Question

A hollow cylinder of radius 1m is rotating with angular velocity \( \omega = 10 \) rad/sec. Find minimum coefficient of friction \( \mu \) so that the block remains at rest w.r.t. the cylinder.

• A. 0.1
• B. 0.2
• C. 0.15
• D. 0.25

Hint:

In an Atwood machine, the heavier mass causes the system to accelerate in its direction. The center of mass moves accordingly, and the displacement can be calculated using kinematic equations.

Answer 1

The Correct Option is A

Step 1: Understanding the system.
The hollow cylinder is rotating with angular velocity \( \omega \). To prevent the block from slipping, friction must counteract the centrifugal force acting on the block due to the rotation of the cylinder. The frictional force provides the centripetal force needed for the block to remain at rest relative to the cylinder.

Step 2: Frictional force and centrifugal force.
The centrifugal force acting on the block is given by:
\[ F_{\text{centrifugal}} = m \cdot r \cdot \omega^2 \] Where:
- \( m \) is the mass of the block,
- \( r = 1 \, \text{m} \) is the radius of the cylinder,
- \( \omega = 10 \, \text{rad/sec} \) is the angular velocity.
The frictional force required to keep the block at rest is given by:
\[ F_{\text{friction}} = \mu \cdot m \cdot g \] Where:
- \( \mu \) is the coefficient of friction,
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Step 3: Equating forces.
For the block to remain at rest, the frictional force must balance the centrifugal force:
\[ \mu \cdot m \cdot g = m \cdot r \cdot \omega^2 \] Canceling \( m \) from both sides:
\[ \mu \cdot g = r \cdot \omega^2 \] Substituting the known values \( r = 1 \, \text{m} \), \( \omega = 10 \, \text{rad/sec} \), and \( g = 9.8 \, \text{m/s}^2 \):
\[ \mu \cdot 9.8 = 1 \cdot (10)^2 \] \[ \mu \cdot 9.8 = 100 \] Solving for \( \mu \):
\[ \mu = \frac{100}{9.8} \approx 0.1 \]
Step 4: Conclusion.
Therefore, the minimum coefficient of friction required for the block to remain at rest with respect to the cylinder is \( \mu \approx 0.1 \).
Final Answer: 0.1