A solid cylinder having radius $R$ and length $L$ is slipping on a rough horizontal plane. At time $t = 0$ the cylinder has a translational velocity $v_0 = 49\ \text{m/s}$, perpendicular to its axis and a rotational velocity $v_0/4R$ about the centre. The time taken by the cylinder to start rolling is ______ seconds. (coefficient of kinetic friction $\mu_K = 0.25$ and $g = 9.8\ \text{m/s}^2$)
Answer: (B) $5$
Concept. While slipping, kinetic friction $\mu mg$ acts backwards at the contact point. It decelerates the centre and increases the spin. Rolling starts when $v=\omega R$.
Step 1. Translation. $a=-\mu g=-0.25\times9.8=-2.45$ m s$^{-2}$, so $v=49-2.45t$.
Step 2. Rotation. Torque $\mu mgR=\tfrac12mR^2\alpha$ gives $\alpha R=2\mu g=4.9$ m s$^{-2}$. Starting from $\omega_0R=\dfrac{v_0}{4}=12.25$ m s$^{-1}$:
$$\omega R=12.25+4.9t$$
Step 3. Rolling condition.
$$49-2.45t=12.25+4.9t\ \Rightarrow\ 36.75=7.35t\ \Rightarrow\ t=5\ \text{s}$$
Answer. $5$ s (Option B).
Solution by Sreeraj P, M.Sc Physics
