IB Diploma Physics · topic A.4
Rigid body mechanics
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.
The load that tips a mobile platform
A 20 kg platform has supports at x = 0 and x = 1.2 m. Its centre of mass is at x = 0.60 m. A 30 kg load is placed at variable x. Use g = 10 N kg⁻¹. The graph supplies the right-support reaction while both supports remain in contact. Supplied data (Load position x / m; Right-support reaction / N): (0, 100), (0.4, 200), (0.8, 300), (1.2, 400), (1.6, 500). (a) Find both support forces for x = 1.0 m and determine the load position at which tipping begins. [3 marks] (b) Find the combined centre of mass at the tipping threshold and assess lowering the platform's centre of mass on this ideal level-surface threshold. [3 marks]
Right-support reaction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Right-support reaction / N
Horizontal axis (x): Load position x / m
Read the exact data as a table
| Load position x / m | Right-support reaction / N |
|---|---|
| 0 | 100 |
| 0.4 | 200 |
| 0.8 | 300 |
| 1.2 | 400 |
| 1.6 | 500 |
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 both support forces for x = 1.0 m and determine the load position at which tipping begins.
Reveal the worked solution
- Taking moments about the left support: RR(1.2) = 200(0.60)+300x. At x = 1.0, RR = 350 N.
- Vertical force balance gives RL = 500−RR = 150 N.
- Tipping begins when the left reaction reaches zero, so RR = 500 N and 500(1.2) = 120+300x, giving x = 1.6 m. Beyond that point the two-contact static model fails.
- At xload = 1.6 m, xCM = [20(0.60)+30(1.6)]/(20+30) = 1.2 m, directly above the right support.
- At the threshold, the combined weight's line of action reaches the support boundary. A support cannot provide a downward reaction to maintain this contact model.
- Lowering the centre of mass can improve resistance to tipping during tilt or lateral acceleration, but it does not change the horizontal moment arms of vertical weights in this strictly level static setup.
Answer (a) At x = 1.0 m: right support = 350 N, left = 150 N; tipping threshold x = 1.6 m. (b) Combined centre of mass lies at x = 1.2 m; lowering it does not change this ideal horizontal overhang threshold.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.