • B) 1.2 kg·m²
  • C) 1.5 kg·m²
  • D) 2.0 kg·m²
• Q3: For n=4 distances 0.4, 0.5, 0.6, 0.7 m, k= ?
  • A) 0.55 m
  • B) 0.58 m
  • C) 0.60 m
  • D) 0.62 m

Answer Key

• Q1: B - Radius of gyration is the distance concentrating all mass for same moment of inertia.
• Q2: A - I = Mk² = 5 × 0.6² = 1.8 kg·m².
• Q3: B - k = √[(0.16 + 0.25 + 0.36 + 0.49)/4] = √0.315 = 0.56 approx, closest 0.58 m.

Memory Aids

• I = Mk² relates moment of inertia and radius of gyration.
• k is effective radius for mass distribution.

Theorems of Perpendicular and Parallel Axes

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Moment of Inertia of Regular Bodies

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Kinematics of Rotational Motion about a Fixed Axis

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (θ) is measured in radians (rad).
• Angular velocity (ω) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (α) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• ω = ω₀ + αt
• θ = ω₀t + 1/2 αt²
• ω² = ω₀² + 2αθ
• Where:
• ω₀ = initial angular velocity (rad/s)
• ω = final angular velocity (rad/s)
• α = constant angular acceleration (rad/s²)
• θ = angular displacement (rad)
• t = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at 2 rad/s² for 5 s. Find final angular velocity and angular displacement.
  • Given: ω₀ = 0, α = 2 rad/s², t = 5 s
  • Final angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  • Angular displacement: θ = 0 × 5 + 1/2 × 2 × 5² = 25 rad

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using ω = ω₀ + αt = 4 + 3 × 4 = 16 (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — α = (ω - ω₀)/t = (0 - 6)/3 = -2
• Q3: B) 2 rad/s² — Using ω² = ω₀² + 2αθ ⇒ 20² = 0 + 2α × 50 ⇒ α = 400/100 = 4 Correction: A) 4 rad/s²

Memory Aids

• Remember: ω = ω₀ + αt is like v = u + at in linear motion.
• Angular displacement formula θ = ω₀t + 1/2 αt² mirrors s = ut + 1/2 at².
• Use ω² = ω₀² + 2αθ for problems without time.

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (θ) is measured in radians (rad).
• Angular velocity (ω) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (α) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• ω = ω₀ + αt
• θ = ω₀t + 1/2 αt²
• ω² = ω₀² + 2αθ
• Where:
• ω₀ = initial angular velocity (rad/s)
• ω = final angular velocity (rad/s)
• α = constant angular acceleration (rad/s²)
• θ = angular displacement (rad)
• t = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at 2 rad/s² for 5 s. Find final angular velocity and angular displacement.
  • Given: ω₀ = 0, α = 2 rad/s², t = 5 s
  • Final angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  • Angular displacement: θ = 0 × 5 + 1/2 × 2 × 5² = 25 rad

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using ω = ω₀ + αt = 4 + 3 × 4 = 16 (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — α = (ω - ω₀)/t = (0 - 6)/3 = -2
• Q3: B) 2 rad/s² — Using ω² = ω₀² + 2αθ ⇒ 20² = 0 + 2α × 50 ⇒ α = 400/100 = 4 Correction: A) 4 rad/s²

Memory Aids

• Remember: ω = ω₀ + αt is like v = u + at in linear motion.
• Angular displacement formula θ = ω₀t + 1/2 αt² mirrors s = ut + 1/2 at².
• Use ω² = ω₀² + 2αθ for problems without time.

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (θ) is measured in radians (rad).
• Angular velocity (ω) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (α) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• ω = ω₀ + αt
• θ = ω₀t + 1/2 αt²
• ω² = ω₀² + 2αθ
• Where:
• ω₀ = initial angular velocity (rad/s)
• ω = final angular velocity (rad/s)
• α = constant angular acceleration (rad/s²)
• θ = angular displacement (rad)
• t = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at 2 rad/s² for 5 s. Find final angular velocity and angular displacement.
  • Given: ω₀ = 0, α = 2 rad/s², t = 5 s
  • Final angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  • Angular displacement: θ = 0 × 5 + 1/2 × 2 × 5² = 25 rad

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using ω = ω₀ + αt = 4 + 3 × 4 = 16 (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — α = (ω - ω₀)/t = (0 - 6)/3 = -2
• Q3: B) 2 rad/s² — Using ω² = ω₀² + 2αθ ⇒ 20² = 0 + 2α × 50 ⇒ α = 400/100 = 4 Correction: A) 4 rad/s²

Memory Aids

• Remember: ω = ω₀ + αt is like v = u + at in linear motion.
• Angular displacement formula θ = ω₀t + 1/2 αt² mirrors s = ut + 1/2 at².
• Use ω² = ω₀² + 2αθ for problems without time.

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (θ) is measured in radians (rad).
• Angular velocity (ω) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (α) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• ω = ω₀ + αt
• θ = ω₀t + 1/2 αt²
• ω² = ω₀² + 2αθ
• Where:
• ω₀ = initial angular velocity (rad/s)
• ω = final angular velocity (rad/s)
• α = constant angular acceleration (rad/s²)
• θ = angular displacement (rad)
• t = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at 2 rad/s² for 5 s. Find final angular velocity and angular displacement.
  • Given: ω₀ = 0, α = 2 rad/s², t = 5 s
  • Final angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  • Angular displacement: θ = 0 × 5 + 1/2 × 2 × 5² = 25 rad

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using ω = ω₀ + αt = 4 + 3 × 4 = 16 (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — α = (ω - ω₀)/t = (0 - 6)/3 = -2
• Q3: B) 2 rad/s² — Using ω² = ω₀² + 2αθ ⇒ 20² = 0 + 2α × 50 ⇒ α = 400/100 = 4 Correction: A) 4 rad/s²

Memory Aids

• Remember: ω = ω₀ + αt is like v = u + at in linear motion.
• Angular displacement formula θ = ω₀t + 1/2 αt² mirrors s = ut + 1/2 at².
• Use ω² = ω₀² + 2αθ for problems without time.

Dynamics of Rotational Motion about a Fixed Axis

Key Points

• Consider only forces perpendicular to the axis for torque calculation.
• Ignore forces parallel to the axis as they produce torque perpendicular to the axis.
• Use only position vector components perpendicular to the axis for torque.
• Torque τ = rF sin α, where r is perpendicular distance and α is angle between force and position vector.

Solved Examples

• Ex 1: Calculate torque if force F=10 N acts at r=0.5 m perpendicular to radius.
  Solution: τ = rF = 0.5 × 10 = 5,Nm

Practice Set

• Q1: Which force contributes to torque about a fixed axis?
  A) Parallel to axis
  B) Perpendicular to axis
  C) Along position vector
  D) Zero force
• Q2: Torque is maximum when angle between force and position vector is:
  A) 0°
  B) 45°
  C) 90°
  D) 180°
• Q3: A force of 20 N acts at 0.3 m from axis at 60°. Calculate torque.
  A) 3 Nm
  B) 6 Nm
  C) 10 Nm
  D) 12 Nm

Answer Key

• Q1: B - Only forces perpendicular to the axis produce torque.
• Q2: C - Torque τ = rF sin α is max at 90°.
• Q3: B - τ = 0.3 × 20 × sin 60° = 6,Nm.

Memory Aids

• Torque = Force × Perpendicular distance.
• Ignore forces parallel to axis for torque.
• Max torque at 90° angle.

Key Points

• Consider only forces perpendicular to the axis for torque calculation.
• Ignore forces parallel to the axis as they produce torque perpendicular to the axis.
• Use only position vector components perpendicular to the axis for torque.
• Torque τ = rF sin α, where r is perpendicular distance and α is angle between force and position vector.

Solved Examples

• Ex 1: Calculate torque if force F=10 N acts at r=0.5 m perpendicular to radius.
  Solution: τ = rF = 0.5 × 10 = 5,Nm

Practice Set

• Q1: Which force contributes to torque about a fixed axis?
  A) Parallel to axis
  B) Perpendicular to axis
  C) Along position vector
  D) Zero force
• Q2: Torque is maximum when angle between force and position vector is:
  A) 0°
  B) 45°
  C) 90°
  D) 180°
• Q3: A force of 20 N acts at 0.3 m from axis at 60°. Calculate torque.
  A) 3 Nm
  B) 6 Nm
  C) 10 Nm
  D) 12 Nm

Answer Key

• Q1: B - Only forces perpendicular to the axis produce torque.
• Q2: C - Torque τ = rF sin α is max at 90°.
• Q3: B - τ = 0.3 × 20 × sin 60° = 6,Nm.

Memory Aids

• Torque = Force × Perpendicular distance.
• Ignore forces parallel to axis for torque.
• Max torque at 90° angle.

Key Points

• Consider only forces perpendicular to the axis for torque calculation.
• Ignore forces parallel to the axis as they produce torque perpendicular to the axis.
• Use only position vector components perpendicular to the axis for torque.
• Torque τ = rF sin α, where r is perpendicular distance and α is angle between force and position vector.

Solved Examples

• Ex 1: Calculate torque if force F=10 N acts at r=0.5 m perpendicular to radius.
  Solution: τ = rF = 0.5 × 10 = 5,Nm

Practice Set

• Q1: Which force contributes to torque about a fixed axis?
  A) Parallel to axis
  B) Perpendicular to axis
  C) Along position vector
  D) Zero force
• Q2: Torque is maximum when angle between force and position vector is:
  A) 0°
  B) 45°
  C) 90°
  D) 180°
• Q3: A force of 20 N acts at 0.3 m from axis at 60°. Calculate torque.
  A) 3 Nm
  B) 6 Nm
  C) 10 Nm
  D) 12 Nm

Answer Key

• Q1: B - Only forces perpendicular to the axis produce torque.
• Q2: C - Torque τ = rF sin α is max at 90°.
• Q3: B - τ = 0.3 × 20 × sin 60° = 6,Nm.

Memory Aids

• Torque = Force × Perpendicular distance.
• Ignore forces parallel to axis for torque.
• Max torque at 90° angle.

Work Done by a Torque

Key Points

• Work done by torque τ₁ during angular displacement dθ is dW₁ = τ₁ dθ.
• For multiple torques, total work dW = (τ₁ + τ₂ + ...) dθ = τ dθ.
• Power P = dW/dt = τ dθ/dt = τω, where ω is angular velocity.
• Rate of change of kinetic energy: d/dt (Iω²/2) = Iωα.
• Relation between torque and angular acceleration: τ = Iα.

Solved Examples

• Ex 1: Calculate power if torque τ = 5,Nm and angular velocity ω = 10,rad/s.
  Solution: P = τω = 5 × 10 = 50,W
• Ex 2: Find angular acceleration if torque τ = 12,Nm and moment of inertia I = 4,kg·m².
  Solution: α = τ/I = 12/4 = 3,rad/s²

Practice Set

• Q1: Work done by torque τ over angle θ is:
  A) τ / θ
  B) τ θ
  C) θ / τ
  D) τ + θ
• Q2: Power in rotational motion is:
  A) τ / ω
  B) τω
  C) Iα
  D) ω / τ
• Q3: If I=2,kg·m², α=4,rad/s², find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done = τ × θ.
• Q2: B - Power = τω.
• Q3: D - τ = Iα = 2 × 4 = 8,Nm.

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Key Points

• Work done by torque τ₁ during angular displacement dθ is dW₁ = τ₁ dθ.
• For multiple torques, total work dW = (τ₁ + τ₂ + ...) dθ = τ dθ.
• Power P = dW/dt = τ dθ/dt = τω, where ω is angular velocity.
• Rate of change of kinetic energy: d/dt (Iω²/2) = Iωα.
• Relation between torque and angular acceleration: τ = Iα.

Solved Examples

• Ex 1: Calculate power if torque τ = 5,Nm and angular velocity ω = 10,rad/s.
  Solution: P = τω = 5 × 10 = 50,W
• Ex 2: Find angular acceleration if torque τ = 12,Nm and moment of inertia I = 4,kg·m².
  Solution: α = τ/I = 12/4 = 3,rad/s²

Practice Set

• Q1: Work done by torque τ over angle θ is:
  A) τ / θ
  B) τ θ
  C) θ / τ
  D) τ + θ
• Q2: Power in rotational motion is:
  A) τ / ω
  B) τω
  C) Iα
  D) ω / τ
• Q3: If I=2,kg·m², α=4,rad/s², find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done = τ × θ.
• Q2: B - Power = τω.
• Q3: D - τ = Iα = 2 × 4 = 8,Nm.

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Key Points

• Work done by torque τ₁ during angular displacement dθ is dW₁ = τ₁ dθ.
• For multiple torques, total work dW = (τ₁ + τ₂ + ...) dθ = τ dθ.
• Power P = dW/dt = τ dθ/dt = τω, where ω is angular velocity.
• Rate of change of kinetic energy: d/dt (Iω²/2) = Iωα.
• Relation between torque and angular acceleration: τ = Iα.

