Q3 Rigid Body Mechanics, Torque, Angular Acceleration, and Angular Momentum

Rigid Body Mechanics, Torque, Angular Acceleration, and Angular Momentum

IB Physics HL • Paper 2 • Long-Answer Question • Rotational Motion, Moment of Inertia, Angular Impulse, and Rotational Energy

Lesson Overview

In this Paper 2 long-answer question, we study the rotational motion of a rigid body and apply the main ideas of rigid body mechanics in one linked context. The question begins with torque and rotational equilibrium, then develops into angular acceleration, uniform angular acceleration, moment of inertia, angular momentum, angular impulse, and rotational kinetic energy.

This is a classic HL-style question because it connects force, motion, and energy in rotational form. To score well, you should identify the axis of rotation clearly, use torque arguments carefully, and distinguish between linear and angular quantities throughout.

⭐ Key Concepts

The torque of a force about an axis is

$\tau=Fr\sin\theta$

A body is in rotational equilibrium when the resultant torque is zero.
For uniform angular acceleration, the rotational equations of motion are

$\omega_f=\omega_i+\alpha t$

$\Delta\theta=\omega_i t+\frac12\alpha t^2$

$\omega_f^2=\omega_i^2+2\alpha\Delta\theta$

The moment of inertia of a system of point masses is

$I=\sum mr^2$

Newton’s second law for rotation is

$\tau=I\alpha$

The angular momentum of a rotating body is

$L=I\omega$

The angular impulse is

$\Delta L=\tau\Delta t$

The rotational kinetic energy is

$E_k=\frac12 I\omega^2$

📘 Clear IB-Style Explanation

Rigid body mechanics is often easiest to understand by comparing it with linear motion. Force corresponds to torque, mass corresponds to moment of inertia, linear acceleration corresponds to angular acceleration, and momentum corresponds to angular momentum.

A strong IB answer should make these ideas explicit. In rotational questions, students often lose marks by using the correct equation with the wrong axis or by confusing distance from the pivot with the total length of the object.

In longer Paper 2 questions, always decide first:

what axis the body rotates about,
which forces produce clockwise or anticlockwise torque,
whether the motion is in equilibrium, uniformly accelerated, or impulsive.

📌 Worked Example 1 — Torque About a Pivot

A uniform horizontal beam of length

$2.4\text{ m}$

is pivoted at one end. A downward force of

$18\text{ N}$

acts at the other end of the beam, perpendicular to it.

(a) Calculate the torque of the force about the pivot.

Solution

The torque is given by

$\tau=Fr\sin\theta$

Here,

$F=18\text{ N},\qquad r=2.4\text{ m},\qquad \theta=90^\circ$

$\tau=(18)(2.4)\sin90^\circ$

$\tau=43.2\text{ N m}$

Therefore, the torque about the pivot is

$43.2\text{ N m}$

📊 Marks: M1 A1

📌 Worked Example 2 — Rotational Equilibrium

The same beam is now held horizontal by a second vertical force applied at a distance of

$1.2\text{ m}$

from the pivot, on the opposite side of the beam’s centre of mass. The beam is in rotational equilibrium.

(b) Calculate the magnitude of the second force.

Solution

For rotational equilibrium, the resultant torque must be zero.

So the clockwise torque equals the anticlockwise torque.

From part (a), the torque of the 18 N force is

$43.2\text{ N m}$

If the second force is $F$ , then

$F(1.2)=43.2$

$F=\frac{43.2}{1.2}$

$F=36\text{ N}$

Therefore, the second force has magnitude

$36\text{ N}$

📊 Marks: M1 A1

📌 Worked Example 3 — Moment of Inertia of Point Masses

Four small masses are attached to a light rigid frame and rotate about a fixed axis. Their masses and distances from the axis are:

$0.40\text{ kg at }0.30\text{ m}$

$0.25\text{ kg at }0.40\text{ m}$

$0.30\text{ kg at }0.20\text{ m}$

$0.15\text{ kg at }0.50\text{ m}$

(c) Calculate the total moment of inertia of the system.

Solution

For point masses,

$I=\sum mr^2$

$I=(0.40)(0.30)^2+(0.25)(0.40)^2+(0.30)(0.20)^2+(0.15)(0.50)^2$

$I=(0.40)(0.09)+(0.25)(0.16)+(0.30)(0.04)+(0.15)(0.25)$

$I=0.036+0.040+0.012+0.0375$

$I=0.1255\text{ kg m}^2$

Therefore, the total moment of inertia is

$0.126\text{ kg m}^2$

📊 Marks: M1 M1 A1

📌 Worked Example 4 — Angular Acceleration from a Resultant Torque

A resultant torque of

$2.8\text{ N m}$

acts on the rotating system in part (c).

