A block of mass m is at rest w.r.t. hollow cylinder which is rotating with angular speed . Radius of cylinder is R. Find minimum coefficient of friction between block and cylinder.

Correct Option: g/ω²R

Step-by-Step Explanation:

As seen in the provided diagram, a block of mass $m$ is pressed against the inner vertical wall of a hollow cylinder rotating with angular speed $\omega$ about its central axis. The radius of the cylinder is $R$.

1. Analyzing the Forces acting on the block:

• Vertical direction: The downward gravitational force acting on the block is $mg$. To prevent the block from sliding down, an upward static frictional force $f_s$ acts between the inner wall of the cylinder and the block.

• Horizontal/Radial direction: The normal force $N$ exerted horizontally inward by the wall of the cylinder provides the necessary centripetal acceleration for circular motion of radius $R$.

2. Setting up the Equations of Motion:

For horizontal circular motion with angular speed $\omega$:

$N=m{\omega }^{2}R$

For vertical equilibrium (so the block does not slip down):

$f_s=mg$

3. Condition for Static Friction:

The maximum available static friction force (limiting friction) is given by:

$f_{s,\text{max}}=\mu N$

For the block to remain at rest relative to the cylinder, the required static friction force must not exceed the maximum static friction force:

$f_s\le f_{s,\text{max}}$

$mg\le \mu N$

Substituting $N=m{\omega }^{2}R$ into the inequality:

$mg\le \mu \left(m{\omega }^{2}R\right)$

Canceling mass $m$ from both sides:

$g\le \mu {\omega }^{2}R$

Solving for the coefficient of friction $\mu$:

$\mu \ge \frac{g}{{\omega }^{2}R}$

Thus, the minimum coefficient of friction required between the block and cylinder is:

${\mu }_{\text{min}}=\frac{g}{{\omega }^{2}R}$