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Date	May 2017	Marks available	1	Reference code	17M.3.SL.TZ1.5
Level	Standard level	Paper	Paper 3	Time zone	1
Command term	Hence and Show that	Question number	5	Adapted from	N/A

Question

A horizontal rigid bar of length 2R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

A torque of 0.010 N m brings the system to rest after a number of revolutions. For this case R = 0.50 m, M = 0.70 kg and v = 2.1 m s^–1.

Write down an expression, in terms of M, v and R, for the angular momentum of the system about the vertical axis just before the collision.

[1]

a.i.

Just after the collision the system begins to rotate about the vertical axis with angular velocity ω. Show that the angular momentum of the system is equal to $\frac{4}{3}M{R^2}\omega$ .

[1]

a.ii.

Hence, show that $\omega = \frac{v}{{4R}}$ .

[1]

a.iii.

Determine in terms of M and v the energy lost during the collision.

[3]

a.iv.

Show that the angular deceleration of the system is 0.043 rad $\,$ s^–2.

[1]

b.i.

Calculate the number of revolutions made by the system before it comes to rest.

[3]

b.ii.

Markscheme

$\frac{M}{3}vR$

[1 mark]

a.i.

evidence of use of: $L = I\omega = \left( {M{R^2} + \frac{M}{3}{R^2}} \right)\omega$

[1 mark]

a.ii.

evidence of use of conservation of angular momentum, $\frac{{MvR}}{3} = \frac{4}{3}M{R^2}\omega$

«rearranging to get $\omega = \frac{v}{{4R}}$ »

[1 mark]

a.iii.

initial KE = $\frac{{M{v^2}}}{6}$

final KE = $\frac{{M{v^2}}}{{24}}$

energy loss = $\frac{{M{v^2}}}{8}$

[3 marks]

a.iv.

$\alpha$ «= $\frac{3}{4}\frac{\Gamma }{{M{R^2}}}$ » = $\frac{3}{4}\frac{{0.01}}{{0.7 \times {{0.5}^2}}}$

«to give $\alpha$ = 0.04286 rad $\,$ s⁻²»

Working OR answer to at least 3 SF must be shown

[1 mark]

b.i.

$\theta = \frac{{\omega _i^2}}{{2\alpha }}$ «from $\omega _f^2 = \omega _i^2 + 2\alpha \theta$ »

$\theta$ « $= \frac{{{v^2}}}{{32{R^2}\alpha }} = \frac{{{{2.1}^2}}}{{32 \times {{0.5}^2} \times 0.043}}$ » = 12.8 OR 12.9 «rad»

number of rotations «= $\frac{{12.9}}{{2\pi }}$ » = 2.0 revolutions

[3 marks]

b.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

a.iv.

[N/A]

b.i.

[N/A]

b.ii.

Syllabus sections

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

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