Question 2(c) Merit to Excellence

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.

Jane (52.0 kg) runs at 6.39 m s⁻¹ and jumps onto the STATIONARY roundabout, landing 1.20 m from the axis. On the diagram her velocity is at right angles to that radius. The roundabout and Rīhari have I = 430 kg m²; for a point mass, I = mr².

Calculate the angular velocity of the roundabout after she has jumped on.

The 2025 paper, page 6 ↗ Marking schedule, page 4 ↗

Work it out

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

1. 1 Before she lands, the roundabout is still. So where is all the angular momentum?

   With Jane. She is moving at 6.39 m s⁻¹ at right angles to a radius 1.20 m long, so about the axis her angular momentum is L = mvr = 52.0 × 6.39 × 1.20 = 398.74 kg m² s⁻¹.

2. 2 Once she is on, what is turning, and what is its total rotational inertia?

   The roundabout with Rīhari, plus Jane as a point mass at 1.20 m: I = 430 + 52.0 × 1.20² = 430 + 74.88 = 504.88 kg m².

3. 3 No outside torque acts while she lands. What stays the same?

   The total angular momentum: L before = L after, so 398.74 = 504.88 × ω.

4. 4 Solve for ω.

   ω = 398.736 / 504.88 = 0.789 76 rad s⁻¹.

What the marker wanted

• Achieved One of: Jane’s angular momentum; her rotational inertia; ω = 5.325 rad s⁻¹, from using only Jane’s own L and I; ω = 0.927 rad s⁻¹, from using I = 430 kg m² alone; or saying that angular momentum is conserved.
• Merit A right method with one slip in it, and the inertia after she lands must include both the roundabout and Jane.
• Excellence Right working all the way to the right answer, 0.790 rad s⁻¹.

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

Common mistake

How did it go?

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