Question

A 17 g audio compact disk has a diameter of 12 cm. The disk spins under a laser that reads encoded data. The first track to be read is 2.3 cm from the axis; as the disk plays, the laser scans tracks farther and farther from the center. The part of the disk directly under the read head moves at a constant 1.2 m/s. When a disk is inserted, it takes 2.4 s to spin up from rest.

1. What is the torque of the motor?

Answer 1

0.00066518 Nm

Step-by-step explanation:

v = Velocity = 1.2 m/s

r = Distance to head = 2.3 cm

`\omega_f` = Final angular velocity

`\omega_i` = Initial angular velocity = 0

`\alpha` = Angular acceleration

t = Time taken = 2.4 s

Angular speed is given by

`\omega=(v)/(r)\\\Rightarrow \omega=(1.2)/(0.023)\\\Rightarrow \omega=52.17391\ rad/s`

From equation of rotational motion

`\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=(\omega_f-\omega_i)/(t)\\\Rightarrow \alpha=(52.17391-0)/(2.4)\\\Rightarrow \alpha=21.73912\ rad/s^2`

Torque

`\tau=I\alpha\\\Rightarrow \tau=(1)/(2)mR^2\alpha\\\Rightarrow \tau=(1)/(2)0.017* 0.06^2* 21.73912\\\Rightarrow \tau=0.00066518\ Nm`

The torque of the motor is 0.00066518 Nm