AS 91524 · 2023 paper · rotational motion
Question 2(c) Merit
Practises Rotational dynamics
Tane works weekends unloading barrels. He rolls an empty barrel of mass 5.50 kg and radius 0.280 m down a ramp 1.34 m high. The linear speed of the barrel at the bottom of the ramp is 3.40 m s⁻¹. Parts (c) and (d) move to Tane on a swivelling stool holding a full bottle of water in each hand: he spins more slowly with his arms outstretched than with his arms in.
| m | 5.50 kg |
|---|---|
| r | 0.280 m |
| h | 1.34 m |
| v | 3.40 m s⁻¹ |
Tane sits on a swivelling stool holding a full bottle of water in each hand. He spins more slowly with his arms outstretched than when he brings his arms in. Explain the reason for this observation.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 While Tane spins on the stool, is anything outside applying a torque? So what is conserved?
No outside torque acts, since the stool turns freely, so his angular momentum, L = Iω, is conserved. -
2 How does stretching his arms out, a bottle in each hand, change his rotational inertia?
It moves mass further from the axis. Each bit of mass counts as mr², so with the bottles far out his rotational inertia is much bigger. -
3 If L stays the same and I is bigger, what must ω be?
Smaller, since L = Iω. With his arms out he turns more slowly; pulling them in lowers I, and he speeds up.
What the marker wanted
- Achieved One of: angular momentum is conserved; his rotational inertia goes up with his arms out; or a bigger I means a smaller ω.
- Merit I linked to where the mass is (I ∝ mr², or a bigger r; the formula itself is not needed), and then a bigger I giving a smaller ω, through L = Iω and the conservation of angular momentum.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.