(d) Calculate the angular acceleration of the system.

Solution

Newton’s second law for rotation is

$\tau=I\alpha$

$\alpha=\frac{\tau}{I}$

Substitute the values:

$\alpha=\frac{2.8}{0.1255}$

$\alpha=22.3\text{ rad s}^{-2}$

Therefore, the angular acceleration is

$22\text{ rad s}^{-2}$

📊 Marks: M1 A1

📌 Worked Example 5 — Uniform Angular Acceleration

The system starts from rest and rotates with constant angular acceleration

$22.3\text{ rad s}^{-2}$

for

$3.0\text{ s}$

(e)(i) Calculate the final angular speed.

(e)(ii) Determine the angular displacement during this time.

Solution

(e)(i) Final angular speed

Use

$\omega_f=\omega_i+\alpha t$

Since the system starts from rest,

$\omega_i=0$

$\omega_f=(22.3)(3.0)$

$\omega_f=66.9\text{ rad s}^{-1}$

Therefore, the final angular speed is

$66.9\text{ rad s}^{-1}$

📊 Marks: M1 A1

(e)(ii) Angular displacement

Use

$\Delta\theta=\omega_i t+\frac12\alpha t^2$

$\Delta\theta=0+\frac12(22.3)(3.0)^2$

$\Delta\theta=11.15\times9$

$\Delta\theta=100.35\text{ rad}$

Therefore, the angular displacement is approximately

$1.00\times10^2\text{ rad}$

📊 Marks: M1 A1

📌 Worked Example 6 — Angular Momentum

(f) Calculate the angular momentum of the rotating system after $3.0\text{ s}$ .

Solution

Angular momentum is given by

$L=I\omega$

Using

$I=0.1255\text{ kg m}^2 \qquad\text{and}\qquad \omega=66.9\text{ rad s}^{-1}$

gives

$L=(0.1255)(66.9)$

$L=8.40\text{ kg m}^2\text{ s}^{-1}$

Therefore, the angular momentum is

$8.4\text{ kg m}^2\text{ s}^{-1}$

📊 Marks: M1 A1

📌 Worked Example 7 — Angular Impulse

A braking torque of

$1.6\text{ N m}$

acts in the opposite direction for

$2.5\text{ s}$

(g) Calculate the change in angular momentum of the system.

Solution

Angular impulse is

$\Delta L=\tau\Delta t$

$\Delta L=(1.6)(2.5)$

$\Delta L=4.0\text{ kg m}^2\text{ s}^{-1}$

Since the torque opposes the motion, the change in angular momentum is a decrease of

$4.0\text{ kg m}^2\text{ s}^{-1}$

📊 Marks: M1 A1

📌 Worked Example 8 — Rotational Kinetic Energy

(h) Calculate the rotational kinetic energy of the system at angular speed $66.9\text{ rad s}^{-1}$ .

Solution

The rotational kinetic energy is

$E_k=\frac12 I\omega^2$

Substitute:

$E_k=\frac12(0.1255)(66.9)^2$

$E_k=0.06275\times4475.6$

$E_k=280.8\text{ J}$

Therefore, the rotational kinetic energy is approximately

$2.8\times10^2\text{ J}$

📊 Marks: M1 A1

📌 Worked Example 9 — Why Mass Distribution Matters

(i) Explain why the moment of inertia depends not only on the total mass of a body but also on how that mass is distributed about the axis.

Solution

The moment of inertia is given by

$I=\sum mr^2$

This shows that mass farther from the axis contributes much more strongly because the distance is squared.

So two bodies with the same total mass can have different moments of inertia if their mass is distributed differently about the axis.

📊 Marks: A1 A1

⚠ Common Mistakes

Using the full length of the object instead of the perpendicular distance from the axis when calculating torque.
Forgetting the $\sin\theta$ factor in

$\tau=Fr\sin\theta$

Mixing up linear equations of motion with angular equations of motion.
Using mass instead of moment of inertia in rotational Newton’s second law.
For moment of inertia, forgetting to square the distance from the axis.
Confusing angular impulse with torque or with angular momentum itself.
Leaving out units such as $\text{N m}$ , $\text{rad s}^{-1}$ , or $\text{kg m}^2\text{ s}^{-1}$ .

