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:

In rotational motion, the angular velocity is given by \( \omega = \alpha t \), where \( \alpha \) is the angular acceleration and \( t \) is the time.

Answer 1

The Correct Option is D

The work done by the applied force is converted into rotational kinetic energy of the disc. The torque \( \tau \) acting on the disc is given by: \[ \tau = F \cdot R \] Where: - \( F = 4 \, \text{N} \) is the force, - \( R = 0.5 \, \text{m} \) is the radius of the disc. The angular acceleration \( \alpha \) is given by: \[ \alpha = \frac{\tau}{I} \] Where \( I \) is the moment of inertia of the disc, and for a solid disc: \[ I = \frac{1}{2} m R^2 \] Substituting values: \[ I = \frac{1}{2} \times 2 \times 0.5^2 = 0.5 \, \text{kg m}^2 \] Now, calculate the torque: \[ \tau = 4 \times 0.5 = 2 \, \text{N m} \] Thus, the angular acceleration is: \[ \alpha = \frac{2}{0.5} = 4 \, \text{rad/s}^2 \] The angular velocity \( \omega \) at the end of 2 seconds is: \[ \omega = \alpha t = 4 \times 2 = 8 \, \text{rad/s} \] Thus, the angular velocity acquired by the disc at the end of 2 seconds is 16 rad/s.