DP Physics (last assessment 2024)

Question 19N.3.SL.TZ0.5

Date	November 2019	Marks available	[Maximum mark: 10]	Reference code	19N.3.SL.TZ0.5
Level	SL	Paper	3	Time zone	TZ0
Command term	Calculate, Distinguish, Identify, Show that, State	Question number	5	Adapted from	N/A

[Maximum mark: 10]

19N.3.SL.TZ0.5

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 $I = \frac{1}{2} M R^{2}$ .

(a)

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

[1]

Markscheme

zero ✔

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.

(b(i))

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

[1]

Markscheme

«change in» angular momentum ✔

NOTE: Allow angular impulse.

(b(ii))

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

[2]

Markscheme

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(iii))

Calculate the maximum tension in the string.

[1]

Markscheme

$« \frac{0.40}{0.012} = » 33.3 N$ ✔

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

(c(i))

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

[2]

Markscheme

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(ii))

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]

Markscheme

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

Syllabus sections

Option B: Engineering physics

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

Option B: Engineering physics » Option B: Engineering physics (Core topics)

Question 19N.3.SL.TZ0.5

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