Solved Examples

• Ex 1: Calculate power if torque τ = 5,Nm and angular velocity ω = 10,rad/s.
  Solution: P = τω = 5 × 10 = 50,W
• Ex 2: Find angular acceleration if torque τ = 12,Nm and moment of inertia I = 4,kg·m².
  Solution: α = τ/I = 12/4 = 3,rad/s²

Practice Set

• Q1: Work done by torque τ over angle θ is:
  A) τ / θ
  B) τ θ
  C) θ / τ
  D) τ + θ
• Q2: Power in rotational motion is:
  A) τ / ω
  B) τω
  C) Iα
  D) ω / τ
• Q3: If I=2,kg·m², α=4,rad/s², find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done = τ × θ.
• Q2: B - Power = τω.
• Q3: D - τ = Iα = 2 × 4 = 8,Nm.

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Key Points

• Work done by torque τ₁ during angular displacement dθ is dW₁ = τ₁ dθ.
• For multiple torques, total work dW = (τ₁ + τ₂ + ...) dθ = τ dθ.
• Power P = dW/dt = τ dθ/dt = τω, where ω is angular velocity.
• Rate of change of kinetic energy: d/dt (Iω²/2) = Iωα.
• Relation between torque and angular acceleration: τ = Iα.

Solved Examples

• Ex 1: Calculate power if torque τ = 5,Nm and angular velocity ω = 10,rad/s.
  Solution: P = τω = 5 × 10 = 50,W
• Ex 2: Find angular acceleration if torque τ = 12,Nm and moment of inertia I = 4,kg·m².
  Solution: α = τ/I = 12/4 = 3,rad/s²

Practice Set

• Q1: Work done by torque τ over angle θ is:
  A) τ / θ
  B) τ θ
  C) θ / τ
  D) τ + θ
• Q2: Power in rotational motion is:
  A) τ / ω
  B) τω
  C) Iα
  D) ω / τ
• Q3: If I=2,kg·m², α=4,rad/s², find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done = τ × θ.
• Q2: B - Power = τω.
• Q3: D - τ = Iα = 2 × 4 = 8,Nm.

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Comparison of Translational and Rotational Motion

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear = F × d, rotational = τ × θ.
• Kinetic energy: linear = 1/2mv², rotational = 1/2Iω².
• Power: linear = Fv, rotational = τω.
• Momentum: linear = mv, angular = Iω.

Solved Examples

• Ex 1: Find angular momentum if I=3,kg·m² and ω=5,rad/s.
  Solution: L = Iω = 3 × 5 = 15,kg·m²/s
• Ex 2: Calculate kinetic energy for m=2,kg, v=4,m/s and I=2,kg·m², ω=3,rad/s.
  Solution: Linear KE = 1/2 × 2 × 4² = 16,J, Rotational KE = 1/2 × 2 × 3² = 9,J

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity ω is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear = F × d, rotational = τ × θ.
• Kinetic energy: linear = 1/2mv², rotational = 1/2Iω².
• Power: linear = Fv, rotational = τω.
• Momentum: linear = mv, angular = Iω.

Solved Examples

• Ex 1: Find angular momentum if I=3,kg·m² and ω=5,rad/s.
  Solution: L = Iω = 3 × 5 = 15,kg·m²/s
• Ex 2: Calculate kinetic energy for m=2,kg, v=4,m/s and I=2,kg·m², ω=3,rad/s.
  Solution: Linear KE = 1/2 × 2 × 4² = 16,J, Rotational KE = 1/2 × 2 × 3² = 9,J

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity ω is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear = F × d, rotational = τ × θ.
• Kinetic energy: linear = 1/2mv², rotational = 1/2Iω².
• Power: linear = Fv, rotational = τω.
• Momentum: linear = mv, angular = Iω.

Solved Examples

• Ex 1: Find angular momentum if I=3,kg·m² and ω=5,rad/s.
  Solution: L = Iω = 3 × 5 = 15,kg·m²/s
• Ex 2: Calculate kinetic energy for m=2,kg, v=4,m/s and I=2,kg·m², ω=3,rad/s.
  Solution: Linear KE = 1/2 × 2 × 4² = 16,J, Rotational KE = 1/2 × 2 × 3² = 9,J

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity ω is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear = F × d, rotational = τ × θ.
• Kinetic energy: linear = 1/2mv², rotational = 1/2Iω².
• Power: linear = Fv, rotational = τω.
• Momentum: linear = mv, angular = Iω.

Solved Examples

• Ex 1: Find angular momentum if I=3,kg·m² and ω=5,rad/s.
  Solution: L = Iω = 3 × 5 = 15,kg·m²/s
• Ex 2: Calculate kinetic energy for m=2,kg, v=4,m/s and I=2,kg·m², ω=3,rad/s.
  Solution: Linear KE = 1/2 × 2 × 4² = 16,J, Rotational KE = 1/2 × 2 × 3² = 9,J

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity ω is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Angular Momentum Rotation Fixed Axis

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

Definition and Formula

• Rotational analogue of linear momentum is angular momentum \( \vec{L} = \vec{r} \times \vec{p} \).
• Magnitude: \( L = rp \sin \theta = r p_⊥ = r_⊥ p \), where \( \vec{p} = m \vec{v} \).
• Time rate of change: \( \frac{d\vec{L}}{dt} = \vec{τ} \) (torque).
• For system of particles: \( \vec{L} = \sum \vec{L}_i = \sum \vec{r}_i \times \vec{p}_i \), \( \frac{d\vec{L}}{dt} = \sum \vec{τ}_i = \vec{τ}_{ext} \).
• If external torque \( \vec{τ}_{ext} = 0 \), angular momentum is conserved.

Solved Examples

• Ex 1: Particle with mass 2 kg moves with velocity 3 m/s at 60° to position vector 0.5 m. Find angular momentum.
  • \( p = mv = 2 \times 3 = 6\,\mathrm{kg\,m/s} \)
  • \( L = rp \sin \theta = 0.5 \times 6 \times \sin 60^\circ = 2.6\,\mathrm{kg\,m^2/s} \)

Practice Set

• Q1: Angular momentum \( L \) is zero when angle between \( \vec{r} \) and \( \vec{p} \) is?
  • A) 90°
  • B) 0°
  • C) 45°
  • D) 60°
• Q2: For a particle, \( \vec{L} = \vec{r} \times m\vec{v} \). If \( r=2\,m \), \( m=1\,kg \), \( v=3\,m/s \), and \( \theta=90^\circ \), find \( L \).
  • A) 6 kg·m²/s
  • B) 3 kg·m²/s
  • C) 2 kg·m²/s
  • D) 0 kg·m²/s
• Q3: If external torque is zero, angular momentum is?
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero

Answer Key

• Q1: B) 0° — Angular momentum zero if \( \vec{r} \) and \( \vec{p} \) are parallel.
• Q2: A) 6 kg·m²/s — \( L = r m v \sin 90^\circ = 2 \times 1 \times 3 = 6 \)
• Q3: C) Constant — No external torque means angular momentum conserved.

Memory Aids

• Angular momentum \( \vec{L} = \vec{r} \times \vec{p} \)
• \( \frac{d\vec{L}}{dt} = \vec{\tau} \) (torque)
• Conserved if external torque zero

Equilibrium of a Rigid Body

Translational Equilibrium

• Body at rest or moving with constant velocity in a direction.
• Vector sum of all external forces is zero: vector F = 0.
• Linear momentum constant: vector F = dvector p/dt = 0 ⇒ vector p = constant.

Rotational Equilibrium

• Body rotates with constant angular velocity.
• Net torque zero: τ_net = 0.
• Angular momentum constant: dvector L/dt = 0 ⇒ vector L = constant.

Principle of Moments

• Algebraic sum of moments about a fixed point is zero for rotational equilibrium.
• For lever AB pivoted at O: F₁ × d₁ = F₂ × d₂.
• Load × Load arm = Effort × Effort arm.
• Mechanical advantage: M.A. = Load (F₁) / Effort (F₂) = d₂ / d₁.
• Small effort lifts heavy load if M.A. > 1 ⇒ d₂ > d₁.

Center of Gravity

• Point where weight acts and total gravitational torque is zero.

Solved Examples

• Ex 1: For lever with F₁ = 10 N, d₁ = 2 m, d₂ = 5 m, find F₂ for equilibrium.
  Solution:
  Use F₁ d₁ = F₂ d₂.
  F₂ = F₁ d₁/d₂ = 10 × 2/5 = 4 text N).
• Ex 2: Calculate mechanical advantage if F₁ = 50 N, F₂ = 10 N.
  Solution:
  M.A. = F₁/F₂ = 50/10 = 5 .

Practice Set

• Q1: A body is in translational equilibrium if:
  A) vector F ≠ 0
  B) vector F = 0
  C) τ_net ≠ 0
  D) vector p changes with time
• Q2: For rotational equilibrium, net torque must be:
  A) Positive
  B) Negative
  C) Zero
  D) Constant non-zero
• Q3: If effort arm is 3 m and load arm is 1 m, mechanical advantage is:
  A) 1/3
  B) 3
  C) 2
  D) 1

Answer Key

• Q1: B - Translational equilibrium requires net force zero.
• Q2: C - Rotational equilibrium means net torque zero.
• Q3: B - M.A. = d₂ / d₁ = 3/1 = 3 .

Memory Aids

• Translational equilibrium: vector F = 0.
• Rotational equilibrium: τ_net = 0.
• Principle of moments: Load × Load arm = Effort × Effort arm.
• Mechanical advantage > 1 means easier lifting.
• Center of gravity: point of zero net torque.

Rotational Equilibrium

• Body rotates with constant angular velocity.
• Net torque zero: τ_net = 0.
• Angular momentum constant: dvector L/dt = 0 ⇒ vector L = constant.

Principle of Moments

• Algebraic sum of moments about a fixed point is zero for rotational equilibrium.
• For lever AB pivoted at O: F₁ × d₁ = F₂ × d₂.
• Load × Load arm = Effort × Effort arm.
• Mechanical advantage: M.A. = Load (F₁) / Effort (F₂) = d₂ / d₁.
• Small effort lifts heavy load if M.A. > 1 ⇒ d₂ > d₁.

Center of Gravity

• Point where weight acts and total gravitational torque is zero.

Solved Examples

• Ex 1: For lever with F₁ = 10 N, d₁ = 2 m, d₂ = 5 m, find F₂ for equilibrium.
  Solution:
  Use F₁ d₁ = F₂ d₂.
  F₂ = F₁ d₁/d₂ = 10 × 2/5 = 4 text N).
• Ex 2: Calculate mechanical advantage if F₁ = 50 N, F₂ = 10 N.
  Solution:
  M.A. = F₁/F₂ = 50/10 = 5 .

Practice Set

• Q1: A body is in translational equilibrium if:
  A) vector F ≠ 0
  B) vector F = 0
  C) τ_net ≠ 0
  D) vector p changes with time
• Q2: For rotational equilibrium, net torque must be:
  A) Positive
  B) Negative
  C) Zero
  D) Constant non-zero
• Q3: If effort arm is 3 m and load arm is 1 m, mechanical advantage is:
  A) 1/3
  B) 3
  C) 2
  D) 1

Answer Key

• Q1: B - Translational equilibrium requires net force zero.
• Q2: C - Rotational equilibrium means net torque zero.
• Q3: B - M.A. = d₂ / d₁ = 3/1 = 3 .

Memory Aids

• Translational equilibrium: vector F = 0.
• Rotational equilibrium: τ_net = 0.
• Principle of moments: Load × Load arm = Effort × Effort arm.
• Mechanical advantage > 1 means easier lifting.
• Center of gravity: point of zero net torque.

Translational Equilibrium

• Body at rest or moving with constant velocity in a direction.
• Vector sum of all external forces is zero: \( \vec{F} = 0 \).
• Linear momentum constant: \( \vec{F} = \frac{d\vec{p}}{dt} = 0 \Rightarrow \vec{p} = \text{constant} \).

Rotational Equilibrium

• Body rotates with constant angular velocity.
• Net torque zero: \( \tau_{net} = 0 \).
• Angular momentum constant: \( \frac{d\vec{L}}{dt} = 0 \Rightarrow \vec{L} = \text{constant} \).

Principle of Moments

• Algebraic sum of moments about a fixed point is zero for rotational equilibrium.
• For lever AB pivoted at O: \( F_1 \times d_1 = F_2 \times d_2 \).
• Load × Load arm = Effort × Effort arm.
• Mechanical advantage: \( M.A. = \frac{\text{Load } (F_1)}{\text{Effort } (F_2)} = \frac{d_2}{d_1} \).
• Small effort lifts heavy load if \( M.A. > 1 \Rightarrow d_2 > d_1 \).

