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: Understanding the Concept:
A tangential force at the rim gives a torque \(\tau = FR\), which produces angular acceleration \(\alpha\) in a body of moment of inertia \(I\).

Step 2: Key Formula or Approach:
\[ FR = I\alpha, \qquad I_{disc} = \frac{1}{2}MR^2 \]

Step 3: Detailed Explanation:
\[ F = \frac{I\alpha}{R} = \frac{\frac{1}{2}MR^2\alpha}{R} = \frac{1}{2}MR\alpha \]
\[ F = \frac{1}{2}\times 1\times 0.4\times 10 = 2\text{ N} \]

Step 4: Check the options.
1 N, 3 N and 4 N do not fit \(\frac{1}{2}MR\alpha\). The common mistake is to use \(MR\alpha = 4\) N without the factor half.

Final Answer:
The tangential force is 2 N, option (B). \[ \boxed{2\text{ N}} \]