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AS 91524 · 2024 paper · rotational motion (flywheel)
Question 2(b) Merit
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 |
The ball and cup start from rest and fall 0.750 m. Calculate the flywheel's final angular velocity.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 How far round does the flywheel turn while the cup falls 0.750 m?
The string unwinds 0.750 m of rim: θ = d / r = 0.750 / 0.110 = 6.8182 rad. -
2 It starts from rest with a steady α. Which equation gives ω from θ?
ω² = ω0² + 2αθ = 0 + 2 × 16.636 × 6.8182 = 226.86 -
3 Take the square root, then check it another way.
ω = √226.86 = 15.06 rad s⁻¹. Check: the cup’s own speed is v = √(2ad) = √(2 × 1.83 × 0.750) = 1.6568 m s⁻¹, and v / r = 1.6568 / 0.110 = 15.06 rad s⁻¹.
What the marker wanted
- Achieved One of: the angle turned (6.82 rad); the cup’s final speed (1.657 m s⁻¹); or a right ω from a wrong angle.
- Merit The right answer, by angles as above or by finding the cup’s speed first and using v = rω.
In our words, from NZQA’s 2024 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.