Question

A cylinder of mass $m$ is placed on the edge of a long plank of same mass kept on the smooth horizontal surface. $\mu$ is the coefficient of friction between cylinder and plank. The cylinder is given an impulse at $t=0$ which imparts it a velocity $V_0$. Find the time in which pure rolling starts

• A. V_0/2μg
• B. V_0/4μg
• C. V_0/2\sqrt{2}μg
• D. V_0/4\sqrt{2}μg

Answer 1

Answer: (B)

V_0/4μg

Answer 2

Let $V_0$ be the initial velocity of the cylinder. Let $m$ be the mass of the cylinder and the plank. Let $\mu$ be the coefficient of kinetic friction. Let $r$ be the radius of the cylinder. The moment of inertia of a solid cylinder about its axis is $I = \frac{1}{2}mr^2$.

When the cylinder is given an impulse, it starts sliding on the plank. The friction force between the cylinder and the plank is kinetic friction, $f_k = \mu N$, where $N$ is the normal force. For the cylinder, $N = mg$. Thus, $f_k = \mu mg$.

Applying Newton's second law to the cylinder's center of mass: $m a_c = -f_k = -\mu mg$ $a_c = -\mu g$ The velocity of the cylinder's center of mass at time $t$ is: $v_c(t) = V_0 + a_c t = V_0 - \mu gt$

The friction force on the plank is equal and opposite to the friction force on the cylinder. So, the friction force on the plank is to the right: $m a_p = f_k = \mu mg$ $a_p = \mu g$ The velocity of the plank at time $t$ is: $v_p(t) = v_p(0) + a_p t = 0 + \mu gt = \mu gt$

The friction force on the cylinder causes a torque about its center of mass. The friction force is to the left, at a distance $r$ below the center of mass, causing a clockwise torque. $\tau_c = f_k r = (\mu mg) r$ The angular acceleration is: $\alpha_c = \frac{\tau_c}{I} = \frac{\mu mg r}{\frac{1}{2}mr^2} = \frac{2\mu g}{r}$ Assuming clockwise angular velocity as positive. $\omega_c(t) = \omega_c(0) + \alpha_c t = 0 + \frac{2\mu g}{r} t = \frac{2\mu g}{r} t$

Pure rolling starts when the velocity of the point of contact of the cylinder with the plank is the same as the velocity of the plank. The velocity of the point of contact on the cylinder relative to the ground is $v_{contact, c} = v_c(t) - r \omega_c(t)$ (since $\omega_c$ is clockwise). $v_{contact, c} = (V_0 - \mu gt) - r \left(\frac{2\mu g}{r} t\right) = V_0 - \mu gt - 2\mu gt = V_0 - 3\mu gt$

For pure rolling, $v_{contact, c} = v_p(t)$: $V_0 - 3\mu gt = \mu gt$ $V_0 = 4\mu gt$ $t = \frac{V_0}{4\mu g}$