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 applied force's work translates to the disc's rotational kinetic energy. The torque \( \tau \) on the disc is \( \tau = F \cdot R \), where \( F = 4 \, \text{N} \) (force) and \( R = 0.5 \, \text{m} \) (disc radius). The angular acceleration \( \alpha \) is determined by \( \alpha = \frac{\tau}{I} \), with \( I \) being the disc's moment of inertia. For a solid disc, \( I = \frac{1}{2} m R^2 \). Substituting values yields \( I = \frac{1}{2} \times 2 \times 0.5^2 = 0.5 \, \text{kg m}^2 \). The torque calculation is \( \tau = 4 \times 0.5 = 2 \, \text{N m} \). Therefore, the angular acceleration is \( \alpha = \frac{2}{0.5} = 4 \, \text{rad/s}^2 \). After 2 seconds, the angular velocity \( \omega \) is \( \omega = \alpha t = 4 \times 2 = 8 \, \text{rad/s} \). Consequently, the disc attains an angular velocity of 16 rad/s after 2 seconds.