AP Physics 1 · Free-response skill

A weight lands on a spinning turntable

The move. Use conservation of angular momentum to find a new angular speed, use rotational kinetic energy to find what is lost, and predict how a changed setup would compare.

How to do it

  1. Start each rotation problem by asking whether any outside torque acts about the axis; if none does, angular momentum before equals angular momentum after.
  2. A point mass m at distance r from the axis adds mr2mr^2 to the rotational inertia. An object that starts with no motion around the axis brings no angular momentum with it.
  3. Kinetic energy is a separate question: compute 12Iω2\frac{1}{2}I\omega^2 before and after and compare the two numbers.

Practice prompt Try it

A turntable with rotational inertia 0.050 kg·m² spins at 12 rad/s on a bearing with negligible friction. A small 0.25 kg weight, not moving, is lowered gently onto the turntable and sticks 0.20 m from the axis. Treat the weight as a point mass.

Answer each part.

  1. Find the angular speed of the turntable and weight just after the weight sticks.
  2. Find how much rotational kinetic energy is lost when the weight sticks.
  3. Suppose the weight had stuck 0.10 m from the axis instead. Would the final angular speed be greater than, less than or equal to your part a answer?

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 3 points

Model answer, annotated

(a) No outside torque acts about the axis, and the weight brings no angular momentum, so (0.050)(12)=[0.050+(0.25)(0.20)2]ωf=0.060 ωf(0.050)(12) = [0.050 + (0.25)(0.20)^2]\omega_f = 0.060\,\omega_f, giving ωf=0.60/0.060=10\omega_f = 0.60/0.060 = 10 rad/s.

  • Point 1 Adds the weight’s mr2mr^2 and conserves angular momentum.

(b) Before: 12(0.050)(12)2=3.6\frac{1}{2}(0.050)(12)^2 = 3.6 J. After: 12(0.060)(10)2=3.0\frac{1}{2}(0.060)(10)^2 = 3.0 J. The loss is 3.6−3.0=0.603.6 - 3.0 = 0.60 J.

  • Point 2 Computes both kinetic energies and subtracts.

(c) Greater. At 0.10 m the weight adds only (0.25)(0.10)2=0.0025(0.25)(0.10)^2 = 0.0025 kg·m², so the total rotational inertia is smaller and the same angular momentum gives ωf=0.60/0.0525≈11.4\omega_f = 0.60/0.0525 \approx 11.4 rad/s.

  • Point 3 Uses the smaller added rotational inertia to compare.