AS 91524 · 2024 paper · circular motion of a ball bearing
Question 1(a) Achieved
Practises Energy methods for rolling and rotating bodies
A solid steel ball bearing is released from different heights on a steel track. It rolls down and travels over a series of bumps and dips, each of radius 35.0 mm. Point X is the bottom of the first dip; Point Y is the top of a bump. As a solid sphere, I = ⅖mr².
| r ball | 7.00 mm |
|---|---|
| m | 11.2 g |
| bump/dip radius | 35.0 mm |
In one trial the ball reaches the bottom of the first dip (Point X) at 0.850 m s⁻¹. Calculate its angular velocity there.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 The ball rolls without slipping. What links its speed to how fast it spins?
v = rω, where r is the ball’s own radius, 7.00 mm. (The dip’s 35.0 mm sets how the ball’s path curves, not how fast the ball spins.) -
2 Put the radius in metres, then rearrange.
r = 7.00 mm = 0.00700 m, so ω = v / r = 0.850 / 0.00700 = 121.43 rad s⁻¹.
What the marker wanted
- Achieved The right answer.
In our words, from NZQA’s 2024 marking schedule.
✕Common mistake
Leaving the radius in millimetres: 0.850 / 7.00 = 0.121 rad s⁻¹, a thousand times too small. NZQA’s 2024 report says SI prefixes were very poorly known across that year’s physics papers; convert to metres first.How did it go?
Got it moves it further out. Review brings it back tomorrow.