IB · A · extended response
A rotating platform catches a mass
Space, time and motion · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
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(a)
Find angular speed immediately after capture and the change in rotational kinetic energy. Explain the appropriate conservation law.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
With negligible external torque impulse about the axle during capture, initial angular momentum is Iω = 2(6) = 12 kg m² s⁻¹.
- 2
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.
- 3
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.
(b)
Find angular speed after the braking pulse and the rotational energy removed. Distinguish angular impulse from work.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
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.
- 2
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.
- 3
The graph area gives change in angular momentum, not work. Work is the torque–angle integral; the energy difference independently gives 18 J removed.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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