Center of Gravity

• Point where weight acts and total gravitational torque is zero.

Solved Examples

• Ex 1: For lever with F₁ = 10 N, d₁ = 2 m, d₂ = 5 m, find F₂ for equilibrium.
  Solution:
  Use \( F_1 d_1 = F_2 d_2 \).
  \( F_2 = \frac{F_1 d_1}{d_2} = \frac{10 \times 2}{5} = 4 \text{ N} \).
• Ex 2: Calculate mechanical advantage if F₁ = 50 N, F₂ = 10 N.
  Solution:
  \( M.A. = \frac{F_1}{F_2} = \frac{50}{10} = 5 \).

Practice Set

• Q1: A body is in translational equilibrium if:
  A) \( \vec{F} \neq 0 \)
  B) \( \vec{F} = 0 \)
  C) \( \tau_{net} \neq 0 \)
  D) \( \vec{p} \) changes with time
• Q2: For rotational equilibrium, net torque must be:
  A) Positive
  B) Negative
  C) Zero
  D) Constant non-zero
• Q3: If effort arm is 3 m and load arm is 1 m, mechanical advantage is:
  A) 1/3
  B) 3
  C) 2
  D) 1

Answer Key

• Q1: B - Translational equilibrium requires net force zero.
• Q2: C - Rotational equilibrium means net torque zero.
• Q3: B - \( M.A. = \frac{d_2}{d_1} = \frac{3}{1} = 3 \).

Memory Aids

• Translational equilibrium: \( \vec{F} = 0 \).
• Rotational equilibrium: \( \tau_{net} = 0 \).
• Principle of moments: Load × Load arm = Effort × Effort arm.
• Mechanical advantage > 1 means easier lifting.
• Center of gravity: point of zero net torque.

Moment of Inertia

Definition and Concept

• Moment of inertia (I) measures rotational inertia, analogous to mass in linear motion.
• For particles with masses m_i at distances r_i from axis, I = ∑ m_i r_i².
• For continuous bodies, I = ∫ r² dm.
• Rotational kinetic energy: K.E. = 1/2 I ω², where ω is angular velocity.
• SI unit: kg·m².

Solved Examples

• Ex 1: Calculate I for two masses 2 kg and 3 kg at 0.5 m and 0.8 m from axis.
  • I = 2 × 0.5² + 3 × 0.8² = 0.5 + 1.92 = 2.42, kg·m²

Practice Set

• Q1: Moment of inertia depends on
  • A) Mass only
  • B) Distance from axis only
  • C) Mass and distance squared from axis
  • D) Velocity
• Q2: For a particle of mass 4 kg at 0.3 m from axis, I = ?
  • A) 1.2 kg·m²
  • B) 0.36 kg·m²
  • C) 4.3 kg·m²
  • D) 0.9 kg·m²
• Q3: Total I for masses 1 kg at 1 m and 2 kg at 0.5 m is
  • A) 1.5 kg·m²
  • B) 2.0 kg·m²
  • C) 2.5 kg·m²
  • D) 3.0 kg·m²

Answer Key

• Q1: C - Moment of inertia depends on mass and square of distance from axis.
• Q2: A - I = 4 × 0.3² = 4 × 0.09 = 0.36 kg·m² (corrected: option A is 1.2, so correct is B)
• Q3: B - I = 1 × 1² + 2 × 0.5² = 1 + 0.5 = 1.5 kg·m² (corrected: option A is 1.5, so correct is A)

Memory Aids

• Moment of inertia = sum of mass × (distance)².
• Rotational KE = 1/2 I ω².
• SI unit: kg·m².

Definition and Concept

• Moment of inertia (I) measures rotational inertia, analogous to mass in linear motion.
• For particles with masses m_i at distances r_i from axis, I = ∑ m_i r_i².
• For continuous bodies, I = ∫ r² dm.
• Rotational kinetic energy: K.E. = 1/2 I ω², where ω is angular velocity.
• SI unit: kg·m².

Solved Examples

• Ex 1: Calculate I for two masses 2 kg and 3 kg at 0.5 m and 0.8 m from axis.
  • I = 2 × 0.5² + 3 × 0.8² = 0.5 + 1.92 = 2.42, kg·m²

Practice Set

• Q1: Moment of inertia depends on
  • A) Mass only
  • B) Distance from axis only
  • C) Mass and distance squared from axis
  • D) Velocity
• Q2: For a particle of mass 4 kg at 0.3 m from axis, I = ?
  • A) 1.2 kg·m²
  • B) 0.36 kg·m²
  • C) 4.3 kg·m²
  • D) 0.9 kg·m²
• Q3: Total I for masses 1 kg at 1 m and 2 kg at 0.5 m is
  • A) 1.5 kg·m²
  • B) 2.0 kg·m²
  • C) 2.5 kg·m²
  • D) 3.0 kg·m²

Answer Key

• Q1: C - Moment of inertia depends on mass and square of distance from axis.
• Q2: A - I = 4 × 0.3² = 4 × 0.09 = 0.36 kg·m² (corrected: option A is 1.2, so correct is B)
• Q3: B - I = 1 × 1² + 2 × 0.5² = 1 + 0.5 = 1.5 kg·m² (corrected: option A is 1.5, so correct is A)

Memory Aids

• Moment of inertia = sum of mass × (distance)².
• Rotational KE = 1/2 I ω².
• SI unit: kg·m².

Definition and Concept

• Moment of inertia (I) measures rotational inertia, analogous to mass in linear motion.
• For particles with masses \(m_i\) at distances \(r_i\) from axis, \(I = \sum m_i r_i^2\).
• For continuous bodies, \(I = \int r^2 dm\).
• Rotational kinetic energy: \(K.E. = \frac{1}{2} I \omega^2\), where \(\omega\) is angular velocity.
• SI unit: kg·m².

Solved Examples

• Ex 1: Calculate \(I\) for two masses 2 kg and 3 kg at 0.5 m and 0.8 m from axis.
  • \(I = 2 \times 0.5^2 + 3 \times 0.8^2 = 0.5 + 1.92 = 2.42\, \text{kg·m}^2\)

Practice Set

• Q1: Moment of inertia depends on
  • A) Mass only
  • B) Distance from axis only
  • C) Mass and distance squared from axis
  • D) Velocity
• Q2: For a particle of mass 4 kg at 0.3 m from axis, \(I = ?\)
  • A) 1.2 kg·m²
  • B) 0.36 kg·m²
  • C) 4.3 kg·m²
  • D) 0.9 kg·m²
• Q3: Total \(I\) for masses 1 kg at 1 m and 2 kg at 0.5 m is
  • A) 1.5 kg·m²
  • B) 2.0 kg·m²
  • C) 2.5 kg·m²
  • D) 3.0 kg·m²

Answer Key

• Q1: C - Moment of inertia depends on mass and square of distance from axis.
• Q2: A - \(I = 4 \times 0.3^2 = 4 \times 0.09 = 0.36\) kg·m² (corrected: option A is 1.2, so correct is B)
• Q3: B - \(I = 1 \times 1^2 + 2 \times 0.5^2 = 1 + 0.5 = 1.5\) kg·m² (corrected: option A is 1.5, so correct is A)

Memory Aids

• Moment of inertia = sum of mass × (distance)².
• Rotational KE = \(\frac{1}{2} I \omega^2\).
• SI unit: kg·m².

Radius of Gyration

Definition and Formula

• Distance from axis where entire mass can be assumed to concentrate to give same I.
• Formula: k = √[(r₁² + r₂² + ... + r_n²)/n].
• Relation: I = Mk², where M is total mass.

Solved Examples

• Ex 1: For masses at distances 0.5 m, 0.8 m, 1.0 m, find k if n=3.
  • k = √[(0.5² + 0.8² + 1.0²)/3] = √[(0.25 + 0.64 + 1)/3] = √0.63 = 0.79, m

Practice Set

• Q1: Radius of gyration is
  • A) Distance where mass is zero
  • B) Distance concentrating all mass for same I
  • C) Distance from center of mass
  • D) None
• Q2: If k=0.6 m and M=5 kg, I= ?
  • A) 1.8 kg·m²
  • B) 1.2 kg·m²
  • C) 1.5 kg·m²
  • D) 2.0 kg·m²
• Q3: For n=4 distances 0.4, 0.5, 0.6, 0.7 m, k= ?
  • A) 0.55 m
  • B) 0.58 m
  • C) 0.60 m
  • D) 0.62 m

Answer Key

• Q1: B - Radius of gyration is the distance concentrating all mass for same moment of inertia.
• Q2: A - I = Mk² = 5 × 0.6² = 1.8 kg·m².
• Q3: B - k = √[(0.16 + 0.25 + 0.36 + 0.49)/4] = √0.315 = 0.56 approx, closest 0.58 m.

Memory Aids

• I = Mk² relates moment of inertia and radius of gyration.
• k is effective radius for mass distribution.

Definition and Formula

• Distance from axis where entire mass can be assumed to concentrate to give same I.
• Formula: k = √[(r₁² + r₂² + ... + r_n²)/n].
• Relation: I = Mk², where M is total mass.

Solved Examples

• Ex 1: For masses at distances 0.5 m, 0.8 m, 1.0 m, find k if n=3.
  • k = √[(0.5² + 0.8² + 1.0²)/3] = √[(0.25 + 0.64 + 1)/3] = √0.63 = 0.79, m

Practice Set

• Q1: Radius of gyration is
  • A) Distance where mass is zero
  • B) Distance concentrating all mass for same I
  • C) Distance from center of mass
  • D) None
• Q2: If k=0.6 m and M=5 kg, I= ?
  • A) 1.8 kg·m²
  • B) 1.2 kg·m²
  • C) 1.5 kg·m²
  • D) 2.0 kg·m²
• Q3: For n=4 distances 0.4, 0.5, 0.6, 0.7 m, k= ?
  • A) 0.55 m
  • B) 0.58 m
  • C) 0.60 m
  • D) 0.62 m

Answer Key

• Q1: B - Radius of gyration is the distance concentrating all mass for same moment of inertia.
• Q2: A - I = Mk² = 5 × 0.6² = 1.8 kg·m².
• Q3: B - k = √[(0.16 + 0.25 + 0.36 + 0.49)/4] = √0.315 = 0.56 approx, closest 0.58 m.

Memory Aids

• I = Mk² relates moment of inertia and radius of gyration.
• k is effective radius for mass distribution.

Definition and Formula

• Distance from axis where entire mass can be assumed to concentrate to give same I.
• Formula: k = √[(r₁² + r₂² + ... + r_n²)/n].
• Relation: I = Mk², where M is total mass.

Solved Examples

• Ex 1: For masses at distances 0.5 m, 0.8 m, 1.0 m, find k if n=3.
  • k = √[(0.5² + 0.8² + 1.0²)/3] = √[(0.25 + 0.64 + 1)/3] = √0.63 = 0.79, m

Practice Set

• Q1: Radius of gyration is
  • A) Distance where mass is zero
  • B) Distance concentrating all mass for same I
  • C) Distance from center of mass
  • D) None
• Q2: If k=0.6 m and M=5 kg, I= ?
  • A) 1.8 kg·m²
  • B) 1.2 kg·m²
  • C) 1.5 kg·m²
  • D) 2.0 kg·m²
• Q3: For n=4 distances 0.4, 0.5, 0.6, 0.7 m, k= ?
  • A) 0.55 m
  • B) 0.58 m
  • C) 0.60 m
  • D) 0.62 m

Answer Key

• Q1: B - Radius of gyration is the distance concentrating all mass for same moment of inertia.
• Q2: A - I = Mk² = 5 × 0.6² = 1.8 kg·m².
• Q3: B - k = √[(0.16 + 0.25 + 0.36 + 0.49)/4] = √0.315 = 0.56 approx, closest 0.58 m.

Memory Aids

• I = Mk² relates moment of inertia and radius of gyration.
• k is effective radius for mass distribution.

Definition and Formula

• Distance from axis where entire mass can be assumed to concentrate to give same \(I\).
• Formula: \(k = \sqrt{\frac{r_1^2 + r_2^2 + \cdots + r_n^2}{n}}\).
• Relation: \(I = M k^2\), where \(M\) is total mass.

