Rotational Motion

JEE Main level, single correct, medium

243 of 280

Question 243 of 280

A solid cylinder having radius RR and length LL is slipping on a rough horizontal plane. At time t=0t=0 the cylinder has a translational velocity v0=49v_0=49 m/s, perpendicular to its axis and a rotational velocity v0/4Rv_0/4R about the centre. The time taken by the cylinder to start rolling is ________ seconds. (coefficient of kinetic friction μK=0.25\mu_K=0.25 and g=9.8g=9.8 m/s2^2)

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Answer: (B) 55

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

Initial angular velocity, ω0=v04R\omega_0 = \dfrac{v_0}{4R}.

Velocity of the lowest point of the cylinder is vbottom=v0−ω0R=v0−v04=3v04>0v_{bottom} = v_0 - \omega_0 R = v_0 - \dfrac{v_0}{4} = \dfrac{3v_0}{4} \gt 0.

Since the lowest point is slipping forward, kinetic friction fk=μkmgf_k = \mu_k m g acts in the backward direction.

Translational acceleration, a=−fkm=−μkga = -\dfrac{f_k}{m} = -\mu_k g.

Angular acceleration, α=fkRI=μkmgR12mR2=2μkgR\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 tt, the translational velocity is v=v0−μkgtv = v_0 - \mu_k g t.

The angular velocity is ω=ω0+αt=v04R+2μkgRt\omega = \omega_0 + \alpha t = \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t.

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

v0−μkgt=(v04R+2μkgRt)Rv_0 - \mu_k g t = \left( \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t \right) R

v0−μkgt=v04+2μkgtv_0 - \mu_k g t = \dfrac{v_0}{4} + 2\mu_k g t

3μkgt=3v043\mu_k g t = \dfrac{3v_0}{4}

t=v04μkgt = \dfrac{v_0}{4\mu_k g}

Substituting the given values v0=49v_0 = 49 m/s, μk=0.25\mu_k = 0.25, and g=9.8g = 9.8 m/s2^2:

t=494×0.25×9.8=499.8=5t = \dfrac{49}{4 \times 0.25 \times 9.8} = \dfrac{49}{9.8} = 5 s.

Answer: 55

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