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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 Time zone 1
Command term Show that 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.

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

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.

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

[2]
a.

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]
c.

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]

a.

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

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

\({\omega _t} = 1.7\) «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]

c.

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.

ωt = 1.66 from bi AND W = ΔEk

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]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[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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