JEE Main 2026 — Physics Mechanics

Initial translational velocity, $v_0 = 49$v0​=49 m/s.

Initial angular velocity, $\omega_0 = \dfrac{v_0}{4R}$ω0​=4Rv0​​.

Velocity of the lowest point of the cylinder is $v_{bottom} = v_0 - \omega_0 R = v_0 - \dfrac{v_0}{4} = \dfrac{3v_0}{4} > 0$vbottom​=v0​−ω0​R=v0​−4v0​​=43v0​​>0.

Since the lowest point is slipping forward, kinetic friction $f_k = \mu_k m g$fk​=μk​mg acts in the backward direction.

Translational acceleration, $a = -\dfrac{f_k}{m} = -\mu_k g$a=−mfk​​=−μk​g.

Angular acceleration, $\alpha = \dfrac{f_k R}{I} = \dfrac{\mu_k m g R}{\dfrac{1}{2} m R^2} = \dfrac{2\mu_k g}{R}$α=Ifk​R​=21​mR2μk​mgR​=R2μk​g​.

At time $t$t, the translational velocity is $v = v_0 - \mu_k g t$v=v0​−μk​gt.

The angular velocity is $\omega = \omega_0 + \alpha t = \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t$ω=ω0​+αt=4Rv0​​+R2μk​g​t.

For pure rolling to start, $v = \omega R$v=ωR.

$$v_0 - \mu_k g t = \left( \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t \right) R$$v0​−μk​gt=(4Rv0​​+R2μk​g​t)R

$$v_0 - \mu_k g t = \dfrac{v_0}{4} + 2\mu_k g t$$v0​−μk​gt=4v0​​+2μk​gt

$$3\mu_k g t = \dfrac{3v_0}{4}$$3μk​gt=43v0​​

$$t = \dfrac{v_0}{4\mu_k g}$$t=4μk​gv0​​

Substituting the given values $v_0 = 49$v0​=49 m/s, $\mu_k = 0.25$μk​=0.25, and $g = 9.8$g=9.8 m/s$^2$2:

$t = \dfrac{49}{4 \times 0.25 \times 9.8} = \dfrac{49}{9.8} = 5$t=4×0.25×9.849​=9.849​=5 s.

Answer: $5$5