Question

Starting from rest, a 12.32-cm-diameter compact disk takes 7.9 s to reach its operating angular velocity of 2298 rpm. Assume that the angular acceleration is constant. The disk's moment of inertia is 1.00×10-5 kg·m2. How much torque is applied to the disk?

Answer 1

Answer:

T = 0.00030462 Nm

Explanation:

Given:-

- The initial angular velocity, wi = 0 rad/s  (rest)

- The final angular velocity, wf = 2298 rpm = 240.646 rad/s

- The time taken to reach wf, t = 7.9 s

- The moment of inertia of disc, I = 1.00 *10^-5 kg.m^2

Find:-

How much torque is applied to the disk?

Solution:-

- We will first determine the constant angular acceleration (α) of the disc when it starts from rest and reaches the final angular velocity in time t = 7.9 s.

- We will use the first rotational kinematic equation of motion as follows:

                                         wf = wi + α*t

                                         α = ( wf - wi ) / t

                                         α = ( 240.646 - 0 ) / 7.9

                                         α = 30.462 rad/s^2

- The amount of torque T required to accelerate (α) the disc uniformly in time t = 7.9s is given by the relationship as follows:

                                         T = I*α

                                         T = ( 10^-5 ) * ( 30.462 )

                                         T = 0.00030462 Nm