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) $5$
Initially $v_0 > \omega_0R = \dfrac{v_0}{4}$, so the bottom point slides forward and kinetic friction $\mu mg$ acts backward.
Linear: $v = v_0 - \mu g t$.
Angular: $\alpha = \dfrac{\mu mgR}{\frac{1}{2}mR^2} = \dfrac{2\mu g}{R}$, so $\omega R = \dfrac{v_0}{4} + 2\mu g t$.
Rolling starts when $v = \omega R$:
$$v_0 - \mu g t = \frac{v_0}{4} + 2\mu g t \;\Rightarrow\; t = \frac{3v_0}{4 \times 3\mu g} = \frac{v_0}{4\mu g} = \frac{49}{4 \times 0.25 \times 9.8} = 5\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics