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Date	November 2019	Marks available	3	Reference code	19N.3.SL.TZ0.5
Level	Standard level	Paper	Paper 3	Time zone	0 - no time zone
Command term	Calculate	Question number	5	Adapted from	N/A

Question

A flywheel is made of a solid disk with a mass M of 5.00 kg mounted on a small radial axle. The mass of the axle is negligible. The radius R of the disk is 6.00 cm and the radius r of the axle is 1.20 cm.

A string of negligible thickness is wound around the axle. The string is pulled by an electric motor that exerts a vertical tension force T on the flywheel. The diagram shows the forces acting on the flywheel. W is the weight and N is the normal reaction force from the support of the flywheel.

The moment of inertia of the flywheel about the axis is $I = \frac{1}{2} M R^{2}$ .

The flywheel is initially at rest. At time t = 0 the motor is switched on and a time-varying tension force acts on the flywheel. The torque $Γ$ exerted on the flywheel by the tension force in the string varies with t as shown on the graph.

At t = 5.00 s the string becomes fully unwound and it disconnects from the flywheel. The flywheel remains spinning around the axle.

State the torque provided by the force W about the axis of the flywheel.

[1]

Identify the physical quantity represented by the area under the graph.

[1]

b(i).

Show that the angular velocity of the flywheel at t = 5.00 s is 200 rad s^–1.

[2]

b(ii).

Calculate the maximum tension in the string.

[1]

b(iii).

The flywheel is in translational equilibrium. Distinguish between translational equilibrium and rotational equilibrium.

[2]

c(i).

At t = 5.00 s the flywheel is spinning with angular velocity 200 rad s^–1. The support bearings exert a constant frictional torque on the axle. The flywheel comes to rest after 8.00 × 10³ revolutions. Calculate the magnitude of the frictional torque exerted on the flywheel.

[3]

c(ii).

Markscheme

zero ✔

«change in» angular momentum ✔

NOTE: Allow angular impulse.

b(i).

use of L = lω = area under graph = 1.80 «kg m²s^–1» ✔

rearranges «to give ω= area/I» 1.80 = 0.5 × 5.00 × 0.060²× ω ✔

«to get ω = 200 rad s^–1»

b(ii).

$« \frac{0.40}{0.012} = » 33.3 N$ ✔

b(iii).

translational equilibrium is when the sum of all the forces on a body is zero ✔

rotational equilibrium is when the sum of all the torques on a body is zero ✔

c(i).

ALTERNATIVE 1

$0 = 200^{2} + 2 \times α \times 2 π \times 8000$ ✔

$α = « - » 0.398 « rad s^{- 2} »$ ✔

torque = $α I = 0.398 \times (0.5 \times 5.00 \times 0 . 060^{2}) = 3.58 \times 10^{- 3} « N m »$ ✔

ALTERNATIVE 2

change in kinetic energy $= « - » 0.5 \times (0.5 \times 5.00 \times 0 . 060^{2}) \times 200^{2} = « - » 180 « J »$ ✔

identifies work done = change in KE ✔

torque = $\frac{W}{θ} = \frac{180}{2 π \times 8000} = 3.58 \times 10^{- 3} « N m »$ ✔

c(ii).

Examiners report

[N/A]

b(i).

[N/A]

b(ii).

[N/A]

b(iii).

[N/A]

c(i).

[N/A]

c(ii).

Syllabus sections

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

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