Question

A uniform circular disc of mass 2 kg and radius 0.5 m is mounted on a frictionless axle. A force of 4 N is applied tangentially at the rim for 2 seconds. Find the angular velocity acquired by the disc at the end of 2 seconds.

• A. 8 rad/s
• B. 10 rad/s
• C. 12 rad/s
• D. 16 rad/s

Hint:

The angular velocity acquired by a disc is proportional to the angular acceleration and the time. Make sure to use the correct formula for the moment of inertia of a disc.

Answer 1

The Correct Option is D

The torque \( \tau \) acting on the disc is given by: \[ \tau = F \times r \] Where: - \( F = 4 \, \text{N} \) (force applied), - \( r = 0.5 \, \text{m} \) (radius of the disc). Thus: \[ \tau = 4 \times 0.5 = 2 \, \text{N.m} \] Now, the angular acceleration \( \alpha \) is given by the relation: \[ \alpha = \frac{\tau}{I} \] Where \( I \) is the moment of inertia of a solid disc: \[ I = \frac{1}{2} M R^2 \] For this disc: - \( M = 2 \, \text{kg} \) (mass of the disc), - \( R = 0.5 \, \text{m} \) (radius of the disc). So, the moment of inertia is: \[ I = \frac{1}{2} \times 2 \times (0.5)^2 = 0.25 \, \text{kg.m}^2 \] Now, the angular acceleration is: \[ \alpha = \frac{2}{0.25} = 8 \, \text{rad/s}^2 \] Finally, the angular velocity \( \omega \) at the end of 2 seconds is given by: \[ \omega = \alpha \times t \] Substituting the values: \[ \omega = 8 \times 2 = 16 \, \text{rad/s} \] Thus, the angular velocity acquired by the disc is 16 rad/s.