Solved Examples

• Ex 1: For masses at distances 0.5 m, 0.8 m, 1.0 m, find \(k\) if \(n=3\).
  • \(k = \sqrt{(0.5^2 + 0.8^2 + 1.0^2)/3} = \sqrt{(0.25 + 0.64 + 1)/3} = \sqrt{0.63} = 0.79\, m\)

Practice Set

• Q1: Radius of gyration is
  • A) Distance where mass is zero
  • B) Distance concentrating all mass for same \(I\)
  • C) Distance from center of mass
  • D) None
• Q2: If \(k=0.6\) m and \(M=5\) kg, \(I= ?\)
  • A) 1.8 kg·m²
  • B) 1.2 kg·m²
  • C) 1.5 kg·m²
  • D) 2.0 kg·m²
• Q3: For \(n=4\) distances 0.4, 0.5, 0.6, 0.7 m, \(k= ?\)
  • A) 0.55 m
  • B) 0.58 m
  • C) 0.60 m
  • D) 0.62 m

Answer Key

• Q1: B - Radius of gyration is the distance concentrating all mass for same moment of inertia.
• Q2: A - \(I = M k^2 = 5 \times 0.6^2 = 1.8\) kg·m².
• Q3: B - \(k = \sqrt{(0.16 + 0.25 + 0.36 + 0.49)/4} = \sqrt{0.315} = 0.56\) approx, closest 0.58 m.

Memory Aids

• \(I = M k^2\) relates moment of inertia and radius of gyration.
• \(k\) is effective radius for mass distribution.

Theorems of Perpendicular and Parallel Axes

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, I_z = I_x + I_y, where z is perpendicular to plane, x,y in plane.
• Used to find I about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis I' = I + Md².
• I: about parallel axis through center of mass.
• M: mass, d: distance between axes.

Solved Examples

• Ex 1: Find I' if I=10 kg·m², M=5 kg, d=2 m.
  • I' = 10 + 5 × 2² = 10 + 20 = 30 kg·m²

Practice Set

• Q1: I_z = I_x + I_y applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If I=4 kg·m², M=3 kg, d=1 m, find I'
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) I about any axis and I about center of mass axis
  • B) I and mass only
  • C) I and velocity
  • D) I and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - I' = 4 + 3 × 1² = 7 kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus Md².

Memory Aids

• Perpendicular axes: I_z = I_x + I_y.
• Parallel axes: I' = I + Md².

Theorem of Perpendicular Axes

• For planar body, \(I_z = I_x + I_y\), where \(z\) is perpendicular to plane, \(x,y\) in plane.
• Used to find \(I\) about axis perpendicular to plane from known axes.

Theorem of Parallel Axes

• Moment of inertia about any axis \(I' = I + M d^2\).
• \(I\): about parallel axis through center of mass.
• \(M\): mass, \(d\): distance between axes.

Solved Examples

• Ex 1: Find \(I'\) if \(I=10\) kg·m², \(M=5\) kg, \(d=2\) m.
  • \(I' = 10 + 5 \times 2^2 = 10 + 20 = 30\) kg·m²

Practice Set

• Q1: \(I_z = I_x + I_y\) applies to
  • A) 3D bodies
  • B) Planar bodies
  • C) Spheres
  • D) Cylinders
• Q2: If \(I=4\) kg·m², \(M=3\) kg, \(d=1\) m, find \(I'\)
  • A) 7 kg·m²
  • B) 10 kg·m²
  • C) 4 kg·m²
  • D) 3 kg·m²
• Q3: Parallel axis theorem relates
  • A) \(I\) about any axis and \(I\) about center of mass axis
  • B) \(I\) and mass only
  • C) \(I\) and velocity
  • D) \(I\) and radius

Answer Key

• Q1: B - Theorem of perpendicular axes applies to planar bodies.
• Q2: A - \(I' = 4 + 3 \times 1^2 = 7\) kg·m².
• Q3: A - It relates moment of inertia about any axis to that about center of mass axis plus \(Md^2\).

Memory Aids

• Perpendicular axes: \(I_z = I_x + I_y\).
• Parallel axes: \(I' = I + Md^2\).

Moment of Inertia of Regular Bodies

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Formulas for Common Shapes

• Ring about central axis: I = MR²
• Solid disc about central axis: I = 1/2 MR²
• Solid sphere about diameter: I = 2/5 MR²
• Cylinder about central axis: I = 1/2 MR²
• Rod about center: I = 1/12 ML²
• Rectangular lamina about center: I = 1/12 M(a² + b²)

Solved Examples

• Ex 1: Calculate I of a solid disc with mass 4 kg and radius 0.5 m.
  • I = 1/2 × 4 × 0.5² = 0.5 kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: I of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - I = MR² = 3 × 0.4² = 0.48 kg·m².
• Q2: B - I = 1/12 ML² = 1/12 × 6 × 2² = 2 kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - I = 2/5 MR² = 2/5 × 5 × 0.3² = 0.18 kg·m².

Memory Aids

• Ring: I = MR².
• Disc/Cylinder: I = 1/2 MR².
• Sphere: I = 2/5 MR².
• Rod: I = 1/12 ML².

Formulas for Common Shapes

• Ring about central axis: \(I = MR^2\)
• Solid disc about central axis: \(I = \frac{1}{2} MR^2\)
• Solid sphere about diameter: \(I = \frac{2}{5} MR^2\)
• Cylinder about central axis: \(I = \frac{1}{2} MR^2\)
• Rod about center: \(I = \frac{1}{12} ML^2\)
• Rectangular lamina about center: \(I = \frac{1}{12} M(a^2 + b^2)\)

Solved Examples

• Ex 1: Calculate \(I\) of a solid disc with mass 4 kg and radius 0.5 m.
  • \(I = \frac{1}{2} \times 4 \times 0.5^2 = 0.5\) kg·m²

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 0.4 m is
  • A) 0.48 kg·m²
  • B) 0.24 kg·m²
  • C) 0.12 kg·m²
  • D) 0.36 kg·m²
• Q2: \(I\) of a rod of length 2 m and mass 6 kg about center is
  • A) 2 kg·m²
  • B) 1 kg·m²
  • C) 3 kg·m²
  • D) 4 kg·m²
• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.3 m is
  • A) 0.18 kg·m²
  • B) 0.09 kg·m²
  • C) 0.27 kg·m²
  • D) 0.36 kg·m²

Answer Key

• Q1: A - \(I = MR^2 = 3 \times 0.4^2 = 0.48\) kg·m².
• Q2: B - \(I = \frac{1}{12} ML^2 = \frac{1}{12} \times 6 \times 2^2 = 2\) kg·m² (corrected: option A is 2, so correct is A)
• Q3: A - \(I = \frac{2}{5} MR^2 = \frac{2}{5} \times 5 \times 0.3^2 = 0.18\) kg·m².

Memory Aids

• Ring: \(I = MR^2\).
• Disc/Cylinder: \(I = \frac{1}{2} MR^2\).
• Sphere: \(I = \frac{2}{5} MR^2\).
• Rod: \(I = \frac{1}{12} ML^2\).

Kinematics of Rotational Motion about a Fixed Axis

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (θ) is measured in radians (rad).
• Angular velocity (ω) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (α) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• ω = ω₀ + αt
• θ = ω₀t + 1/2 αt²
• ω² = ω₀² + 2αθ
• Where:
• ω₀ = initial angular velocity (rad/s)
• ω = final angular velocity (rad/s)
• α = constant angular acceleration (rad/s²)
• θ = angular displacement (rad)
• t = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at 2 rad/s² for 5 s. Find final angular velocity and angular displacement.
  • Given: ω₀ = 0, α = 2 rad/s², t = 5 s
  • Final angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  • Angular displacement: θ = 0 × 5 + 1/2 × 2 × 5² = 25 rad

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using ω = ω₀ + αt = 4 + 3 × 4 = 16 (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — α = (ω - ω₀)/t = (0 - 6)/3 = -2
• Q3: B) 2 rad/s² — Using ω² = ω₀² + 2αθ ⇒ 20² = 0 + 2α × 50 ⇒ α = 400/100 = 4 Correction: A) 4 rad/s²

Memory Aids

• Remember: ω = ω₀ + αt is like v = u + at in linear motion.
• Angular displacement formula θ = ω₀t + 1/2 αt² mirrors s = ut + 1/2 at².
• Use ω² = ω₀² + 2αθ for problems without time.

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (θ) is measured in radians (rad).
• Angular velocity (ω) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (α) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• ω = ω₀ + αt
• θ = ω₀t + 1/2 αt²
• ω² = ω₀² + 2αθ
• Where:
• ω₀ = initial angular velocity (rad/s)
• ω = final angular velocity (rad/s)
• α = constant angular acceleration (rad/s²)
• θ = angular displacement (rad)
• t = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at 2 rad/s² for 5 s. Find final angular velocity and angular displacement.
  • Given: ω₀ = 0, α = 2 rad/s², t = 5 s
  • Final angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  • Angular displacement: θ = 0 × 5 + 1/2 × 2 × 5² = 25 rad

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using ω = ω₀ + αt = 4 + 3 × 4 = 16 (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — α = (ω - ω₀)/t = (0 - 6)/3 = -2
• Q3: B) 2 rad/s² — Using ω² = ω₀² + 2αθ ⇒ 20² = 0 + 2α × 50 ⇒ α = 400/100 = 4 Correction: A) 4 rad/s²

Memory Aids

• Remember: ω = ω₀ + αt is like v = u + at in linear motion.
• Angular displacement formula θ = ω₀t + 1/2 αt² mirrors s = ut + 1/2 at².
• Use ω² = ω₀² + 2αθ for problems without time.

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (θ) is measured in radians (rad).
• Angular velocity (ω) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (α) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• ω = ω₀ + αt
• θ = ω₀t + 1/2 αt²
• ω² = ω₀² + 2αθ
• Where:
• ω₀ = initial angular velocity (rad/s)
• ω = final angular velocity (rad/s)
• α = constant angular acceleration (rad/s²)
• θ = angular displacement (rad)
• t = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at 2 rad/s² for 5 s. Find final angular velocity and angular displacement.
  • Given: ω₀ = 0, α = 2 rad/s², t = 5 s
  • Final angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  • Angular displacement: θ = 0 × 5 + 1/2 × 2 × 5² = 25 rad

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using ω = ω₀ + αt = 4 + 3 × 4 = 16 (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — α = (ω - ω₀)/t = (0 - 6)/3 = -2
• Q3: B) 2 rad/s² — Using ω² = ω₀² + 2αθ ⇒ 20² = 0 + 2α × 50 ⇒ α = 400/100 = 4 Correction: A) 4 rad/s²

Memory Aids

• Remember: ω = ω₀ + αt is like v = u + at in linear motion.
• Angular displacement formula θ = ω₀t + 1/2 αt² mirrors s = ut + 1/2 at².
• Use ω² = ω₀² + 2αθ for problems without time.

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.
• Angular displacement (\( \theta \)) is measured in radians (rad).
• Angular velocity (\( \omega \)) is the rate of change of angular displacement, in rad/s.
• Angular acceleration (\( \alpha \)) is the rate of change of angular velocity, in rad/s².
• These quantities have direct analogues in linear motion: displacement, velocity, and acceleration.

Rotational Kinematic Equations

• \( \omega = \omega_0 + \alpha t \)
• \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
• \( \omega^2 = \omega_0^2 + 2 \alpha \theta \)
• Where:
• \( \omega_0 \) = initial angular velocity (rad/s)
• \( \omega \) = final angular velocity (rad/s)
• \( \alpha \) = constant angular acceleration (rad/s²)
• \( \theta \) = angular displacement (rad)
• \( t \) = time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at \( 2 \; \text{rad/s}^2 \) for 5 s. Find final angular velocity and angular displacement.
  • Given: \( \omega_0 = 0, \alpha = 2 \; \text{rad/s}^2, t = 5 \; s \)
  • Final angular velocity: \( \omega = 0 + 2 \times 5 = 10 \; \text{rad/s} \)
  • Angular displacement: \( \theta = 0 \times 5 + \frac{1}{2} \times 2 \times 5^2 = 25 \; \text{rad} \)

Practice Set

• Q1: A rotating disc has an initial angular velocity of 4 rad/s and accelerates at 3 rad/s² for 4 s. What is its final angular velocity?
  • A) 10 rad/s
  • B) 16 rad/s
  • C) 12 rad/s
  • D) 7 rad/s
• Q2: A wheel rotating at 6 rad/s slows down uniformly to rest in 3 s. What is its angular acceleration?
  • A) -2 rad/s²
  • B) -3 rad/s²
  • C) -1 rad/s²
  • D) -0.5 rad/s²
• Q3: A wheel starts from rest and reaches an angular velocity of 20 rad/s after rotating through 50 rad. Find the angular acceleration.
  • A) 4 rad/s²
  • B) 2 rad/s²
  • C) 8 rad/s²
  • D) 5 rad/s²

Answer Key

• Q1: C) 12 rad/s — Using \( \omega = \omega_0 + \alpha t = 4 + 3 \times 4 = 16 \) (Check calculation: 4 + 12 = 16, so correct answer is B) Correction: B) 16 rad/s
• Q2: A) -2 rad/s² — \( \alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 6}{3} = -2 \)
• Q3: B) 2 rad/s² — Using \( \omega^2 = \omega_0^2 + 2 \alpha \theta \Rightarrow 20^2 = 0 + 2 \alpha \times 50 \Rightarrow \alpha = \frac{400}{100} = 4 \) Correction: A) 4 rad/s²

Memory Aids

• Remember: \( \omega = \omega_0 + \alpha t \) is like \( v = u + at \) in linear motion.
• Angular displacement formula \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \) mirrors \( s = ut + \frac{1}{2} at^2 \).
• Use \( \omega^2 = \omega_0^2 + 2 \alpha \theta \) for problems without time.

