A solid cylinder having radius … and…

Single correct+4 / -1Answer matches NTA’s final key

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 $v_0 = 49$ 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 $g = 9.8$ m/s²)

Solution

At the start $R\omega_0 = \frac{v_0}{4} < v_0$ , so the contact point slides forward and kinetic friction $\mu mg$ acts backward. It slows the centre and spins the cylinder up:

$v = v_0 - \mu g\,t$
$\alpha = \dfrac{\mu mgR}{\frac12 mR^2} = \dfrac{2\mu g}{R}$ , so $R\omega = \dfrac{v_0}{4} + 2\mu g\,t$

Rolling begins when $v = R\omega$ :

$v_0 - \mu g t = \frac{v_0}{4} + 2\mu g t \;\Rightarrow\; t = \frac{v_0}{4\mu g} = \frac{49}{4 \times 0.25 \times 9.8} = 5 \text{ s}.$

Answer: 5 s.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Rotational Motion

Solution

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