Question

A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 4 m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is 5 rad/s. The coefficient of friction between the drum's inner wall surface and mass is _______. (Take \( g = 10 \, \text{m/s}^2 \))}

• A. 0.1
• B. 0.5
• C. 0.7
• D. 0.3

Answer 1

The Correct Option is D

For the mass to stay on the rotating drum without falling, the centripetal force must be equal to the frictional force. The frictional force is given by: \[ f = \mu mg, \] where \( \mu \) is the coefficient of friction, \( m = 0.5 \, \text{kg} \), and \( g = 10 \, \text{m/s}^2 \). The centripetal force required to keep the mass on the drum is: \[ F_c = m \omega^2 r, \] where \( \omega = 5 \, \text{rad/s} \) is the angular velocity and \( r = 4 \, \text{m} \) is the radius of the drum. Equating the frictional force and centripetal force: \[ \mu mg = m \omega^2 r. \] Substitute the known values: \[ \mu (0.5)(10) = (0.5)(5^2)(4). \] Simplifying: \[ \mu (5) = (0.5)(25)(4), \] \[ 5\mu = 50, \] \[ \mu = 0.3. \]
Final Answer: 0.3