Torque and Work — Worked Solutions

These solutions belong to Torque and Work. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

A graph plots torque vertically against angular position horizontally. What does its signed area represent?

1. Inspect the horizontal axis: it is angular position. Why: The choice between work and angular impulse depends on whether the axis is angle or time.

2. The signed area is τΔθ, which is work in joules. Why: Torque–time area would instead have units N·m·s.

Back to this question.

Question 2

A wheel stays at a fixed angle while a torque is exerted on it. What rotational work does that torque do on the wheel?

1. The wheel’s angular displacement is zero. Why: A force or torque can act while its application point has no rotational motion.

2. W = τΔθ = 0. Why: Elapsed time alone does not imply work or rotational kinetic-energy transfer.

Back to this question.

Question 3

A fixed-axis wheel starts from rest. Its net work doubles with the same inertia. What happens to its final angular speed?

1. Starting from rest gives W_net = ½Iω_f². Why: There is no initial rotational kinetic energy to subtract.

2. At fixed I, doubling W multiplies ω_f by √2. Why: Angular speed depends on the square root of supplied net work.

Back to this question.

Question 4

A wheel with I = 0.050 kg·m² turns at 12 rad/s. A constant opposing torque of magnitude 0.30 N·m brings it to rest. Find the angle it turns through while stopping.

1. Find K_i = ½(0.050)(12²) = 3.6 J and K_f = 0. Why: The wheel’s stated final condition is rest.

2. Use W_net = ΔK = −3.6 J. Why: The opposing torque removes the initial rotational kinetic energy.

3. Write −0.30Δθ = −3.6 to get Δθ = 12 rad. Why: Work uses angular displacement; this calculation does not supply stopping time.

Back to this question.

Question 5

A sensor records driving torque 2.0 N·m at angle 0, 4.0 N·m at 1.0 rad, and 4.0 N·m at 2.0 rad. Assume linear interpolation between measurements. A constant opposing bearing torque has magnitude 0.50 N·m. The wheel begins with 1.0 J of rotational kinetic energy and has a fixed axle. Decide whether a torque–time or torque–angle graph can directly determine the work here, then use the supplied data to predict final kinetic energy. State the energy-transfer assumption needed for that prediction.

1. Use torque vertically against angle horizontally. Why: Work is torque–angle area. A torque–time graph would determine angular impulse, and no elapsed-time measurements are supplied here.

2. Driving work is ½(2.0 + 4.0)(1.0) + 4.0(1.0) = 7.0 J. Bearing work is −0.50(2.0) = −1.0 J, so net work is +6.0 J. Why: Add the first trapezoid and second rectangle, then include the opposing torque through the same angular displacement.

3. K_f = K_i + W_net = 1.0 + 6.0 = 7.0 J, assuming the stated torque work accounts for all energy transfers and no other energy storage changes. Why: The fixed-axis work–energy theorem connects net torque work to rotational kinetic energy only when the ledger includes all relevant work and energy changes.

Back to this question.