Q57 / 75 · JEE Main 2026 (08 April Shift 2) · Physics
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/s$^2$)
- A$15$
- Correct answerB$5$
- C$10$
- D$7.5$
Solution
Initial translational velocity, $v_0 = 49$ m/s.
Initial angular velocity, $\omega_0 = \dfrac{v_0}{4R}$.
Velocity of the lowest point of the cylinder is $v_{bottom} = v_0 - \omega_0 R = v_0 - \dfrac{v_0}{4} = \dfrac{3v_0}{4} > 0$.
Since the lowest point is slipping forward, kinetic friction $f_k = \mu_k m g$ acts in the backward direction.
Translational acceleration, $a = -\dfrac{f_k}{m} = -\mu_k g$.
Angular acceleration, $\alpha = \dfrac{f_k R}{I} = \dfrac{\mu_k m g R}{\dfrac{1}{2} m R^2} = \dfrac{2\mu_k g}{R}$.
At time $t$, the translational velocity is $v = v_0 - \mu_k g t$.
The angular velocity is $\omega = \omega_0 + \alpha t = \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t$.
For pure rolling to start, $v = \omega R$.
$v_0 - \mu_k g t = \left( \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t \right) R$
$v_0 - \mu_k g t = \dfrac{v_0}{4} + 2\mu_k g t$
$3\mu_k g t = \dfrac{3v_0}{4}$
$t = \dfrac{v_0}{4\mu_k g}$
Substituting the given values $v_0 = 49$ m/s, $\mu_k = 0.25$, and $g = 9.8$ m/s$^2$:
$t = \dfrac{49}{4 \times 0.25 \times 9.8} = \dfrac{49}{9.8} = 5$ s.
Answer: $5$
Initial angular velocity, $\omega_0 = \dfrac{v_0}{4R}$.
Velocity of the lowest point of the cylinder is $v_{bottom} = v_0 - \omega_0 R = v_0 - \dfrac{v_0}{4} = \dfrac{3v_0}{4} > 0$.
Since the lowest point is slipping forward, kinetic friction $f_k = \mu_k m g$ acts in the backward direction.
Translational acceleration, $a = -\dfrac{f_k}{m} = -\mu_k g$.
Angular acceleration, $\alpha = \dfrac{f_k R}{I} = \dfrac{\mu_k m g R}{\dfrac{1}{2} m R^2} = \dfrac{2\mu_k g}{R}$.
At time $t$, the translational velocity is $v = v_0 - \mu_k g t$.
The angular velocity is $\omega = \omega_0 + \alpha t = \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t$.
For pure rolling to start, $v = \omega R$.
$v_0 - \mu_k g t = \left( \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t \right) R$
$v_0 - \mu_k g t = \dfrac{v_0}{4} + 2\mu_k g t$
$3\mu_k g t = \dfrac{3v_0}{4}$
$t = \dfrac{v_0}{4\mu_k g}$
Substituting the given values $v_0 = 49$ m/s, $\mu_k = 0.25$, and $g = 9.8$ m/s$^2$:
$t = \dfrac{49}{4 \times 0.25 \times 9.8} = \dfrac{49}{9.8} = 5$ s.
Answer: $5$
Hence, the correct answer is option: 'B'.