AS 91524 · 2025 paper · rotational motion

Question 2(a) Achieved

Rīhari and Jane are on a playground roundabout of radius 1.30 m. Rīhari stands on it while Jane pushes; from rest she steadily accelerates it so it takes 12.0 s to make 4.00 rotations.

r	1.30 m
4.00 rotations in 12.0 s
I (roundabout + Rīhari)	430 kg m²

Show that the roundabout's angular velocity after the 12.0 s is 4.19 rad s⁻¹.

The 2025 paper, page 5 ↗ Marking schedule, page 3 ↗

Work it out

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

1 Turn 4.00 rotations into an angle in radians.
θ = 4.00 × 2π = 25.133 rad.
2 It starts from rest and speeds up steadily. Which equation links θ, t and α when ω₀ = 0?
θ = ω₀t + ½αt² with ω₀ = 0: 25.133 = ½ × α × 12.0², so α = 0.349 07 rad s⁻².
3 Now the angular velocity at 12.0 s.
ω = ω₀ + αt = 0 + 0.349 07 × 12.0 = 4.1888 rad s⁻¹. A shortcut gives the same: the average is 25.133 / 12.0 = 2.0944 rad s⁻¹, and a steady speed-up from rest ends at twice its average.

Answer ω = 4.1888 rad s⁻¹, which rounds to the 4.19 rad s⁻¹ asked for.

What the marker wanted

Achieved The equation with the values put in, or the equation and the unrounded result.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

Dividing the angle by the time: 25.133 / 12.0 = 2.09 rad s⁻¹. That is the average angular velocity, which would only be the answer if it had turned at a constant rate the whole time; this roundabout starts from rest. NZQA’s 2025 report lists constant-speed formulas used on things speeding up from rest among the common reasons for Not Achieved.

How did it go?

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