📘 IB Exam Tips

Always identify the axis of rotation before calculating torque or moment of inertia.
For equilibrium questions, state clearly that clockwise torque equals anticlockwise torque.
Use angular symbols carefully: $\theta$ , $\omega$ , and $\alpha$ are not interchangeable.
When using

$\tau=I\alpha$

check that the torque is the resultant torque.
In longer questions, keep track of units carefully so each stage stays physically meaningful.
For explanation questions, connect the equation to the physical idea, not just the algebra.

🧪 Challenge Problem

A light rigid bar rotates about one end. A force of $12\text{ N}$ acts perpendicular to the bar at a distance of $1.8\text{ m}$ from the pivot.

(a) Calculate the torque about the pivot.

(b) The bar has two attached point masses: $0.50\text{ kg}$ at $0.40\text{ m}$ and $0.30\text{ kg}$ at $0.70\text{ m}$ . Calculate the total moment of inertia.

(c) Hence determine the angular acceleration.

(d) If the system starts from rest and rotates for $2.0\text{ s}$ with constant angular acceleration, calculate the final angular speed.

✅ Self-Practice

A rotating platform has moment of inertia $0.80\text{ kg m}^2$ . A resultant torque of $4.0\text{ N m}$ acts on it for $3.0\text{ s}$ . The platform starts from rest.

(a) Calculate the angular acceleration.

(b) Determine the final angular speed.

(c) Calculate the angular momentum after $3.0\text{ s}$ .

(d) Calculate the rotational kinetic energy after $3.0\text{ s}$ .

Show Solutions

Challenge Problem

(a)

$\tau=Fr$

$\tau=(12)(1.8)=21.6\text{ N m}$

(b)

$I=\sum mr^2$

$I=(0.50)(0.40)^2+(0.30)(0.70)^2$

$I=(0.50)(0.16)+(0.30)(0.49)$

$I=0.080+0.147$

$I=0.227\text{ kg m}^2$

(c)

$\alpha=\frac{\tau}{I}=\frac{21.6}{0.227}$

$\alpha=95.2\text{ rad s}^{-2}$

(d)

$\omega=\omega_i+\alpha t$

$\omega=0+(95.2)(2.0)$

$\omega=190.4\text{ rad s}^{-1}$

Self-Practice

(a)

$\alpha=\frac{\tau}{I}=\frac{4.0}{0.80}=5.0\text{ rad s}^{-2}$

(b)

$\omega=\omega_i+\alpha t$

$\omega=0+(5.0)(3.0)=15\text{ rad s}^{-1}$

(c)

$L=I\omega=(0.80)(15)=12\text{ kg m}^2\text{ s}^{-1}$

(d)

$E_k=\frac12 I\omega^2$

$E_k=\frac12(0.80)(15)^2$

$E_k=0.40\times225$

$E_k=90\text{ J}$

Rigid Body Mechanics, Torque, Angular Acceleration, and Angular Momentum

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Torque About a Pivot

Solution

📌 Worked Example 2 — Rotational Equilibrium

Solution

📌 Worked Example 3 — Moment of Inertia of Point Masses

Solution

📌 Worked Example 4 — Angular Acceleration from a Resultant Torque

Solution

📌 Worked Example 5 — Uniform Angular Acceleration

Solution

📌 Worked Example 6 — Angular Momentum

Solution

📌 Worked Example 7 — Angular Impulse

Solution

📌 Worked Example 8 — Rotational Kinetic Energy

Solution

📌 Worked Example 9 — Why Mass Distribution Matters

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

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Q3 Rigid Body Mechanics, Torque, Angular Acceleration, and Angular Momentum

Rigid Body Mechanics, Torque, Angular Acceleration, and Angular Momentum

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Torque About a Pivot

Solution

📌 Worked Example 2 — Rotational Equilibrium

Solution

📌 Worked Example 3 — Moment of Inertia of Point Masses

Solution

📌 Worked Example 4 — Angular Acceleration from a Resultant Torque

Solution

📌 Worked Example 5 — Uniform Angular Acceleration

Solution

📌 Worked Example 6 — Angular Momentum

Solution

📌 Worked Example 7 — Angular Impulse

Solution

📌 Worked Example 8 — Rotational Kinetic Energy

Solution

📌 Worked Example 9 — Why Mass Distribution Matters

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

Related

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