A.4 Rigid body mechanics

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Torque

\tau = Fr\sin\theta

Rotational second law

\tau = I\alpha

Moment of inertia (point masses)

I = \Sigma mr^{2}

Rotational kinematics

\omega_{f} = \omega_{i} + \alpha t

Angular displacement

\Delta\theta = \omega_{i}t + \tfrac{1}{2}\alpha t^{2}

Angular velocity–displacement

\omega_{f}^{2} = \omega_{i}^{2} + 2\alpha\Delta\theta

Angular momentum

L = I\omega

Angular impulse

\Delta L = \tau\Delta t

Rotational kinetic energy

E_{k} = \tfrac{1}{2}I\omega^{2} = \frac{L^{2}}{2I}

Section A — Multiple choice

A1. An ice skater spinning with arms outstretched pulls her arms in. Which row is correct?

A Angular momentum increases; kinetic energy constant
B Angular momentum constant; kinetic energy increases
C Both angular momentum and kinetic energy constant
D Angular momentum constant; kinetic energy decreases

A2. Two children of equal mass sit on a seesaw, one at 2.0 m and one at 1.5 m from the pivot. For balance, an extra downward force must act on the shorter side. If each child weighs 400 N, what extra torque is needed?

A $100\ \text{N m}$
B $200\ \text{N m}$
C $400\ \text{N m}$
D $800\ \text{N m}$

A3. A solid sphere and a hollow sphere of equal mass and radius roll from rest down the same incline. Which reaches the bottom first?

A The solid sphere
B The hollow sphere
C They arrive together
D It depends on the angle of the incline

Section B — Short answer

B1. A flywheel of moment of inertia $0.80\ \text{kg m}^2$ accelerates uniformly from rest to $300\ \text{rad s}^{-1}$ in 12 s. Calculate the torque applied and the number of revolutions completed. [4 marks]

B2. Explain why a tightrope walker carries a long pole. [3 marks]

B3. State the rotational analogue of each quantity: force, mass, linear momentum, and Newton's second law. [2 marks]

Answer key

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Section A

A1: Angular momentum constant; kinetic energy increases — No external torque acts, so $L = I\omega$ is conserved: $I$ falls, $\omega$ rises. But $E_k = L^2/2I$ , so a smaller $I$ at fixed $L$ means more kinetic energy — supplied by the work her muscles do pulling the arms inward.

A2: $200\ \text{N m}$ — Torques about the pivot: $400 \times 2.0 = 800\ \text{N m}$ versus $400 \times 1.5 = 600\ \text{N m}$ . The deficit that must be supplied on the shorter side is $200\ \text{N m}$ .

A3: The solid sphere — The hollow sphere has more mass far from the axis, hence a larger moment of inertia, so a bigger share of its potential energy becomes rotational rather than translational kinetic energy. Less translational speed means it arrives later — independent of mass, radius or angle.

Section B

B1: $\alpha = 300/12 = 25\ \text{rad s}^{-2}$ , so $\tau = I\alpha = 0.80 \times 25 = 20\ \text{N m}$ . Angular displacement: $\Delta\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(25)(144) = 1800\ \text{rad}$ , which is $1800/2\pi \approx 286$ revolutions.

B2: The pole is massive and extends far from the rotation axis, greatly increasing the walker's moment of inertia about the wire. For a given unbalancing torque, $\alpha = \tau/I$ is much smaller, so the walker tips more slowly and has time to correct. Bending the pole ends downward lowers the centre of mass too.

B3: Force → torque $\tau$ ; mass → moment of inertia $I$ ; linear momentum $p = mv$ → angular momentum $L = I\omega$ ; $F = ma$ → $\tau = I\alpha$ (or $F = \Delta p/\Delta t$ → $\tau = \Delta L/\Delta t$ ).