A person is in contact with inner wall of a vertical hollow cylinder of radius 2m without floor under his feet, remains in equilibrium without slipping down when the cylinder is rotated about it's own vertical axis. If the coefficient of static friction between person and wall of cylinder is 0.4, the minimum angular velocity of cylinder should be

Options

A. `1.5 rad//s`
B. `2 rad//s`
C. `3 rad//s`
D. `3.5 rad//s`

Answer

To solve the problem, we need to find the minimum angular velocity (\( \omega \)) of a vertical hollow cylinder such that a person in contact with the inner wall remains in equilibrium without slipping down. The given parameters are:

- Radius of the cylinder (\( r \)) = 2 m
- Coefficient of static friction (\( \mu \)) = 0.4
- Acceleration due to gravity (\( g \)) = 9.8 m/s²

### Step-by-Step Solution:

1. **Identify Forces Acting on the Person**:
- The weight of the person (\( W \)) acts downward and is given by:
\[ W = m \cdot g \]
- The frictional force (\( F_f \)) acts upward and is given by:
\[ F_f = \mu \cdot N \]
- Here, \( N \) is the normal force exerted by the wall of the cylinder on the person.

2. **Centripetal Force Requirement**:
- For the person to remain in circular motion without slipping, the normal force must provide the necessary centripetal force. The centripetal force (\( F_c \)) is given by:
\[ F_c = m \cdot \omega^2 \cdot r \]

3. **Equilibrium Condition**:
- In equilibrium, the upward frictional force must balance the downward weight of the person:
\[ F_f = W \]
- Therefore, we can write:
\[ \mu \cdot N = m \cdot g \]

4. **Relate Normal Force to Centripetal Force**:
- The normal force can also be expressed in terms of the centripetal force:
\[ N = m \cdot \omega^2 \cdot r \]
- Substituting this into the equilibrium condition gives:
\[ \mu \cdot (m \cdot \omega^2 \cdot r) = m \cdot g \]

5. **Cancel Mass (\( m \))**:
- Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \mu \cdot \omega^2 \cdot r = g \]

6. **Solve for Angular Velocity (\( \omega \))**:
- Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{g}{\mu \cdot r} \]
- Taking the square root to find \( \omega \):
\[ \omega = \sqrt{\frac{g}{\mu \cdot r}} \]

7. **Substituting Values**:
- Substitute \( g = 9.8 \, \text{m/s}^2 \), \( \mu = 0.4 \), and \( r = 2 \, \text{m} \):
\[ \omega = \sqrt{\frac{9.8}{0.4 \cdot 2}} = \sqrt{\frac{9.8}{0.8}} = \sqrt{12.25} = 3.5 \, \text{rad/s} \]

### Final Answer:
The minimum angular velocity of the cylinder should be \( \omega = 3.5 \, \text{rad/s} \).