Question

A large drum having radius $R$ is spinning around its axis with angular velocity $\omega$, as shown in figure. The minimum value of $\omega$ so that a body of mass $M$ remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass $M$ as $\mu$, is :

• A. $\sqrt{\frac{\mu g}{R}}$
• B. $\sqrt{\frac{g}{\mu R}}$
• C. $\sqrt{\frac{2g}{\mu R}}$
• D. $\sqrt{\frac{g}{2\mu R}}$

Hint:

In "Rotor" or "Drum" problems, the normal force is the centripetal force ($m\omega^2R$). The friction force required to prevent sliding is simply $mg$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For the mass to remain stuck to the vertical wall, the upward frictional force must balance the weight of the mass. The normal force required for friction is provided by the centripetal acceleration due to the drum's rotation.

Step 2: Key Formula or Approach:
1. Weight $W = Mg$ (downwards).
2. Normal force $N = M \omega^2 R$ (towards center).
3. Static friction $f_s \le \mu N$.

Step 3: Detailed Explanation:
For vertical equilibrium:
\[ f_s = Mg \]
Since $f_s \le \mu N$, we have:
\[ Mg \le \mu (M \omega^2 R) \]
Cancelling $M$ from both sides:
\[ g \le \mu \omega^2 R \]
\[ \omega^2 \ge \frac{g}{\mu R} \]
The minimum angular velocity $\omega_{min}$ is:
\[ \omega_{min} = \sqrt{\frac{g}{\mu R}} \]

Step 4: Final Answer:
The minimum angular velocity is $\sqrt{\frac{g}{\mu R}}$.