Question

A disc has mass \( M \) and radius \( R \). How much tangential force should be applied to the rim of the disc so as to rotate with angular velocity \( \omega \) in time \( t \)?

• A. \( \frac{MR^2 \omega}{2t} \)
• B. \( \frac{MR^2 \omega}{t} \)
• C. \( \frac{MR \omega}{2t} \)
• D. \( \frac{MR \omega}{t} \)

Hint:

The force needed to rotate an object can be calculated using the moment of inertia, angular acceleration, and the relationship between torque and force.

Answer 1

The Correct Option is C

Step 1: Understanding the problem.
The torque required to rotate the disc is related to the moment of inertia \( I \) and the angular acceleration \( \alpha \). The moment of inertia of a solid disc is given by: \[ I = \frac{1}{2} MR^2 \] The angular acceleration \( \alpha \) is related to the angular velocity \( \omega \) and the time \( t \) by the equation: \[ \alpha = \frac{\omega}{t} \] The tangential force \( F \) applied at the rim of the disc is related to the torque \( \tau \) by: \[ \tau = F R = I \alpha \] Substituting \( I = \frac{1}{2} MR^2 \) and \( \alpha = \frac{\omega}{t} \), we get: \[ F R = \frac{1}{2} MR^2 \cdot \frac{\omega}{t} \] Solving for \( F \), we find: \[ F = \frac{MR \omega}{2t} \] Step 2: Conclusion.
The tangential force required to rotate the disc is \( \frac{MR \omega}{2t} \). Therefore, the correct answer is option (C).