Question

A solid cylinder of mass \(m\) and radius \(R\) is projected on a rough surface having kinetic friction coefficient \( \mu_k \) with velocity \(v_0\) and angular velocity \( \omega_0 \) as shown in the figure. Find out time after which rolling starts. \((\omega_0 = \frac{v_0}{4R})\)

• A. \( \dfrac{3\omega_0 R}{\mu_k g} \)
• B. \( \dfrac{2\omega_0 R}{\mu_k g} \)
• C. \( \dfrac{\omega_0 R}{\mu_k g} \)
• D. \( \dfrac{\omega_0 R}{3\mu_k g} \)

Answer 1

The Correct Option is C

Concept: When a body slides on a rough surface, kinetic friction acts opposite to motion. For a rolling body: • Linear deceleration is caused by friction. • Friction also produces torque that changes angular velocity. • Pure rolling begins when \[ v = \omega R \] For a solid cylinder: \[ I = \frac{1}{2}MR^2 \]
Step 1:Find linear acceleration due to friction. Friction force: \[ f = \mu_k mg \] Using Newton's second law: \[ a = \frac{f}{m} = \mu_k g \] Since friction opposes motion, \[ v = v_0 - \mu_k g t \]
Step 2:Find angular acceleration produced by friction. Torque due to friction: \[ \tau = \mu_k mgR \] Using rotational equation: \[ \tau = I\alpha \] \[ \mu_k mgR = \frac{MR^2}{2}\alpha \] \[ \alpha = \frac{2\mu_k g}{R} \] Thus angular velocity becomes \[ \omega = \omega_0 + \frac{2\mu_k g}{R}t \]
Step 3:Apply rolling condition. Pure rolling begins when \[ v = \omega R \] Substitute expressions: \[ v_0 - \mu_k g t = (\omega_0 + \frac{2\mu_k g}{R}t)R \] \[ v_0 - \mu_k g t = \omega_0 R + 2\mu_k g t \]
Step 4:Use given relation \( \omega_0 = \frac{v_0}{4R} \). \[ v_0 = 4\omega_0 R \] Substitute: \[ 4\omega_0 R - \omega_0 R = 3\mu_k g t \] \[ 3\omega_0 R = 3\mu_k g t \] \[ t = \frac{\omega_0 R}{\mu_k g} \] \[ \boxed{t = \frac{\omega_0 R}{\mu_k g}} \]