Question

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

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

Hint:

For objects sticking to the wall of a rotating cylinder:

Normal reaction provides centripetal force
Friction balances weight
Always equate friction to \(mg\) for minimum coefficient

Answer 1

The Correct Option is B

To determine the minimum coefficient of friction between the block and the cylinder that keeps the block at rest relative to the rotating cylinder, we need to consider the forces acting on the block.

The block experiences two main forces: 

1. The gravitational force acting downward: \(mg\)
2. The frictional force providing the necessary centripetal force to keep the block in circular motion: \(f\)

The required centripetal force for circular motion is given by:

\(F_{\text{centripetal}} = \dfrac{mv^{2}}{R}\)

Since \(v = \omega R\), the centripetal force becomes:

\(F_{\text{centripetal}} = m\omega^{2}R\)

To keep the block from sliding, the frictional force must satisfy:

\(f = \mu N \geq m\omega^{2}R\)

Where \(\mu\) is the coefficient of friction and \(N = mg\) is the normal force.

The maximum friction that can be provided is static friction, so:

\(\mu mg \geq m\omega^{2}R\)

Canceling \(m\) from both sides gives:

\(\mu g \geq \omega^{2}R\)

Solving for \(\mu\) gives the minimum coefficient of friction:

\(\mu \geq \dfrac{\omega^{2}R}{g}\)

Therefore, the minimum coefficient of friction required is:

\(\mu = \dfrac{g}{\omega^{2}R}\)

Thus, the correct option is:

\(\dfrac{g}{\omega^{2}R}\)