Question 1
The rotating part of an electric motor is called the rotor.
Figure 1 shows an end view of a rotor turning clockwise due to a driving torque from the motor. In this question, the clockwise direction is treated as positive.
Figure 1
Figure 2
The rotor can be brought to rest rapidly by reversing the electrical supply connections to the motor. Figure 2 shows the rotor at time t = 0 when the supply connections are reversed. The rotor then slows down due to a constant anticlockwise retarding torque so that it
The angular velocity of the rotor at t = 0 is 98.0 rad s−1 clockwise. The applied torque on the rotor at t = 0 is anticlockwise. The applied torque produces a constant angular acceleration of −303 rad s−2. Friction torque is negligible.
The electrical supply remains connected and the rotor now accelerates uniformly anticlockwise with an acceleration of magnitude 303 rad s−2. At a later time t = t2, the angular velocity of the rotor is −120 rad s−1.
Determine the number of anticlockwise revolutions made by the rotor between
[2 marks]
Another way of quickly stopping the motor is to use a disc brake. The motor is brought to rest by switching off the power supply and then forcing two brake pads against the disc.
Figure 3 shows the two brake pads each applying a constant retarding force of 182 N to the disc.
moment of inertia of rotor and disc about the axis of rotation = 0.101 kg m2 diameter of disc = 0.160 m diameter of pad = 22 mm
Compare the angular deceleration produced by the disc brake system with the angular deceleration produced by reversing the electrical connections.
[3 marks]