Question

A block of mass \(10 kg\) is in contact against the inner wall of a hollow cylindrical drum of radius \( 1 m.\) 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: (\(g=10 \frac{m}{s^2}\)

• A. \(\sqrt10\ rad/s\)
• B. \(\frac{10}{2\pi}\ rad/s\)
• C. \(10\ rad/s\)
• D. \(10\pi\ rad/s\)

Answer 1

The Correct Option is C

To solve this physics problem, we need to find the minimum angular velocity required to keep a block stationary against the inner wall of a rotating cylinder. Let's go through the steps to find the solution:

1. Understand the Forces: When the cylinder is rotating, the block experiences a centrifugal force that pushes it outward. To prevent the block from sliding down, the frictional force between the block and the cylinder wall must balance the gravitational force. The forces involved are:
   • Centrifugal Force, \( F_c = m \omega^2 r \)
   • Gravitational Force, \( F_g = mg \)
   • Frictional Force, \( F_f = \mu N = \mu m \omega^2 r \)
   where \( m = 10 \, \text{kg} \), \( r = 1 \, \text{m} \), \( \mu = 0.1 \), and \( g = 10 \, \text{m/s}^2 \).
2. Condition for Static Friction: For the block to remain stationary, the frictional force must at least equal the gravitational force:
   \[ \mu m \omega^2 r = mg \]
3. Set the Forces Equal: Substitute the known values into the equation:
   \[ 0.1 \times 10 \times \omega^2 \times 1 = 10 \times 10 \]
   \[ \omega^2 = \frac{100}{1} = 100 \]
4. Solve for Angular Velocity: Take the square root to find the minimum angular velocity:
   \[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

Therefore, the minimum angular velocity required to keep the block stationary is \( 10 \, \text{rad/s} \). This matches with the correct option: \(10\ rad/s\).

Conclusion: The correct answer is \( 10 \, \text{rad/s} \) which ensures the frictional force is sufficient to counteract gravitational pull, keeping the block in place.