A block of mass $10 ext{ kg}$ is in contact with the inner wall of a hollow cylindrical drum of radius $1

Question

A block of mass $10 	ext{ kg}$ is in contact with the inner wall of a hollow cylindrical drum of radius $1 	ext{ m}$. The coefficient of friction between the block and the inner wall of the cylinder is $0.1$. The minimum angular velocity needed for the cylinder, which is vertical and rotating about its axis, will be: ($g=10 	ext{ m/s}^2$)

Accepted Answer

1. **System Analysis:** The block is rotating with the cylinder. The normal reaction ($N$) from the wall provides the necessary centripetal force. The friction force ($f$) acts vertically upwards to balance the weight ($mg$) of the block and prevent it from sliding down.
2. **Equations of Motion:**
   - Horizontal (Centripetal) Force: $N = m \omega^2 R$
   - Vertical Equilibrium: $f = mg$
3. **Condition for No Sliding:** The static friction must be sufficient to hold the weight. The limiting friction is $f_{max} = \mu N$. For the block to remain stationary relative to the wall, required friction $f \le f_{max}$.
   $$ mg \le \mu N $$
   Substitute $N = m \omega^2 R$:
   $$ mg \le \mu (m \omega^2 R) $$
   $$ g \le \mu \omega^2 R $$
4. **Minimum Angular Velocity:**
   $$ \omega^2 \ge \frac{g}{\mu R} $$
   $$ \omega_{min} = \sqrt{\frac{g}{\mu R}} $$
5. **Calculation:** Given $g = 10 	ext{ m/s}^2$, $\mu = 0.1$, $R = 1 	ext{ m}$.
   $$ \omega_{min} = \sqrt{\frac{10}{0.1 	imes 1}} = \sqrt{100} = 10 	ext{ rad/s} $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Section 5.10 Circular Motion and Section 5.9 Friction).

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