Dynamics of Rotational Motion about a Fixed Axis

Key Points

• Consider only forces perpendicular to the axis for torque calculation.
• Ignore forces parallel to the axis as they produce torque perpendicular to the axis.
• Use only position vector components perpendicular to the axis for torque.
• Torque τ = rF sin α, where r is perpendicular distance and α is angle between force and position vector.

Solved Examples

• Ex 1: Calculate torque if force F=10 N acts at r=0.5 m perpendicular to radius.
  Solution: τ = rF = 0.5 × 10 = 5,Nm

Practice Set

• Q1: Which force contributes to torque about a fixed axis?
  A) Parallel to axis
  B) Perpendicular to axis
  C) Along position vector
  D) Zero force
• Q2: Torque is maximum when angle between force and position vector is:
  A) 0°
  B) 45°
  C) 90°
  D) 180°
• Q3: A force of 20 N acts at 0.3 m from axis at 60°. Calculate torque.
  A) 3 Nm
  B) 6 Nm
  C) 10 Nm
  D) 12 Nm

Answer Key

• Q1: B - Only forces perpendicular to the axis produce torque.
• Q2: C - Torque τ = rF sin α is max at 90°.
• Q3: B - τ = 0.3 × 20 × sin 60° = 6,Nm.

Memory Aids

• Torque = Force × Perpendicular distance.
• Ignore forces parallel to axis for torque.
• Max torque at 90° angle.

Key Points

• Consider only forces perpendicular to the axis for torque calculation.
• Ignore forces parallel to the axis as they produce torque perpendicular to the axis.
• Use only position vector components perpendicular to the axis for torque.
• Torque τ = rF sin α, where r is perpendicular distance and α is angle between force and position vector.

Solved Examples

• Ex 1: Calculate torque if force F=10 N acts at r=0.5 m perpendicular to radius.
  Solution: τ = rF = 0.5 × 10 = 5,Nm

Practice Set

• Q1: Which force contributes to torque about a fixed axis?
  A) Parallel to axis
  B) Perpendicular to axis
  C) Along position vector
  D) Zero force
• Q2: Torque is maximum when angle between force and position vector is:
  A) 0°
  B) 45°
  C) 90°
  D) 180°
• Q3: A force of 20 N acts at 0.3 m from axis at 60°. Calculate torque.
  A) 3 Nm
  B) 6 Nm
  C) 10 Nm
  D) 12 Nm

Answer Key

• Q1: B - Only forces perpendicular to the axis produce torque.
• Q2: C - Torque τ = rF sin α is max at 90°.
• Q3: B - τ = 0.3 × 20 × sin 60° = 6,Nm.

Memory Aids

• Torque = Force × Perpendicular distance.
• Ignore forces parallel to axis for torque.
• Max torque at 90° angle.

Key Points

• Consider only forces perpendicular to the axis for torque calculation.
• Ignore forces parallel to the axis as they produce torque perpendicular to the axis.
• Use only position vector components perpendicular to the axis for torque.
• Torque τ = rF sin α, where r is perpendicular distance and α is angle between force and position vector.

Solved Examples

• Ex 1: Calculate torque if force F=10 N acts at r=0.5 m perpendicular to radius.
  Solution: τ = rF = 0.5 × 10 = 5,Nm

Practice Set

• Q1: Which force contributes to torque about a fixed axis?
  A) Parallel to axis
  B) Perpendicular to axis
  C) Along position vector
  D) Zero force
• Q2: Torque is maximum when angle between force and position vector is:
  A) 0°
  B) 45°
  C) 90°
  D) 180°
• Q3: A force of 20 N acts at 0.3 m from axis at 60°. Calculate torque.
  A) 3 Nm
  B) 6 Nm
  C) 10 Nm
  D) 12 Nm

Answer Key

• Q1: B - Only forces perpendicular to the axis produce torque.
• Q2: C - Torque τ = rF sin α is max at 90°.
• Q3: B - τ = 0.3 × 20 × sin 60° = 6,Nm.

Memory Aids

• Torque = Force × Perpendicular distance.
• Ignore forces parallel to axis for torque.
• Max torque at 90° angle.

Key Points

• Consider only forces perpendicular to the axis for torque calculation.
• Ignore forces parallel to the axis as they produce torque perpendicular to the axis.
• Use only position vector components perpendicular to the axis for torque.
• Torque \( \tau = rF \sin \alpha \), where \(r\) is perpendicular distance and \(\alpha\) is angle between force and position vector.

Solved Examples

• Ex 1: Calculate torque if force \(F=10\,N\) acts at \(r=0.5\,m\) perpendicular to radius.
  Solution: \(\tau = rF = 0.5 \times 10 = 5\,Nm\)

Practice Set

• Q1: Which force contributes to torque about a fixed axis?
  A) Parallel to axis
  B) Perpendicular to axis
  C) Along position vector
  D) Zero force
• Q2: Torque is maximum when angle between force and position vector is:
  A) 0°
  B) 45°
  C) 90°
  D) 180°
• Q3: A force of 20 N acts at 0.3 m from axis at 60°. Calculate torque.
  A) 3 Nm
  B) 6 Nm
  C) 10 Nm
  D) 12 Nm

Answer Key

• Q1: B - Only forces perpendicular to the axis produce torque.
• Q2: C - Torque \(\tau = rF \sin \alpha\) is max at 90°.
• Q3: B - \(\tau = 0.3 \times 20 \times \sin 60^\circ = 6\,Nm\).

Memory Aids

• Torque = Force × Perpendicular distance.
• Ignore forces parallel to axis for torque.
• Max torque at 90° angle.

Work Done by a Torque

Key Points

• Work done by torque τ₁ during angular displacement dθ is dW₁ = τ₁ dθ.
• For multiple torques, total work dW = (τ₁ + τ₂ + ...) dθ = τ dθ.
• Power P = dW/dt = τ dθ/dt = τω, where ω is angular velocity.
• Rate of change of kinetic energy: d/dt (Iω²/2) = Iωα.
• Relation between torque and angular acceleration: τ = Iα.

Solved Examples

• Ex 1: Calculate power if torque τ = 5,Nm and angular velocity ω = 10,rad/s.
  Solution: P = τω = 5 × 10 = 50,W
• Ex 2: Find angular acceleration if torque τ = 12,Nm and moment of inertia I = 4,kg·m².
  Solution: α = τ/I = 12/4 = 3,rad/s²

Practice Set

• Q1: Work done by torque τ over angle θ is:
  A) τ / θ
  B) τ θ
  C) θ / τ
  D) τ + θ
• Q2: Power in rotational motion is:
  A) τ / ω
  B) τω
  C) Iα
  D) ω / τ
• Q3: If I=2,kg·m², α=4,rad/s², find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done = τ × θ.
• Q2: B - Power = τω.
• Q3: D - τ = Iα = 2 × 4 = 8,Nm.

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Key Points

• Work done by torque τ₁ during angular displacement dθ is dW₁ = τ₁ dθ.
• For multiple torques, total work dW = (τ₁ + τ₂ + ...) dθ = τ dθ.
• Power P = dW/dt = τ dθ/dt = τω, where ω is angular velocity.
• Rate of change of kinetic energy: d/dt (Iω²/2) = Iωα.
• Relation between torque and angular acceleration: τ = Iα.

Solved Examples

• Ex 1: Calculate power if torque τ = 5,Nm and angular velocity ω = 10,rad/s.
  Solution: P = τω = 5 × 10 = 50,W
• Ex 2: Find angular acceleration if torque τ = 12,Nm and moment of inertia I = 4,kg·m².
  Solution: α = τ/I = 12/4 = 3,rad/s²

Practice Set

• Q1: Work done by torque τ over angle θ is:
  A) τ / θ
  B) τ θ
  C) θ / τ
  D) τ + θ
• Q2: Power in rotational motion is:
  A) τ / ω
  B) τω
  C) Iα
  D) ω / τ
• Q3: If I=2,kg·m², α=4,rad/s², find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done = τ × θ.
• Q2: B - Power = τω.
• Q3: D - τ = Iα = 2 × 4 = 8,Nm.

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Key Points

• Work done by torque τ₁ during angular displacement dθ is dW₁ = τ₁ dθ.
• For multiple torques, total work dW = (τ₁ + τ₂ + ...) dθ = τ dθ.
• Power P = dW/dt = τ dθ/dt = τω, where ω is angular velocity.
• Rate of change of kinetic energy: d/dt (Iω²/2) = Iωα.
• Relation between torque and angular acceleration: τ = Iα.

Solved Examples

• Ex 1: Calculate power if torque τ = 5,Nm and angular velocity ω = 10,rad/s.
  Solution: P = τω = 5 × 10 = 50,W
• Ex 2: Find angular acceleration if torque τ = 12,Nm and moment of inertia I = 4,kg·m².
  Solution: α = τ/I = 12/4 = 3,rad/s²

Practice Set

• Q1: Work done by torque τ over angle θ is:
  A) τ / θ
  B) τ θ
  C) θ / τ
  D) τ + θ
• Q2: Power in rotational motion is:
  A) τ / ω
  B) τω
  C) Iα
  D) ω / τ
• Q3: If I=2,kg·m², α=4,rad/s², find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done = τ × θ.
• Q2: B - Power = τω.
• Q3: D - τ = Iα = 2 × 4 = 8,Nm.

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Key Points

• Work done by torque \(\tau_1\) during angular displacement \(d\theta\) is \(dW_1 = \tau_1 d\theta\).
• For multiple torques, total work \(dW = (\tau_1 + \tau_2 + ...) d\theta = \tau d\theta\).
• Power \(P = \frac{dW}{dt} = \tau \frac{d\theta}{dt} = \tau \omega\), where \(\omega\) is angular velocity.
• Rate of change of kinetic energy: \(\frac{d}{dt} \left( \frac{I \omega^2}{2} \right) = I \omega \alpha\).
• Relation between torque and angular acceleration: \(\tau = I \alpha\).

Solved Examples

• Ex 1: Calculate power if torque \(\tau = 5\,Nm\) and angular velocity \(\omega = 10\,rad/s\).
  Solution: \(P = \tau \omega = 5 \times 10 = 50\,W\)
• Ex 2: Find angular acceleration if torque \(\tau = 12\,Nm\) and moment of inertia \(I = 4\,kg\cdot m^2\).
  Solution: \(\alpha = \frac{\tau}{I} = \frac{12}{4} = 3\,rad/s^2\)

Practice Set

• Q1: Work done by torque \(\tau\) over angle \(\theta\) is:
  A) \(\tau / \theta\)
  B) \(\tau \theta\)
  C) \(\theta / \tau\)
  D) \(\tau + \theta\)
• Q2: Power in rotational motion is:
  A) \(\tau / \omega\)
  B) \(\tau \omega\)
  C) \(I \alpha\)
  D) \(\omega / \tau\)
• Q3: If \(I=2\,kg\cdot m^2\), \(\alpha=4\,rad/s^2\), find torque.
  A) 2 Nm
  B) 4 Nm
  C) 6 Nm
  D) 8 Nm

Answer Key

• Q1: B - Work done \(= \tau \times \theta\).
• Q2: B - Power \(= \tau \omega\).
• Q3: D - \(\tau = I \alpha = 2 \times 4 = 8\,Nm\).

Memory Aids

• Power = Torque × Angular velocity.
• Torque = Moment of inertia × Angular acceleration.
• Work done = Torque × Angular displacement.

Comparison of Translational and Rotational Motion

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear = F × d, rotational = τ × θ.
• Kinetic energy: linear = 1/2mv², rotational = 1/2Iω².
• Power: linear = Fv, rotational = τω.
• Momentum: linear = mv, angular = Iω.

Solved Examples

• Ex 1: Find angular momentum if I=3,kg·m² and ω=5,rad/s.
  Solution: L = Iω = 3 × 5 = 15,kg·m²/s
• Ex 2: Calculate kinetic energy for m=2,kg, v=4,m/s and I=2,kg·m², ω=3,rad/s.
  Solution: Linear KE = 1/2 × 2 × 4² = 16,J, Rotational KE = 1/2 × 2 × 3² = 9,J

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity ω is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear (= F × d), rotational (= τ × θ).
• Kinetic energy: linear (= 1/2mv²), rotational (= 1/2Iω²).
• Power: linear (= Fv), rotational (= τω).
• Momentum: linear (= mv), angular (= Iω).

