Question

A circular disc of diameter 0.8 m and mass 4 kg is rolling on a smooth horizontal plane. If 2.56 N torque is acting on the disc, then its angular acceleration is:

• A. 8 rad s\(^{-2}\)
• B. 4 rad s\(^{-2}\)
• C. 2 rad s\(^{-2}\)
• D. 16 rad s\(^{-2}\)

Hint:

The moment of inertia of a solid disc is \(\frac{1}{2} m r^2\), and angular acceleration is calculated using \(\alpha = \frac{\tau}{I}\).

Answer 1

The Correct Option is A

The angular acceleration is related to the torque and moment of inertia by: \[ \tau = I \alpha, \] where \(\tau = 2.56 \, \text{N.m}\) is the torque and \(I = \frac{1}{2} m r^2\) is the moment of inertia for a solid disc. Using \(m = 4 \, \text{kg}\) and \(r = 0.4 \, \text{m}\), we calculate the moment of inertia: \[ I = \frac{1}{2} \times 4 \times (0.4)^2 = 0.32 \, \text{kg.m}^2. \] Then, the angular acceleration \(\alpha\) is: \[ \alpha = \frac{\tau}{I} = \frac{2.56}{0.32} = 8 \, \text{rad/s}^2. \]