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

Question 1(d) Merit to Excellence

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

By considering energy changes, calculate the maximum height h above Point Y that the ball can be released from if it is to stay in contact with the track over the bumps (a solid sphere, I = ⅖mr²).

The 2024 paper, page 3 ↗ Marking schedule, page 2 ↗

Work it out

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

1 To stay on the track at Y, what is the most speed the ball can have there?
From (b), v = √(gr), so v² = gr = 9.81 × 0.0350 = 0.343 35 m² s⁻².
2 Released from rest a height h above Y, where has its gravitational potential energy gone by the time it gets to Y?
Into kinetic energy of two kinds, because it rolls: mgh = ½mv² + ½Iω².
3 Put in I = ⅖mr² and ω = v / r. What happens to the ball’s radius?
½Iω² = ½ × ⅖mr² × v² / r² = ⅕mv². So mgh = ½mv² + ⅕mv² = 0.7mv². The ball’s mass and its radius both cancel.
4 Solve for h.
h = 0.7v² / g = 0.7 × 0.343 35 / 9.81 = 0.0245 m.
5 Check it with the energies themselves.
With m = 0.0112 kg: mgh = 0.0112 × 9.81 × 0.0245 = 2.692 × 10⁻³ J, which splits into ½mv² = 1.923 × 10⁻³ J and ½Iω² = 0.769 × 10⁻³ J (the ball spins at ω = 0.586 / 0.00700 = 83.7 rad s⁻¹).

Answer h = 0.0245 m (24.5 mm) above Point Y. Released any higher, it reaches Y faster than 0.586 m s⁻¹ and leaves the track.

What the marker wanted

Achieved One of: an attempt by conservation of energy (E_gp = E_k(lin) + E_k(rot)); one energy right (E_k(lin) = 1.92 × 10⁻³ J or E_k(rot) = 7.69 × 10⁻⁴ J); or the spin at Y, ω = 83.7 rad s⁻¹.
Merit The right process with one error, such as a slip while simplifying; or two of the three energies right, having tried all three; or h = 7.00 × 10⁻³ m.
Excellence The right answer, 0.0245 m.

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

Common mistake

Counting only one kind of kinetic energy. Using the rotational energy alone gives h = 7.00 × 10⁻³ m, which the schedule holds at Merit. The ball both moves along and spins, so it needs ½mv² and ½Iω² at Y.

How did it go?

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