Solved Examples

• Ex 1: Find angular momentum if I=3,kg·m² and ω=5,rad/s.
  Solution: L = Iω = 3 × 5 = 15,kg·m²/s
• Ex 2: Calculate kinetic energy for m=2,kg, v=4,m/s and I=2,kg·m², ω=3,rad/s.
  Solution: Linear KE = 1/2 × 2 × 4² = 16,J, Rotational KE = 1/2 × 2 × 3² = 9,J

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity ω is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear = F × d, rotational = τ × θ.
• Kinetic energy: linear = 1/2mv², rotational = 1/2Iω².
• Power: linear = Fv, rotational = τω.
• Momentum: linear = mv, angular = Iω.

Solved Examples

• Ex 1: Find angular momentum if I=3,kg·m² and ω=5,rad/s.
  Solution: L = Iω = 3 × 5 = 15,kg·m²/s
• Ex 2: Calculate kinetic energy for m=2,kg, v=4,m/s and I=2,kg·m², ω=3,rad/s.
  Solution: Linear KE = 1/2 × 2 × 4² = 16,J, Rotational KE = 1/2 × 2 × 3² = 9,J

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity ω is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Key Points

• Linear motion: displacement, velocity, acceleration; Rotational motion: angular displacement, angular velocity, angular acceleration.
• Mass in linear motion corresponds to moment of inertia in rotation.
• Force causes linear acceleration; torque causes angular acceleration.
• Work done: linear \(= F \times d\), rotational \(= \tau \times \theta\).
• Kinetic energy: linear \(= \frac{1}{2}mv^2\), rotational \(= \frac{1}{2}I\omega^2\).
• Power: linear \(= Fv\), rotational \(= \tau \omega\).
• Momentum: linear \(= mv\), angular \(= I\omega\).

Solved Examples

• Ex 1: Find angular momentum if \(I=3\,kg\cdot m^2\) and \(\omega=5\,rad/s\).
  Solution: \(L = I \omega = 3 \times 5 = 15\,kg\cdot m^2/s\)
• Ex 2: Calculate kinetic energy for \(m=2\,kg\), \(v=4\,m/s\) and \(I=2\,kg\cdot m^2\), \(\omega=3\,rad/s\).
  Solution: Linear KE = \(\frac{1}{2} \times 2 \times 4^2 = 16\,J\), Rotational KE = \(\frac{1}{2} \times 2 \times 3^2 = 9\,J\)

Practice Set

• Q1: Moment of inertia corresponds to:
  A) Mass
  B) Force
  C) Torque
  D) Acceleration
• Q2: Work done in rotational motion is:
  A) Force × distance
  B) Torque × angle
  C) Mass × velocity
  D) Force × velocity
• Q3: Angular velocity \(\omega\) is analogous to:
  A) Linear displacement
  B) Linear velocity
  C) Linear acceleration
  D) Force

Answer Key

• Q1: A - Moment of inertia is rotational equivalent of mass.
• Q2: B - Work done rotationally is torque times angle.
• Q3: B - Angular velocity corresponds to linear velocity.

Memory Aids

• Mass ↔ Moment of inertia
• Force ↔ Torque
• Velocity ↔ Angular velocity
• Work done: Force × distance ↔ Torque × angle

Angular Momentum Rotation Fixed Axis

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum ⃗L = ∑_i=1^N ⃗r_i × ⃗p_i
• ⃗r_i: position vector of particle i
• ⃗p_i = m_i ⃗v_i: linear momentum of particle i
• Cross product gives vector perpendicular to ⃗r_i and ⃗p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: ⃗r = ⃗OP = ⃗OC + ⃗CP
• Momentum: ⃗p = m ⃗v
• Angular momentum: ⃗l = (⃗OC × m ⃗v) + (⃗CP × m ⃗v)
• ⃗CP × m ⃗v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• ⃗L = ∑ ⃗l_i = ∑ l_z + ∑ (⃗OC_i × m_i ⃗v_i) = ⃗L_z + ⃗L_⊥
• ⃗L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: d⃗L/dt = d⃗L_z/dt + d⃗L_⊥/dt
• Since d⃗L_⊥/dt = 0, d⃗L/dt = d⃗L_z/dt = d/dt (Iω k̂) = ⃗τ = I ⃗α

Solved Examples

• Ex 1: Calculate total angular momentum ⃗L for a particle with mass m, velocity ⃗v, and position vector ⃗r.
  • Use ⃗L = ⃗r × m ⃗v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum ⃗L is given by:
  • A) ∑ m_i ⃗v_i
  • B) ∑ ⃗r_i × m_i ⃗v_i
  • C) ∑ ⃗r_i + m_i ⃗v_i
  • D) ∑ m_i ⃗r_i
• Q2: For a particle at point P, angular momentum ⃗l is:
  • A) ⃗OC × m ⃗v
  • B) ⃗CP × m ⃗v
  • C) Sum of ⃗OC × m ⃗v and ⃗CP × m ⃗v
  • D) m ⃗v
• Q3: The torque ⃗τ is related to angular acceleration ⃗α by:
  • A) ⃗τ = I ⃗α
  • B) ⃗τ = m ⃗a
  • C) ⃗τ = ⃗L
  • D) ⃗τ = ⃗r × ⃗p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum ⃗L = ∑ ⃗r × m ⃗v
• Torque ⃗τ = I ⃗α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum ⃗L = ∑_i=1^N ⃗r_i × ⃗p_i
• ⃗r_i: position vector of particle i
• ⃗p_i = m_i ⃗v_i: linear momentum of particle i
• Cross product gives vector perpendicular to ⃗r_i and ⃗p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: ⃗r = ⃗OP = ⃗OC + ⃗CP
• Momentum: ⃗p = m ⃗v
• Angular momentum: ⃗l = (⃗OC × m ⃗v) + (⃗CP × m ⃗v)
• ⃗CP × m ⃗v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• ⃗L = ∑ ⃗l_i = ∑ l_z + ∑ (⃗OC_i × m_i ⃗v_i) = ⃗L_z + ⃗L_⊥
• ⃗L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: d⃗L/dt = d⃗L_z/dt + d⃗L_⊥/dt
• Since d⃗L_⊥/dt = 0, d⃗L/dt = d⃗L_z/dt = d/dt (Iω k̂) = ⃗τ = I ⃗α

Solved Examples

• Ex 1: Calculate total angular momentum ⃗L for a particle with mass m, velocity ⃗v, and position vector ⃗r.
  • Use ⃗L = ⃗r × m ⃗v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum ⃗L is given by:
  • A) ∑ m_i ⃗v_i
  • B) ∑ ⃗r_i × m_i ⃗v_i
  • C) ∑ ⃗r_i + m_i ⃗v_i
  • D) ∑ m_i ⃗r_i
• Q2: For a particle at point P, angular momentum ⃗l is:
  • A) ⃗OC × m ⃗v
  • B) ⃗CP × m ⃗v
  • C) Sum of ⃗OC × m ⃗v and ⃗CP × m ⃗v
  • D) m ⃗v
• Q3: The torque ⃗τ is related to angular acceleration ⃗α by:
  • A) ⃗τ = I ⃗α
  • B) ⃗τ = m ⃗a
  • C) ⃗τ = ⃗L
  • D) ⃗τ = ⃗r × ⃗p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum ⃗L = ∑ ⃗r × m ⃗v
• Torque ⃗τ = I ⃗α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum ⃗L = ∑_i=1^N ⃗r_i × ⃗p_i
• ⃗r_i: position vector of particle i
• ⃗p_i = m_i ⃗v_i: linear momentum of particle i
• Cross product gives vector perpendicular to ⃗r_i and ⃗p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: ⃗r = ⃗OP = ⃗OC + ⃗CP
• Momentum: ⃗p = m ⃗v
• Angular momentum: ⃗l = (⃗OC × m ⃗v) + (⃗CP × m ⃗v)
• ⃗CP × m ⃗v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• ⃗L = ∑ ⃗l_i = ∑ l_z + ∑ (⃗OC_i × m_i ⃗v_i) = ⃗L_z + ⃗L_⊥
• ⃗L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: d⃗L/dt = d⃗L_z/dt + d⃗L_⊥/dt
• Since d⃗L_⊥/dt = 0, d⃗L/dt = d⃗L_z/dt = d/dt (Iω k̂) = ⃗τ = I ⃗α

Solved Examples

• Ex 1: Calculate total angular momentum ⃗L for a particle with mass m, velocity ⃗v, and position vector ⃗r.
  • Use ⃗L = ⃗r × m ⃗v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum ⃗L is given by:
  • A) ∑ m_i ⃗v_i
  • B) ∑ ⃗r_i × m_i ⃗v_i
  • C) ∑ ⃗r_i + m_i ⃗v_i
  • D) ∑ m_i ⃗r_i
• Q2: For a particle at point P, angular momentum ⃗l is:
  • A) ⃗OC × m ⃗v
  • B) ⃗CP × m ⃗v
  • C) Sum of ⃗OC × m ⃗v and ⃗CP × m ⃗v
  • D) m ⃗v
• Q3: The torque ⃗τ is related to angular acceleration ⃗α by:
  • A) ⃗τ = I ⃗α
  • B) ⃗τ = m ⃗a
  • C) ⃗τ = ⃗L
  • D) ⃗τ = ⃗r × ⃗p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum ⃗L = ∑ ⃗r × m ⃗v
• Torque ⃗τ = I ⃗α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum vector L = ∑_i=1^N vector r_i × vector p_i
• vector r_i: position vector of particle i
• vector p_i = m_i vector v_i: linear momentum of particle i
• Cross product gives vector perpendicular to vector r_i and vector p_i
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: vector r = vector OP = vector OC + vector CP
• Momentum: vector p = m vector v
• Angular momentum: vector l = (vector OC × m vector v) + (vector CP × m vector v)
• vector CP × m vector v = r_⊥ (mv) k̂ = I_z

Total Angular Momentum of Rigid Body

• vector L = ∑ vector l_i = ∑ l_z + ∑ (vector OC_i × m_i vector v_i) = vector L_z + vector L_⊥
• vector L_z = ∑ r_⊥ (mv) k̂ = ∑ (mr²)ω k̂ = Iω k̂
• Time derivative: dvector L/dt = dvector L_z/dt + dvector L_⊥/dt
• Since dvector L_⊥/dt = 0, dvector L/dt = dvector L_z/dt = d/dt (Iω k̂) = vector τ = I vector α

Solved Examples

• Ex 1: Calculate total angular momentum vector L for a particle with mass m, velocity vector v, and position vector vector r.
  • Use vector L = vector r × m vector v
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum vector L is given by:
  • A) ∑ m_i vector v_i
  • B) ∑ vector r_i × m_i vector v_i
  • C) ∑ vector r_i + m_i vector v_i
  • D) ∑ m_i vector r_i
• Q2: For a particle at point P, angular momentum vector l is:
  • A) vector OC × m vector v
  • B) vector CP × m vector v
  • C) Sum of vector OC × m vector v and vector CP × m vector v
  • D) m vector v
• Q3: The torque vector τ is related to angular acceleration vector α by:
  • A) vector τ = I vector α
  • B) vector τ = m vector a
  • C) vector τ = vector L
  • D) vector τ = vector r × vector p

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum vector L = ∑ vector r × m vector v
• Torque vector τ = I vector α
• Cross product gives perpendicular vector

General Expression

• Total angular momentum \( \vec{L} = \sum_{i=1}^N \vec{r_i} \times \vec{p_i} \)
• \( \vec{r_i} \): position vector of particle \(i\)
• \( \vec{p_i} = m_i \vec{v_i} \): linear momentum of particle \(i\)
• Cross product gives vector perpendicular to \( \vec{r_i} \) and \( \vec{p_i} \)
• Sum over all particles gives total angular momentum

Angular Momentum for Particle at P

• Position vector: \( \vec{r} = \vec{OP} = \vec{OC} + \vec{CP} \)
• Momentum: \( \vec{p} = m \vec{v} \)
• Angular momentum: \( \vec{l} = (\vec{OC} \times m \vec{v}) + (\vec{CP} \times m \vec{v}) \)
• \( \vec{CP} \times m \vec{v} = r_\perp (mv) \hat{k} = I_z \)

Total Angular Momentum of Rigid Body

• \( \vec{L} = \sum \vec{l_i} = \sum l_z + \sum (\vec{OC_i} \times m_i \vec{v_i}) = \vec{L_z} + \vec{L_\perp} \)
• \( \vec{L_z} = \sum r_\perp (mv) \hat{k} = \sum (mr^2) \omega \hat{k} = I \omega \hat{k} \)
• Time derivative: \( \frac{d\vec{L}}{dt} = \frac{d\vec{L_z}}{dt} + \frac{d\vec{L_\perp}}{dt} \)
• Since \( \frac{d\vec{L_\perp}}{dt} = 0 \), \( \frac{d\vec{L}}{dt} = \frac{d\vec{L_z}}{dt} = \frac{d}{dt} (I \omega \hat{k}) = \vec{\tau} = I \vec{\alpha} \)

