Question

A disc starts to rotate from rest and after completing 2 revolutions, it angular velocity is 6.88 rad s-1. The moment of inertia of the disc about its axis of rotation is 1.24 kg m2

(a) The angular acceleration of the disc. Does the disc accelerate or decelerate?
(b) torque on the disc.​
A) Angular acceleration is 6.88 rad/s², and the disc accelerates.
B) Angular acceleration is -6.88 rad/s², and the disc decelerates.
C) Angular acceleration is 13.76 rad/s², and the disc accelerates.
D) Angular acceleration is -13.76 rad/s², and the disc decelerates.

Answer 1

The angular acceleration of the disc is 1.09 rad/s² and it accelerates. The torque on the disc is approximately 1.35 Nm.

Step-by-step explanation:

To find the angular acceleration of the disc, we can use the formula: angular acceleration = change in angular velocity/time. The change in angular velocity is given as 6.88 rad/s and the time is the number of revolutions completed, which is 2. Since each revolution is equivalent to 2π radians, the change in angular velocity is 2 x 2π = 4π rad/s. Therefore, the angular acceleration is 6.88 rad/s / 4π rad/s = 1.09 rad/s². Since the angular acceleration is positive, the disc accelerates. To find the torque on the disc, we can use the formula: torque = moment of inertia x angular acceleration. Substituting the given values, we have torque = 1.24 kg m² x 1.09 rad/s² ≈ 1.35 Nm.