A block of mass 10 kg in contact against the inner wall of a hollow cylindrical drum of radius 1m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be `(g=10 m//s^(2))`

Options

A. `10pi rad//s`
B. `sqrt(10) rad//s`
C. `(10)/(2pi) rad//s`
D. `10 rad//s`

Explanation

To solve the problem of finding the minimum angular velocity needed for a block of mass 10 kg to remain stationary against the inner wall of a hollow cylindrical drum of radius 1 m, we can follow these steps:

### Step 1: Identify Forces Acting on the Block
The forces acting on the block are:
1. Gravitational force (downward): \( F_g = mg \)
2. Normal force (perpendicular to the wall): \( N \)
3. Frictional force (acting upward to counteract gravity): \( F_f = \mu N \)

### Step 2: Set Up the Equations
For the block to remain stationary, the frictional force must be equal to or greater than the gravitational force:
\[ F_f \geq F_g \]
This translates to:
\[ \mu N \geq mg \]

### Step 3: Relate Normal Force to Centripetal Force
When the cylinder rotates, the normal force \( N \) can be expressed in terms of the centripetal force required to keep the block moving in a circle of radius \( r \):
\[ N = m r \omega^2 \]
where \( \omega \) is the angular velocity.

### Step 4: Substitute Normal Force in the Friction Equation
Substituting \( N \) into the friction equation gives:
\[ \mu (m r \omega^2) \geq mg \]
This simplifies to:
\[ \mu r \omega^2 \geq g \]

### Step 5: Solve for Angular Velocity \( \omega \)
Rearranging the equation to solve for \( \omega \):
\[ \omega^2 \geq \frac{g}{\mu r} \]
Taking the square root of both sides:
\[ \omega \geq \sqrt{\frac{g}{\mu r}} \]

### Step 6: Substitute Known Values
Given:
- \( m = 10 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( r = 1 \, \text{m} \)
- \( \mu = 0.1 \)

Substituting these values into the equation:
\[ \omega \geq \sqrt{\frac{10}{0.1 \times 1}} \]
\[ \omega \geq \sqrt{100} \]
\[ \omega \geq 10 \, \text{rad/s} \]

### Conclusion
The minimum angular velocity needed for the block to remain stationary is \( \omega = 10 \, \text{rad/s} \).

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