Solved Examples

• Ex 1: Calculate total angular momentum \( \vec{L} \) for a particle with mass \(m\), velocity \(\vec{v}\), and position vector \(\vec{r}\).
  • Use \( \vec{L} = \vec{r} \times m \vec{v} \)
  • Calculate cross product and magnitude

Practice Set

• Q1: Total angular momentum \( \vec{L} \) is given by:
  • A) \( \sum m_i \vec{v_i} \)
  • B) \( \sum \vec{r_i} \times m_i \vec{v_i} \)
  • C) \( \sum \vec{r_i} + m_i \vec{v_i} \)
  • D) \( \sum m_i \vec{r_i} \)
• Q2: For a particle at point P, angular momentum \( \vec{l} \) is:
  • A) \( \vec{OC} \times m \vec{v} \)
  • B) \( \vec{CP} \times m \vec{v} \)
  • C) Sum of \( \vec{OC} \times m \vec{v} \) and \( \vec{CP} \times m \vec{v} \)
  • D) \( m \vec{v} \)
• Q3: The torque \( \vec{\tau} \) is related to angular acceleration \( \vec{\alpha} \) by:
  • A) \( \vec{\tau} = I \vec{\alpha} \)
  • B) \( \vec{\tau} = m \vec{a} \)
  • C) \( \vec{\tau} = \vec{L} \)
  • D) \( \vec{\tau} = \vec{r} \times \vec{p} \)

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.
• Q2: C - Angular momentum at P is sum of two cross products as per formula.
• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum \( \vec{L} = \sum \vec{r} \times m \vec{v} \)
• Torque \( \vec{\tau} = I \vec{\alpha} \)
• Cross product gives perpendicular vector

Rolling Motion

Definition and Conditions

• Rolling = rotation + translation
• Pure rolling: velocity of contact point with surface = 0
• Velocity relation: v_CM = Rω_CM

Velocity of Points on Rolling Body

• Velocity of any point: vector v = vector v_CM + vector ω × vector R
• At point A: vector v_A = 2 vector v_CM
• At point B: vector v_B = 3/2 vector v_CM
• At point C: vector v_C = vector v_CM
• At point D: vector v_D = 3/4 vector v_CM
• At point E (contact point): vector v_E = 0

Kinetic Energy in Rolling

• K = 1/2 Iω² + 1/2 mv_CM² = 1/2 mv_CM² (1 + k²/R²)
• Energy ratio: K_translation : K_rotation : K_rolling = 1 : k²/R² : (1 + k²/R²)

Solved Examples

• Ex 1: Calculate velocity at point A on rolling wheel with v_CM = 5 m/s.
  • v_A = 2 v_CM = 10 m/s

Practice Set

• Q1: Pure rolling means:
  • A) Contact point velocity = v_CM
  • B) Contact point velocity = 0
  • C) Wheel does not rotate
  • D) Wheel slips on surface
• Q2: Velocity at top point A of rolling wheel is:
  • A) v_CM
  • B) 2 v_CM
  • C) 1/2 v_CM
  • D) 0
• Q3: Kinetic energy of rolling is:
  • A) 1/2 mv_CM²
  • B) 1/2 Iω²
  • C) 1/2 mv_CM² (1 + k²/R²)
  • D) mv_CM²

Answer Key

• Q1: B - Pure rolling means zero velocity at contact point.
• Q2: B - Velocity at top is twice the center velocity.
• Q3: C - Rolling kinetic energy includes translation and rotation parts.

Memory Aids

• Pure rolling: v_contact = 0
• Velocity top point = 2 v_CM
• Energy: K = 1/2 mv_CM² (1 + k²/R²)

Definition and Conditions

• Rolling = rotation + translation
• Pure rolling: velocity of contact point with surface = 0
• Velocity relation: v_CM = Rω_CM

Velocity of Points on Rolling Body

• Velocity of any point: vector v = vector v_CM + vector ω × vector R
• At point A: vector v_A = 2 vector v_CM
• At point B: vector v_B = 3/2 vector v_CM
• At point C: vector v_C = vector v_CM
• At point D: vector v_D = 3/4 vector v_CM
• At point E (contact point): vector v_E = 0

Kinetic Energy in Rolling

• K = 1/2 Iω² + 1/2 mv_CM² = 1/2 mv_CM² (1 + k²/R²)
• Energy ratio: K_translation : K_rotation : K_rolling = 1 : k²/R² : (1 + k²/R²)

Solved Examples

• Ex 1: Calculate velocity at point A on rolling wheel with v_CM = 5 m/s.
  • v_A = 2 v_CM = 10 m/s

Practice Set

• Q1: Pure rolling means:
  • A) Contact point velocity = v_CM
  • B) Contact point velocity = 0
  • C) Wheel does not rotate
  • D) Wheel slips on surface
• Q2: Velocity at top point A of rolling wheel is:
  • A) v_CM
  • B) 2 v_CM
  • C) 1/2 v_CM
  • D) 0
• Q3: Kinetic energy of rolling is:
  • A) 1/2 mv_CM²
  • B) 1/2 Iω²
  • C) 1/2 mv_CM² (1 + k²/R²)
  • D) mv_CM²

Answer Key

• Q1: B - Pure rolling means zero velocity at contact point.
• Q2: B - Velocity at top is twice the center velocity.
• Q3: C - Rolling kinetic energy includes translation and rotation parts.

Memory Aids

• Pure rolling: v_contact = 0
• Velocity top point = 2 v_CM
• Energy: K = 1/2 mv_CM² (1 + k²/R²)

Definition and Conditions

• Rolling = rotation + translation
• Pure rolling: velocity of contact point with surface = 0
• Velocity relation: v_CM = Rω_CM

Velocity of Points on Rolling Body

• Velocity of any point: vector v = vector v_CM + vector ω × vector R
• At point A: vector v_A = 2 vector v_CM
• At point B: vector v_B = 3/2 vector v_CM
• At point C: vector v_C = vector v_CM
• At point D: vector v_D = 3/4 vector v_CM
• At point E (contact point): vector v_E = 0

Kinetic Energy in Rolling

• K = 1/2 Iω² + 1/2 mv_CM² = 1/2 mv_CM² (1 + k²/R²)
• Energy ratio: K_translation : K_rotation : K_rolling = 1 : k²/R² : (1 + k²/R²)

Solved Examples

• Ex 1: Calculate velocity at point A on rolling wheel with v_CM = 5 m/s.
  • v_A = 2 v_CM = 10 m/s

Practice Set

• Q1: Pure rolling means:
  • A) Contact point velocity = v_CM
  • B) Contact point velocity = 0
  • C) Wheel does not rotate
  • D) Wheel slips on surface
• Q2: Velocity at top point A of rolling wheel is:
  • A) v_CM
  • B) 2 v_CM
  • C) 1/2 v_CM
  • D) 0
• Q3: Kinetic energy of rolling is:
  • A) 1/2 mv_CM²
  • B) 1/2 Iω²
  • C) 1/2 mv_CM² (1 + k²/R²)
  • D) mv_CM²

Answer Key

• Q1: B - Pure rolling means zero velocity at contact point.
• Q2: B - Velocity at top is twice the center velocity.
• Q3: C - Rolling kinetic energy includes translation and rotation parts.

Memory Aids

• Pure rolling: v_contact = 0
• Velocity top point = 2 v_CM
• Energy: K = 1/2 mv_CM² (1 + k²/R²)

Definition and Conditions

• Rolling = rotation + translation
• Pure rolling: velocity of contact point with surface = 0
• Velocity relation: v_CM = Rω_CM

Velocity of Points on Rolling Body

• Velocity of any point: vector v = vector v_CM + vector ω × vector R
• At point A: vector v_A = 2 vector v_CM
• At point B: vector v_B = 3/2 vector v_CM
• At point C: vector v_C = vector v_CM
• At point D: vector v_D = 3/4 vector v_CM
• At point E (contact point): vector v_E = 0

Kinetic Energy in Rolling

• K = 1/2 Iω² + 1/2 mv_CM² = 1/2 mv_CM² (1 + k²/R²)
• Energy ratio: K_translation : K_rotation : K_rolling = 1 : k²/R² : (1 + k²/R²)

Solved Examples

• Ex 1: Calculate velocity at point A on rolling wheel with v_CM = 5 m/s.
  • v_A = 2 v_CM = 10 m/s

Practice Set

• Q1: Pure rolling means:
  • A) Contact point velocity = v_CM
  • B) Contact point velocity = 0
  • C) Wheel does not rotate
  • D) Wheel slips on surface
• Q2: Velocity at top point A of rolling wheel is:
  • A) v_CM
  • B) 2 v_CM
  • C) 1/2 v_CM
  • D) 0
• Q3: Kinetic energy of rolling is:
  • A) 1/2 mv_CM²
  • B) 1/2 Iω²
  • C) 1/2 mv_CM² (1 + k²/R²)
  • D) mv_CM²

Answer Key

• Q1: B - Pure rolling means zero velocity at contact point.
• Q2: B - Velocity at top is twice the center velocity.
• Q3: C - Rolling kinetic energy includes translation and rotation parts.

Memory Aids

• Pure rolling: v_contact = 0
• Velocity top point = 2 v_CM
• Energy: K = 1/2 mv_CM² (1 + k²/R²)

Definition and Conditions

• Rolling = rotation + translation
• Pure rolling: velocity of contact point with surface = 0
• Velocity relation: v_CM = Rω_CM

Velocity of Points on Rolling Body

• Velocity of any point: vector v = vector v_CM + vector ω × vector R
• At point A: vector v_A = 2 vector v_CM
• At point B: vector v_B = 3/2 vector v_CM
• At point C: vector v_C = vector v_CM
• At point D: vector v_D = 3/4 vector v_CM
• At point E (contact point): vector v_E = 0

Kinetic Energy in Rolling

• K = 1/2 Iω² + 1/2 mv_CM² = 1/2 mv_CM² (1 + k²/R²)
• Energy ratio: K_translation : K_rotation : K_rolling = 1 : k²/R² : (1 + k²/R²)

Solved Examples

• Ex 1: Calculate velocity at point A on rolling wheel with v_CM = 5 m/s.
  • v_A = 2 v_CM = 10 m/s

Practice Set

• Q1: Pure rolling means:
  • A) Contact point velocity = v_CM
  • B) Contact point velocity = 0
  • C) Wheel does not rotate
  • D) Wheel slips on surface
• Q2: Velocity at top point A of rolling wheel is:
  • A) v_CM
  • B) 2 v_CM
  • C) 1/2 v_CM
  • D) 0
• Q3: Kinetic energy of rolling is:
  • A) 1/2 mv_CM²
  • B) 1/2 Iω²
  • C) 1/2 mv_CM² (1 + k²/R²)
  • D) mv_CM²

Answer Key

• Q1: B - Pure rolling means zero velocity at contact point.
• Q2: B - Velocity at top is twice the center velocity.
• Q3: C - Rolling kinetic energy includes translation and rotation parts.

Memory Aids

• Pure rolling: v_contact = 0
• Velocity top point = 2 v_CM
• Energy: K = 1/2 mv_CM² (1 + k²/R²)

Definition and Conditions

• Rolling = rotation + translation
• Pure rolling: velocity of contact point with surface = 0
• Velocity relation: \( v_{CM} = R \omega_{CM} \)

Velocity of Points on Rolling Body

• Velocity of any point: \( \vec{v} = \vec{v}_{CM} + \vec{\omega} \times \vec{R} \)
• At point A: \( \vec{v}_A = 2 \vec{v}_{CM} \)
• At point B: \( \vec{v}_B = \frac{3}{2} \vec{v}_{CM} \)
• At point C: \( \vec{v}_C = \vec{v}_{CM} \)
• At point D: \( \vec{v}_D = \frac{3}{4} \vec{v}_{CM} \)
• At point E (contact point): \( \vec{v}_E = 0 \)

Kinetic Energy in Rolling

• \( K = \frac{1}{2} I \omega^2 + \frac{1}{2} m v_{CM}^2 = \frac{1}{2} m v_{CM}^2 \left(1 + \frac{k^2}{R^2}\right) \)
• Energy ratio: \( K_{translation} : K_{rotation} : K_{rolling} = 1 : \frac{k^2}{R^2} : \left(1 + \frac{k^2}{R^2}\right) \)

Solved Examples

• Ex 1: Calculate velocity at point A on rolling wheel with \( v_{CM} = 5 \mathrm{m/s} \).
  • \( v_A = 2 v_{CM} = 10 \mathrm{m/s} \)

Practice Set

• Q1: Pure rolling means:
  • A) Contact point velocity = \( v_{CM} \)
  • B) Contact point velocity = 0
  • C) Wheel does not rotate
  • D) Wheel slips on surface
• Q2: Velocity at top point A of rolling wheel is:
  • A) \( v_{CM} \)
  • B) \( 2 v_{CM} \)
  • C) \( \frac{1}{2} v_{CM} \)
  • D) 0
• Q3: Kinetic energy of rolling is:
  • A) \( \frac{1}{2} m v_{CM}^2 \)
  • B) \( \frac{1}{2} I \omega^2 \)
  • C) \( \frac{1}{2} m v_{CM}^2 \left(1 + \frac{k^2}{R^2}\right) \)
  • D) \( m v_{CM}^2 \)

Answer Key

• Q1: B - Pure rolling means zero velocity at contact point.
• Q2: B - Velocity at top is twice the center velocity.
• Q3: C - Rolling kinetic energy includes translation and rotation parts.

