AS 91524 · 2024 paper · rotational motion (flywheel)
Question 2(c) Merit to Excellence
Practises Rotational dynamics
A golf ball is placed in a light cup attached to a flywheel of diameter 0.220 m, and begins to accelerate downwards at a constant 1.83 m s⁻².
| m ball | 0.0460 kg |
|---|---|
| flywheel diameter | 0.220 m |
| a | 1.83 m s⁻² |
Calculate the rotational inertia of the flywheel. Begin with the force that makes it accelerate: a free-body diagram helps you find the tension in the string.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Draw the forces on the ball and cup. What are they, and which is bigger?
Its weight, mg, down, and the string’s tension, T, up. The cup speeds up downwards, so the weight is the bigger of the two. -
2 Use Newton’s second law on the ball and cup to find T.
mg − T = ma, so T = m(g − a) = 0.0460 × (9.81 − 1.83) = 0.367 08 N. -
3 The tension pulls on the rim of the flywheel. What torque does it give?
τ = Tr = 0.367 08 × 0.110 = 0.040 379 N m. -
4 Which equation links that torque to the angular acceleration from (a)?
τ = Iα, so I = τ / α = 0.040 379 / 16.636 = 2.427 × 10⁻³ kg m².
What the marker wanted
- Achieved The tension identified (on a diagram is fine), or τ = Fd used, or τ = Iα used.
- Merit The right method with one error, such as the weight used in place of the tension (which gives 2.98 × 10⁻³ kg m²).
- Excellence The right answer, 2.43 × 10⁻³ kg m².
In our words, from NZQA’s 2024 marking schedule.
✕Common mistake
Using the ball’s weight, mg, as the force turning the wheel. That gives 2.98 × 10⁻³ kg m², which the schedule holds at Merit. The cup is accelerating, so the string pulls with less than its weight. NZQA’s 2024 report says many candidates found one of the two forces but missed that the tension is what produces the torque.How did it go?
Got it moves it further out. Review brings it back tomorrow.