Question

A disk of 10 cm radius has a moment of inertia of 0.02 kg.m\(^2\). A force of 15 N is applied tangentially to the periphery of the disk to give it an angular acceleration \( \alpha \) of magnitude:

• A. 25 rad/s\(^2\)
• B. 35 rad/s\(^2\)
• C. 45 rad/s\(^2\)
• D. 75 rad/s\(^2\)

Hint:

The torque is the product of the force applied tangentially to the radius of the disk. The resulting angular acceleration depends on the moment of inertia.

Answer 1

The Correct Option is D

The torque \( \tau \) produced by the force is given by: \[ \tau = F \times r \] where \( r = 0.1 \, \text{m} \) (radius of the disk) and \( F = 15 \, \text{N} \). So, \[ \tau = 15 \times 0.1 = 1.5 \, \text{N m} \] Now, the angular acceleration \( \alpha \) can be found using the relation: \[ \tau = I \alpha \] where \( I = 0.02 \, \text{kg m}^2 \). Solving for \( \alpha \): \[ \alpha = \frac{\tau}{I} = \frac{1.5}{0.02} = 75 \, \text{rad/s}^2 \]