AS 91524 · 2024 paper · circular motion of a ball bearing

Question 1(c) Merit to Excellence

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

Compare the forces acting at the top (Point Y) and bottom (Point X), and hence explain how the net force varies between them. Labelled force diagrams will help.

The 2024 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Which forces act on the ball at X and at Y?
The same two at both: its weight, mg, straight down and the same size everywhere, and the track’s reaction force, N, pushing up.
2 At X the ball is at the bottom of a dip. Where is the centre of that curve, so which way must the net force point?
The centre is above the ball, so the net force (the centripetal force) points up: N − mg = mv² / r. So N = mg + mv² / r: the track pushes up harder than the weight pulls down.
3 At Y, on top of a bump, where is the centre now?
Below the ball, so the net force points down: mg − N = mv² / r. So N = mg − mv² / r: the reaction force is now smaller than the weight.
4 So how does the net force change between the two points, in direction and in size?
It always points to the centre of the curve the ball is on: up at X, down at Y. Its size is mv² / r, and both curves have the same 35.0 mm radius, so it is bigger where the ball is faster. X is lower than Y, so going from Y down to X the ball loses gravitational potential energy and, with energy conserved, gains the same amount of kinetic energy: it is faster at X, and the net force there is larger. At the 0.850 m s⁻¹ from (a), it is 0.0112 × 0.850² / 0.0350 = 0.231 N at X, about twice the ball’s 0.110 N weight.

Answer The weight is the same at both points; the reaction force is what changes. At X (bottom of a dip) the reaction force is bigger than the weight, so the net force points up, towards the centre of the dip. At Y (top of a bump) the reaction force is smaller than the weight, so the net force points down, towards the centre of the bump. Either way the net force is the centripetal force, mv² / r, and it is larger at X: X is lower, so by conservation of energy the ball has more kinetic energy there and is moving faster.

What the marker wanted

Achieved One of: the net force is the centripetal force; the centripetal force is bigger at X than at Y; weight down and reaction up at both points; the weight stays the same; F_c = F_R − F_w at X or F_c = F_w − F_R at Y; a correct force diagram for one point; or a bigger reaction force at X than at Y.
Merit One point explained in full, including the size and direction of the net force there; or the sizes and directions of the reaction force and the weight described at both points. An accurate force diagram can be part of it.
Excellence The full comparison of both points. Accurate force diagrams can be part of it.

In our words, from NZQA’s 2024 marking schedule.

Common mistake

Drawing the reaction force pointing down at Y, or the net force pointing away from the centre. NZQA’s 2024 report says weight and support forces drawn the wrong way round were common below Achieved, and that even Achieved answers often confused the direction of the net force.

How did it go?

Got it moves it further out. Review brings it back tomorrow.