AP Physics · topic 6
Energy and Momentum of Rotating Systems
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A rotating platform catches a mass
A horizontal platform of moment of inertia 2.0 kg m² rotates at 6.0 rad s⁻¹ about a frictionless fixed axle. A 1.0 kg soft mass is dropped vertically and sticks at radius 1.0 m. It has no initial angular momentum about the axle. The supplied torque–time graph describes a later applied braking torque; all values are signed in the initial rotation direction. Supplied data (Time / s; Torque / N m): (0, 0), (1, -6), (2, 0). (a) Find angular speed immediately after capture and the change in rotational kinetic energy. Explain the appropriate conservation law. [3 marks] (b) Find angular speed after the braking pulse and the rotational energy removed. Distinguish angular impulse from work. [3 marks]
Later braking torque
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Torque / N m
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Torque / N m |
|---|---|
| 0 | 0 |
| 1 | -6 |
| 2 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find angular speed immediately after capture and the change in rotational kinetic energy. Explain the appropriate conservation law.
Reveal the worked solution
- With negligible external torque impulse about the axle during capture, initial angular momentum is Iω = 2(6) = 12 kg m² s⁻¹.
- The attached point mass contributes mr² = 1.0 kg m², so the final inertia is 3.0 kg m² and ω = 12/3 = 4.0 rad s⁻¹.
- Initial rotational energy is ½(2)(6²)=36 J and final is ½(3)(4²)=24 J. Sticking is inelastic; angular momentum conservation does not imply kinetic-energy conservation.
- The torque–time area is angular impulse ½(2)(−6) = −6 N m s. Angular momentum falls from 12 to 6 kg m² s⁻¹.
- The inertia stays 3.0 kg m², so final ω = 6/3 = 2.0 rad s⁻¹. Final energy is ½(3)(2²) = 6 J, down from 24 J.
- The graph area gives change in angular momentum, not work. Work is the torque–angle integral; the energy difference independently gives 18 J removed.
Answer (a) ωafter = 4.0 rad s⁻¹; rotational energy falls from 36 J to 24 J, a loss of 12 J. (b) Final angular speed = 2.0 rad s⁻¹; rotational energy removed = 18 J.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.