Memory Aids

• Pure rolling: \( v_{contact} = 0 \)
• Velocity top point = \( 2 v_{CM} \)
• Energy: \( K = \frac{1}{2} m v_{CM}^2 (1 + \frac{k^2}{R^2}) \)

Motion Spherical Body Rough Inclined

Friction and Rolling on Incline

• Minimum friction for pure rolling: μ_min = tan θ / (1 + mR²/I)
• If μ = 0, body slips down with acceleration a = g sin θ
• If μ > μ_min, pure rolling with acceleration a₂ = g sin θ / (1 + I/mR²)
• Friction force magnitude: f = mg sin θ / (1 + mR²/I) (acts upward)
• If μ < μ_min, rolling with slipping, acceleration a₃ = g sin θ - μg cos θ

Comparison Pure Rolling vs Pure Sliding

• Acceleration: Rolling < Sliding
• Velocity: Rolling < Sliding
• Time to cover distance: Rolling > Sliding
• Rolling involves rotational inertia, sliding does not

Solved Examples

• Ex 1: Calculate minimum friction coefficient for pure rolling on incline θ = 30°, m = 2 kg, R = 0.5 m, I = 0.25 m R².
  • Use μ_min = tan θ / (1 + mR²/I)
  • Calculate and substitute values

Practice Set

• Q1: Minimum friction for pure rolling depends on:
  • A) tan θ only
  • B) m, R, I only
  • C) tan θ and mR²/I
  • D) None of these
• Q2: If μ = 0, acceleration of body on incline is:
  • A) 0
  • B) g cos θ
  • C) g sin θ
  • D) g sin θ / (1 + I/mR²)
• Q3: If μ < μ_min, acceleration is:
  • A) g sin θ
  • B) g sin θ - μg cos θ
  • C) g sin θ / (1 + I/mR²)
  • D) 0

Answer Key

• Q1: C - Minimum friction depends on slope angle and inertia ratio mR²/I
• Q2: C - Without friction, body slips with acceleration g sin θ
• Q3: B - Rolling with slipping acceleration includes friction term

Memory Aids

• μ_min = tan θ / (1 + mR²/I)
• Slip if μ = 0, acceleration g sin θ
• Rolling acceleration < sliding acceleration

Friction and Rolling on Incline

• Minimum friction for pure rolling: μ_min = tan θ / (1 + mR²/I)
• If μ = 0, body slips down with acceleration a = g sin θ
• If μ > μ_min, pure rolling with acceleration a₂ = g sin θ / (1 + I/mR²)
• Friction force magnitude: f = mg sin θ / (1 + mR²/I) (acts upward)
• If μ < μ_min, rolling with slipping, acceleration a₃ = g sin θ - μg cos θ

Comparison Pure Rolling vs Pure Sliding

• Acceleration: Rolling < Sliding
• Velocity: Rolling < Sliding
• Time to cover distance: Rolling > Sliding
• Rolling involves rotational inertia, sliding does not

Solved Examples

• Ex 1: Calculate minimum friction coefficient for pure rolling on incline θ = 30°, m = 2 kg, R = 0.5 m, I = 0.25 m R².
  • Use μ_min = tan θ / (1 + mR²/I)
  • Calculate and substitute values

Practice Set

• Q1: Minimum friction for pure rolling depends on:
  • A) tan θ only
  • B) m, R, I only
  • C) tan θ and mR²/I
  • D) None of these
• Q2: If μ = 0, acceleration of body on incline is:
  • A) 0
  • B) g cos θ
  • C) g sin θ
  • D) g sin θ / (1 + I/mR²)
• Q3: If μ < μ_min, acceleration is:
  • A) g sin θ
  • B) g sin θ - μg cos θ
  • C) g sin θ / (1 + I/mR²)
  • D) 0

Answer Key

• Q1: C - Minimum friction depends on slope angle and inertia ratio mR²/I
• Q2: C - Without friction, body slips with acceleration g sin θ
• Q3: B - Rolling with slipping acceleration includes friction term

Memory Aids

• μ_min = tan θ / (1 + mR²/I)
• Slip if μ = 0, acceleration g sin θ
• Rolling acceleration < sliding acceleration

Friction and Rolling on Incline

• Minimum friction for pure rolling: μ_min = tan θ / (1 + mR²/I)
• If μ = 0, body slips down with acceleration a = g sin θ
• If μ > μ_min, pure rolling with acceleration a₂ = g sin θ / (1 + I/mR²)
• Friction force magnitude: f = mg sin θ / (1 + mR²/I) (acts upward)
• If μ < μ_min, rolling with slipping, acceleration a₃ = g sin θ - μg cos θ

Comparison Pure Rolling vs Pure Sliding

• Acceleration: Rolling < Sliding
• Velocity: Rolling < Sliding
• Time to cover distance: Rolling > Sliding
• Rolling involves rotational inertia, sliding does not

Solved Examples

• Ex 1: Calculate minimum friction coefficient for pure rolling on incline θ = 30°, m = 2 kg, R = 0.5 m, I = 0.25 m R².
  • Use μ_min = tan θ / (1 + mR²/I)
  • Calculate and substitute values

Practice Set

• Q1: Minimum friction for pure rolling depends on:
  • A) tan θ only
  • B) m, R, I only
  • C) tan θ and mR²/I
  • D) None of these
• Q2: If μ = 0, acceleration of body on incline is:
  • A) 0
  • B) g cos θ
  • C) g sin θ
  • D) g sin θ / (1 + I/mR²)
• Q3: If μ < μ_min, acceleration is:
  • A) g sin θ
  • B) g sin θ - μg cos θ
  • C) g sin θ / (1 + I/mR²)
  • D) 0

Answer Key

• Q1: C - Minimum friction depends on slope angle and inertia ratio mR²/I
• Q2: C - Without friction, body slips with acceleration g sin θ
• Q3: B - Rolling with slipping acceleration includes friction term

Memory Aids

• μ_min = tan θ / (1 + mR²/I)
• Slip if μ = 0, acceleration g sin θ
• Rolling acceleration < sliding acceleration

Friction and Rolling on Incline

• Minimum friction for pure rolling: μ_min = tan θ / (1 + mR²/I)
• If μ = 0, body slips down with acceleration a = g sin θ
• If μ > μ_min, pure rolling with acceleration a₂ = g sin θ / (1 + I/mR²)
• Friction force magnitude: f = mg sin θ / (1 + mR²/I) (acts upward)
• If μ < μ_min, rolling with slipping, acceleration a₃ = g sin θ - μg cos θ

Comparison Pure Rolling vs Pure Sliding

• Acceleration: Rolling < Sliding
• Velocity: Rolling < Sliding
• Time to cover distance: Rolling > Sliding
• Rolling involves rotational inertia, sliding does not

Solved Examples

• Ex 1: Calculate minimum friction coefficient for pure rolling on incline θ = 30°, m = 2 kg, R = 0.5 m, I = 0.25 m R².
  • Use μ_min = tan θ / (1 + mR²/I)
  • Calculate and substitute values

Practice Set

• Q1: Minimum friction for pure rolling depends on:
  • A) tan θ only
  • B) m, R, I only
  • C) tan θ and mR²/I
  • D) None of these
• Q2: If μ = 0, acceleration of body on incline is:
  • A) 0
  • B) g cos θ
  • C) g sin θ
  • D) g sin θ / (1 + I/mR²)
• Q3: If μ < μ_min, acceleration is:
  • A) g sin θ
  • B) g sin θ - μg cos θ
  • C) g sin θ / (1 + I/mR²)
  • D) 0

Answer Key

• Q1: C - Minimum friction depends on slope angle and inertia ratio mR²/I
• Q2: C - Without friction, body slips with acceleration g sin θ
• Q3: B - Rolling with slipping acceleration includes friction term

Memory Aids

• μ_min = tan θ / (1 + mR²/I)
• Slip if μ = 0, acceleration g sin θ
• Rolling acceleration < sliding acceleration

Friction and Rolling on Incline

• Minimum friction for pure rolling: μ_min = tan θ / (1 + mR²/I)
• If μ = 0, body slips down with acceleration a = g sin θ
• If μ > μ_min, pure rolling with acceleration a₂ = g sin θ / (1 + I/mR²)
• Friction force magnitude: f = mg sin θ / (1 + mR²/I) (acts upward)
• If μ < μ_min, rolling with slipping, acceleration a₃ = g sin θ - μg cos θ

Comparison Pure Rolling vs Pure Sliding

• Acceleration: Rolling < Sliding
• Velocity: Rolling < Sliding
• Time to cover distance: Rolling > Sliding
• Rolling involves rotational inertia, sliding does not

Solved Examples

• Ex 1: Calculate minimum friction coefficient for pure rolling on incline θ = 30°, m = 2 kg, R = 0.5 m, I = 0.25 m R².
  • Use μ_min = tan θ / (1 + mR²/I)
  • Calculate and substitute values

Practice Set

• Q1: Minimum friction for pure rolling depends on:
  • A) tan θ only
  • B) m, R, I only
  • C) tan θ and mR²/I
  • D) None of these
• Q2: If μ = 0, acceleration of body on incline is:
  • A) 0
  • B) g cos θ
  • C) g sin θ
  • D) g sin θ / (1 + I/mR²)
• Q3: If μ < μ_min, acceleration is:
  • A) g sin θ
  • B) g sin θ - μg cos θ
  • C) g sin θ / (1 + I/mR²)
  • D) 0

Answer Key

• Q1: C - Minimum friction depends on slope angle and inertia ratio mR²/I
• Q2: C - Without friction, body slips with acceleration g sin θ
• Q3: B - Rolling with slipping acceleration includes friction term

Memory Aids

• μ_min = tan θ / (1 + mR²/I)
• Slip if μ = 0, acceleration g sin θ
• Rolling acceleration < sliding acceleration

Friction and Rolling on Incline

• Minimum friction for pure rolling: \( \mu_{min} = \frac{\tan \theta}{1 + \frac{mR^2}{I}} \)
• If \( \mu = 0 \), body slips down with acceleration \( a = g \sin \theta \)
• If \( \mu > \mu_{min} \), pure rolling with acceleration \( a_2 = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \)
• Friction force magnitude: \( f = \frac{mg \sin \theta}{1 + \frac{mR^2}{I}} \) (acts upward)
• If \( \mu < \mu_{min} \), rolling with slipping, acceleration \( a_3 = g \sin \theta - \mu g \cos \theta \)

Comparison Pure Rolling vs Pure Sliding

• Acceleration: Rolling \( < \) Sliding
• Velocity: Rolling \( < \) Sliding
• Time to cover distance: Rolling \( > \) Sliding
• Rolling involves rotational inertia, sliding does not

Solved Examples

• Ex 1: Calculate minimum friction coefficient for pure rolling on incline \( \theta = 30^\circ \), \( m = 2 \mathrm{kg} \), \( R = 0.5 \mathrm{m} \), \( I = 0.25 m R^2 \).
  • Use \( \mu_{min} = \frac{\tan \theta}{1 + \frac{mR^2}{I}} \)
  • Calculate and substitute values

Practice Set

• Q1: Minimum friction for pure rolling depends on:
  • A) \( \tan \theta \) only
  • B) \( m, R, I \) only
  • C) \( \tan \theta \) and \( \frac{mR^2}{I} \)
  • D) None of these
• Q2: If \( \mu = 0 \), acceleration of body on incline is:
  • A) 0
  • B) \( g \cos \theta \)
  • C) \( g \sin \theta \)
  • D) \( \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \)
• Q3: If \( \mu < \mu_{min} \), acceleration is:
  • A) \( g \sin \theta \)
  • B) \( g \sin \theta - \mu g \cos \theta \)
  • C) \( \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \)
  • D) 0

Answer Key

• Q1: C - Minimum friction depends on slope angle and inertia ratio \( \frac{mR^2}{I} \)
• Q2: C - Without friction, body slips with acceleration \( g \sin \theta \)
• Q3: B - Rolling with slipping acceleration includes friction term

Memory Aids

• \( \mu_{min} = \frac{\tan \theta}{1 + \frac{mR^2}{I}} \)
• Slip if \( \mu = 0 \), acceleration \( g \sin \theta \)
• Rolling acceleration \( < \) sliding acceleration