A 70 kg man stands in contact against the wall of a cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. 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 ?

Explanation

To solve the problem, we need to determine the minimum rotational speed of the cylindrical drum that will allow the man to remain stuck to the wall when the floor is suddenly removed.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man:**
When the floor is removed, the only force that can prevent the man from falling is the frictional force between his clothing and the wall of the drum. The gravitational force acting on him is \( F_g = mg \), where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity.

2. **Frictional Force:**
The frictional force can be expressed as:
\[ F_f = \mu N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force acting on the man.

3. **Normal Force in Circular Motion:**
In circular motion, the normal force \( N \) is provided by the centripetal force, which is given by:
\[ N = m \omega^2 r \]
where \( \omega \) is the angular velocity in radians per second and \( r \) is the radius of the drum.

4. **Setting Up the Equation:**
To prevent the man from falling, the frictional force must be equal to or greater than the gravitational force:
\[ F_f \geq F_g \implies \mu N \geq mg \]
Substituting for \( N \):
\[ \mu (m \omega^2 r) \geq mg \]

5. **Canceling Mass:**
Since \( m \) appears on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)):
\[ \mu \omega^2 r \geq g \]

6. **Solving for Angular Velocity \( \omega \):**
Rearranging the inequality gives:
\[ \omega^2 \geq \frac{g}{\mu r} \]
Taking the square root:
\[ \omega \geq \sqrt{\frac{g}{\mu r}} \]

7. **Substituting Values:**
Given:
- \( m = 70 \, \text{kg} \) (not needed for calculation)
- \( r = 3 \, \text{m} \)
- \( \mu = 0.15 \)
- \( g = 10 \, \text{m/s}^2 \)

We substitute these values into the equation:
\[ \omega \geq \sqrt{\frac{10}{0.15 \times 3}} \]

8. **Calculating the Right Side:**
\[ \omega \geq \sqrt{\frac{10}{0.45}} = \sqrt{\frac{1000}{45}} = \sqrt{22.22} \approx 4.72 \, \text{rad/s} \]

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

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