AS 91524 · 2023 paper · rotational motion

Question 2(d) Merit to Excellence

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 spins with an angular velocity of 3.00 rad s⁻¹ with his arms outstretched. Bringing his arms in, he reaches 7.00 rad s⁻¹ in 4.50 s. Calculate his angular acceleration and the number of revolutions made in this time.

The 2023 paper, page 5 ↗ Marking schedule, page 4 ↗

Work it out

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

1 Which equation gives α from the starting and final angular velocities and the time?
ω = ω₀ + αt: 7.00 = 3.00 + α × 4.50, so α = 4.00 / 4.50 = 0.889 rad s⁻².
2 How far round does he turn in those 4.50 s?
ω² = ω₀² + 2αθ: 49.0 = 9.00 + 2 × 0.8889 × θ, so θ = 40.0 / 1.7778 = 22.5 rad. (Or θ = ω₀t + ½αt² = 13.5 + 9.0 = 22.5 rad.)
3 The question wants revolutions. How many radians make one?
2π. 22.5 / 2π = 3.58 revolutions.

Answer α = 0.889 rad s⁻², and 3.58 revolutions.

What the marker wanted

Achieved The right angular acceleration; or the right method for the angle with one small slip carried from α, such as the wrong change in ω.
Merit One error in the calculation, such as two of its three stages right; or both α = 0.889 rad s⁻² and θ = 22.5 rad.
Excellence The right number of revolutions, 3.58.

In our words, from NZQA’s 2023 marking schedule.

Common mistake

Stopping at 22.5 rad. The question asks for revolutions, and the schedule holds α and θ without them at Merit: divide by 2π.

How did it go?

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