Question

A ring of radius \( R \) is first rotated with an angular velocity \( \omega \), and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is \( \mu \). Time after which its angular speed is reduced to half is:

• A. \( \frac{2 \omega R}{\mu g} \)
• B. \( \frac{\omega R}{\mu g} \)
• C. \( \frac{2 \omega R}{g} \)
• D. \( \frac{\omega R}{2 \mu g} \)

Hint:

When a rolling object slows down due to friction, the time to reduce its angular speed is inversely proportional to the friction coefficient and directly proportional to the radius.

Answer 1

The Correct Option is B

When a ring rolls on a rough surface, the frictional force acts to reduce its angular velocity.
The equation for the angular deceleration \( \alpha \) due to friction is given by: \[ f = \mu mg \] where \( m \) is the mass of the ring and \( g \) is the acceleration due to gravity. The torque \( \tau \) due to the friction is: \[ \tau = fR = \mu mgR \] The angular acceleration \( \alpha \) is given by: \[ \tau = I \alpha \] where \( I \) is the moment of inertia of the ring.
For a ring, \( I = mR^2 \), so: \[ \mu mgR = mR^2 \alpha \] Simplifying: \[ \alpha = \frac{\mu g}{R} \] Now, to reduce the angular velocity to half, we use the equation: \[ \omega_f = \omega_0 + \alpha t \] Substituting \( \omega_f = \frac{\omega_0}{2} \), we get: \[ \frac{\omega_0}{2} = \omega_0 - \frac{\mu g}{R} t \] Solving for \( t \): \[ t = \frac{\omega_0 R}{\mu g} \]
Thus, the correct answer is (b).