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Date	May 2018	Marks available	2	Reference code	18M.3.SL.TZ1.6
Level	Standard level	Paper	Paper 3	Time zone	1
Command term	Explain	Question number	6	Adapted from	N/A

Question

A constant force of 50.0 N is applied tangentially to the outer edge of a merry-go-round. The following diagram shows the view from above.

M18/4/PHYSI/SP3/ENG/TZ1/06

The merry-go-round has a moment of inertia of 450 kg m² about a vertical axis. The merry-go-round has a diameter of 4.00 m.

A child of mass 30.0 kg is now placed onto the edge of the merry-go-round. No external torque acts on the system.

The child now moves towards the centre.

The merry-go-round starts from rest and the force is applied for one complete revolution.

Show that the angular acceleration of the merry-go-round is 0.2 rad s^–2.

[2]

Calculate, for the merry-go-round after one revolution, the angular speed.

[1]

b.i.

Calculate, for the merry-go-round after one revolution, the angular momentum.

[1]

b.ii.

Calculate the new angular speed of the rotating system.

[2]

Explain why the angular speed will increase.

[2]

d.i.

Calculate the work done by the child in moving from the edge to the centre.

[2]

d.ii.

Markscheme

Γ «= Fr = 50 × 2» = 100 «Nm»

α « $= \frac{\Gamma }{I} = \frac{{100}}{{450}}$ » =0.22 «rads^–2»

Final value to at least 2 sig figs, OR clear working with substitution required for mark.

[2 marks]

« $\omega _t^2 - \omega _0^2 = 2\alpha \Delta \theta$ »

« $\omega _t^2 - 0 = 2 \times 0.22 \times 2\pi$ »

${\omega _t} = 1.7$ «rads^–1»

Accept BCA, values in the range: 1.57 to 1.70.

[1 mark]

b.i.

«L = Iω = 450 × 1.66»

= 750 «kgm² rads^–1»

Accept BCA, values in the range: 710 to 780.

[1 mark]

b.ii.

«I = 450 + mr²»

I «= 450 + 30 × 2²» = 570 «kgm²»

«L = 570 × ω = 747»

ω = 1.3 «rads^–1»

Watch for ECF from (a) and (b).

Accept BCA, values in the range: 1.25 to 1.35.

[2 marks]

moment of inertia will decrease

angular momentum will be constant «as the system is isolated»

«so the angular speed will increase»

[2 marks]

d.i.

ω_t = 1.66 from bi AND W = ΔE_k

W = $\frac{1}{2}$ × 450 × 1.66² – $\frac{1}{2}$ × 570 × 1.31² = 131 «J»

ECF from 8bi

Accept BCA, value depends on the answers in previous questions.

[2 marks]

d.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Core topics) » B.1 – Rigid bodies and rotational dynamics

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