AS 91524 · 2024 paper · circular motion of a ball bearing
Question 1(b) Merit
Practises Vertical circular motion and apparent weight
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 |
Show that the maximum speed is 0.586 m s⁻¹ for the ball to stay in contact with the track going over the bump at Point Y (radius 35.0 mm).
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 At the top of the bump, where is the centre of the curve, and which forces act on the ball?
The centre is straight below the ball. Its weight, mg, pulls down, towards the centre; the track’s reaction force, N, pushes up, away from it. -
2 The net force towards the centre is the centripetal force. Write that as an equation.
mg − N = mv² / r -
3 What happens to N as the ball goes faster? So what is the fastest it can go and still touch the track?
A faster ball needs more centripetal force, so N gets smaller. At the limit N = 0 and the weight alone is the centripetal force: mg = mv² / r, so v = √(gr). -
4 Put in the bump’s radius, in metres, and keep the unrounded result.
v = √(9.81 × 0.0350) = √0.343 35 = 0.585 96 m s⁻¹.
What the marker wanted
- Achieved The reaction force set to zero, so that the weight alone provides the centripetal force.
- Merit The working carried through to the answer.
In our words, from NZQA’s 2024 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.