A `70kg` man stands in contact against the inner wall of a hollw cylindrical drum of radius `3m` rotating about its verticle axis. The coefficient of friction between the wall and his clothing nis `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?

Explanation

To solve the problem, we need to determine the minimum rotational speed (angular velocity) of the cylindrical drum that allows the man to remain stuck against the wall due to the forces acting on him when the floor is suddenly removed.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man**:
- The weight of the man (W) acting downwards: \( W = mg \)
- The normal force (N) acting towards the center of the drum.
- The frictional force (F) acting upwards, which prevents the man from falling: \( F = \mu N \)

2. **Write the Equations for Forces**:
- The weight of the man must be balanced by the frictional force when the floor is removed:
\[ F = mg \]
- The frictional force can also be expressed in terms of the normal force:
\[ F = \mu N \]

3. **Relate Normal Force to Centripetal Force**:
- The normal force provides the necessary centripetal force for the circular motion of the man:
\[ N = m \frac{v^2}{R} \]
- Here, \( v \) is the tangential speed of the man and \( R \) is the radius of the drum.

4. **Express Tangential Speed in Terms of Angular Velocity**:
- The tangential speed \( v \) can be related to angular velocity \( \omega \) by the equation:
\[ v = \omega R \]
- Substituting this into the centripetal force equation gives:
\[ N = m \frac{(\omega R)^2}{R} = m \omega^2 R \]

5. **Substitute Normal Force into the Friction Equation**:
- From the friction equation:
\[ mg = \mu N \]
- Substitute \( N \):
\[ mg = \mu (m \omega^2 R) \]

6. **Simplify the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g = \mu \omega^2 R \]
- Rearranging gives:
\[ \omega^2 = \frac{g}{\mu R} \]
- Taking the square root:
\[ \omega = \sqrt{\frac{g}{\mu R}} \]

7. **Substitute the Given Values**:
- Given:
- \( m = 70 \, \text{kg} \) (not needed for the final calculation)
- \( R = 3 \, \text{m} \)
- \( \mu = 0.15 \)
- \( g = 10 \, \text{m/s}^2 \)
- Substitute these values into the equation:
\[ \omega = \sqrt{\frac{10}{0.15 \times 3}} \]
- Calculate:
\[ \omega = \sqrt{\frac{10}{0.45}} = \sqrt{22.22} \approx 4.72 \, \text{rad/s} \]

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