Question

A disc of radius 0.4 m and mass 1 kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration is 10 rad/s\(^2\). The tangential force applied to the rim of the disc is

• A. \(1\) N
• B. \(2\) N
• C. \(3\) N
• D. \(4\) N

Hint:

Torque equals moment of inertia times angular acceleration, and torque also equals force times radius.

Answer 1

The Correct Option is B

Step 1: Find I.
$I = \dfrac{1}{2}(1)(0.4)^2 = 0.08\text{ kg m}^2$.

Step 2: Find torque.
$\tau = I\alpha = 0.08 \times 10 = 0.8\text{ N m}$.

Step 3: Find force.
$F = \tau/R = 0.8/0.4 = 2$ N.

Final Answer:
Option (B). \[ \boxed{2\text{ N}} \]