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

Concept: The block is at rest relative to the rotating hollow cylinder. Hence, the block is in a non-inertial (rotating) frame. For equilibrium of the block:

Friction balances the weight of the block.
Normal reaction provides the required centripetal force.

Step 1: Forces acting on the block Forces acting on the block are:

Weight \(mg\) acting vertically downward
Normal reaction \(N\) acting radially inward
Frictional force \(f\) acting upward (to prevent slipping)

Step 2: Radial force balance (centripetal force) Since the block is rotating with angular speed \( \omega \) at radius \(R\), it requires centripetal force: \[ F_c = m\omega^{2}R \] This centripetal force is provided by the normal reaction: \[ N = m\omega^{2}R \]
Step 3: Vertical force balance For the block to remain at rest vertically: \[ f = mg \] The maximum frictional force is: \[ f_{\max} = \mu N \] For limiting equilibrium: \[ \mu N = mg \]
Step 4: Substitute value of normal reaction \[ \mu (m\omega^{2}R) = mg \] \[ \mu = \frac{g}{\omega^{2}R} \] Final Answer: \[ \boxed{\mu_{\min} = \frac{g}{\omega^{2}R}} \]