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 .
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 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.
Identify the physical quantity represented by the area under the graph.
Show that the angular velocity of the flywheel at t = 5.00 s is 200 rad s–1.
Calculate the maximum tension in the string.
The flywheel is in translational equilibrium. Distinguish between translational equilibrium and rotational equilibrium.
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.
Markscheme
zero ✔
«change in» angular momentum ✔
NOTE: Allow angular impulse.
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 »
✔
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 ✔
ALTERNATIVE 1
✔
✔
torque = ✔
ALTERNATIVE 2
change in kinetic energy ✔
identifies work done = change in KE ✔
torque = ✔