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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=12MR2.

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]
a.

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 ✔

a.

«change in» angular momentum ✔

NOTE: Allow angular impulse.

b(i).

use of L = lω = area under graph = 1.80 «kg m2 s–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).

«0.400.012=»33.3N 

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=2002+2×α×2π×8000 

α=«-»0.398«rads-2» 

torque = αI=0.398×0.5×5.00×0.0602=3.58×10-3«Nm» ✔

 

ALTERNATIVE 2

change in kinetic energy =«-»0.5×0.5×5.00×0.0602×2002=«-»180«J» ✔

identifies work done = change in KE 

torque = Wθ=1802π×8000=3.58×10-3«Nm»

c(ii).

Examiners report

[N/A]
a.
[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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Option B: Engineering physics » Option B: Engineering physics (Core topics)
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