Question

A block of mass \( m \) is at rest w.r.t. a hollow cylinder which is rotating with angular speed \( \omega \), radius of the cylinder is \( R \). Find the minimum coefficient of friction between the block and the cylinder.

• A. \( \frac{3g}{2 \omega^2 R} \)
• B. \( \frac{g}{\omega^2 R} \)
• C. \( \frac{4g}{\omega^2 R} \)
• D. \( \frac{2g}{\omega^2 R} \)

Hint:

To prevent slipping in rotating systems, the frictional force must counterbalance the centrifugal force. The relationship between the coefficient of friction, angular speed, and radius can be derived using this balance.

Answer 1

The Correct Option is B

To find the minimum coefficient of friction between the block and the cylinder, we need to consider the forces acting on the block.

1. The block is at rest relative to the cylinder, which means it experiences an inward centripetal force. This force is provided by the friction between the block and the inner wall of the cylinder.
2. The centripetal force required to keep the block moving in a circle of radius \(R\) with angular speed \(\omega\) is given by: \(F_{\text{centripetal}} = m \omega^2 R\)
3. The normal force acting on the block is the gravitational force, \(mg\).
4. The frictional force \((f_{\text{friction}})\) needs to be equal to the centripetal force for the block to remain at rest relative to the cylinder: \(f_{\text{friction}} = \mu mg = m \omega^2 R\)
5. Solving for the coefficient of friction \(\mu\), we have: \(\mu = \frac{m \omega^2 R}{mg}\)
6. By simplifying, we get: \(\mu = \frac{\omega^2 R}{g}\)

Therefore, the minimum coefficient of friction required to keep the block at rest with respect to the rotating cylinder is:

Answer: \(\frac{g}{\omega^2 R}\)