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. \((\omega_0 = \frac{v_0}{4R})\)
Step 1: Understanding the Concept:
Rolling starts when the velocity of the center of mass $v$ and angular velocity $\omega$ satisfy the condition $v = \omega R$.
Friction causes linear deceleration and angular acceleration until this condition is met. Step 2: Key Formula or Approach:
Frictional force is $f = \mu_K mg$.
Linear acceleration is $a = \frac{f}{m} = \mu_K g$.
Torque is $\tau = f \cdot R = I \alpha$, which gives $\alpha = \frac{\mu_K mg R}{mR^2/2} = \frac{2\mu_K g}{R}$.
Kinematic equations: $v = v_0 - at$ and $\omega = \omega_0 + \alpha t$. Step 3: Detailed Explanation:
Linear velocity at time $t$ is $v = v_0 - \mu_K gt$.
Angular velocity at time $t$ is $\omega = \omega_0 + \frac{2\mu_K g}{R} t$.
For rolling without slipping, we need $v = \omega R$.
Substitute the kinematic expressions into the rolling condition:
\[ v_0 - \mu_K gt = \left( \omega_0 + \frac{2\mu_K g}{R} t \right) R \]
\[ v_0 - \mu_K gt = \omega_0 R + 2\mu_K gt \]
Given in the problem, $\omega_0 = \frac{v_0}{4R} \implies v_0 = 4\omega_0 R$.
Substituting $v_0$:
\[ 4\omega_0 R - \mu_K gt = \omega_0 R + 2\mu_K gt \]
\[ 3\omega_0 R = 3\mu_K gt \]
Solving for $t$:
\[ t = \frac{\omega_0 R}{\mu_K g} \] Step 4: Final Answer:
The time after which rolling starts is $\frac{\omega_0 R}{\mu_K g}$.
