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 :
\((g = 10\ m/s^2)\)

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

Answer 1

The Correct Option is A

To solve the problem, we need to find the minimum angular velocity required for the cylinder to keep the block stationary against its inner wall. The block experiences three main forces:

1. \( F_{gravity} = mg \): the gravitational force pulling the block downward, where \( m = 10 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \). Hence, \( F_{gravity} = 100 \, \text{N} \).
2. Normal force \( N \): the force exerted by the wall perpendicular to the block, acting radially outward.
3. Frictional force \( F_{friction} = \mu N \): acts upward to balance the gravitational force, where \( \mu = 0.1 \).

For the block to remain stationary, the frictional force must at least equal the gravitational force:

\[ \mu N = mg \]

Substitute the values given:

\[ 0.1 N = 100 \, \text{N} \]

Thus, the normal force \( N \) is:

\[ N = \frac{100}{0.1} = 1000 \, \text{N} \]

The normal force also provides the centripetal force required to keep the block in circular motion:

\[ N = m \omega^2 r \]

Substitute the values to find the angular velocity \(\omega\):

\[ 1000 = 10 \cdot \omega^2 \cdot 1 \]

Simplifying, we get:

\[ \omega^2 = 100 \] \[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

Therefore, the minimum angular velocity needed for the cylinder to keep the block stationary is 10 rad/s. This makes option 10 rad/s the correct answer.