A `70kg` man stands in contact against the inner wall of a hollow cylindrical drum of radius `3m` rotating about its verticle axis. The coefficient of friction between the wall and his clothing is `0.15` . What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Options

A. (a)4.7 rad/s
B. (b)9.8 rad/s
C. (c)24 rad/s
D. (d)1 rad/s

Explanation

To solve the problem, we need to find the minimum rotational speed of the cylindrical drum so that the man does not fall when the floor is suddenly removed. We will use the concepts of circular motion and friction.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man:**
- The man experiences two main forces:
- The gravitational force acting downwards: \( F_g = mg \)
- The normal force acting perpendicular to the wall of the cylinder: \( F_n \)

2. **Frictional Force:**
- The frictional force that prevents the man from sliding down is given by:
\[ F_f = \mu F_n \]
- Where \( \mu \) is the coefficient of friction.

3. **Centripetal Force Requirement:**
- For the man to remain in circular motion against the wall, the normal force must provide the necessary centripetal force:
\[ F_n = m \frac{v^2}{r} \]
- Where \( v \) is the tangential speed and \( r \) is the radius of the cylinder.

4. **Relate Tangential Speed to Angular Velocity:**
- The relationship between tangential speed \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \]
- Substituting this into the centripetal force equation gives:
\[ F_n = m \frac{(r \omega)^2}{r} = m r \omega^2 \]

5. **Setting Up the Equations:**
- The frictional force must balance the gravitational force for the man to not fall:
\[ F_f = F_g \]
- Therefore:
\[ \mu F_n = mg \]
- Substituting \( F_n \):
\[ \mu (m r \omega^2) = mg \]

6. **Solving for Angular Velocity \( \omega \):**
- We can simplify the equation:
\[ \mu m r \omega^2 = mg \]
- Dividing both sides by \( m \):
\[ \mu r \omega^2 = g \]
- Rearranging gives:
\[ \omega^2 = \frac{g}{\mu r} \]
- Taking the square root:
\[ \omega = \sqrt{\frac{g}{\mu r}} \]

7. **Substituting Known Values:**
- Given:
- \( g = 10 \, \text{m/s}^2 \) (approximating)
- \( \mu = 0.15 \)
- \( r = 3 \, \text{m} \)
- Substitute these values into the equation:
\[ \omega = \sqrt{\frac{10}{0.15 \times 3}} = \sqrt{\frac{10}{0.45}} = \sqrt{22.22} \approx 4.72 \, \text{rad/s} \]

### Final Answer:
The minimum rotational speed of the cylinder is approximately \( \omega \approx 4.72 \, \text{rad/s} \).

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