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.
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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.
Step 1: Understanding the system.
A block is placed against the inner wall of a rotating hollow cylinder with angular velocity \( \omega \). Due to rotation, the block experiences an outward force. The friction between the block and the wall prevents it from sliding downward. Therefore, the frictional force must be sufficient to balance the weight of the block.
Step 2: Forces acting on the block.
The normal reaction provided by the wall supplies the necessary centripetal force required for circular motion. This force is given by:
\[
N = m \cdot r \cdot \omega^2
\]
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 maximum frictional force acting upward is:
\[
F_{\text{friction}} = \mu \cdot N
\]
This frictional force must balance the weight of the block:
\[
\mu \cdot N = m \cdot g
\]
Step 3: Equating the forces.
Substituting the value of the normal force:
\[
\mu \cdot (m \cdot r \cdot \omega^2) = m \cdot g
\]
Canceling \( m \) from both sides:
\[
\mu \cdot r \cdot \omega^2 = g
\]
Solving for \( \mu \):
\[
\mu = \frac{g}{r \cdot \omega^2}
\]
Substituting the known values \( r = 1 \, \text{m} \), \( \omega = 10 \, \text{rad/sec} \), and \( g = 9.8 \, \text{m/s}^2 \):
\[
\mu = \frac{9.8}{1 \cdot (10)^2}
\]
\[
\mu = \frac{9.8}{100}
\]
\[
\mu \approx 0.098 \approx 0.1
\]
Step 4: Conclusion.
Hence, the minimum coefficient of friction required to prevent the block from slipping downward relative to the rotating cylinder is approximately \( 0.1 \).
Final Answer: \( 0.1 \)
