Question

A long plank of mass M rests upon a smooth horizontal surface. A thin circular ring (m,R) slips ( without rotation ) upon plank. The coefficient of friction between the wheel and plank is k. At t = t s , the ring's slipping ceases and pure rolling starts upon the plank, then what is the value of t = ?

Answer 1

Answer:

time taken by the ring to start pure rolling on the plank is given as

$t = \frac{v_o}{\mu g(2 + \frac{m}{M})}$

Explanation:

If plank and ring is combined system then net force on this system is zero

So we can use momentum conservation after ring will start pure rolling

So we will have

$mv_o = Mv_1 + mv_2$

Also for condition of pure rolling we can say

$v_2 - R\omega = v_1$

now we can say that

$mv_o = M(v_2 - R\omega) + mv_2$

$mv_o = (M + m)v_2 - MR\omega$

now we know that friction force between plank and ring is given as

$F_f = \mu mg$

so we will have

$v_2 = v_o - \mu g t$

also for torque equation of the ring

$\tau = \mu mg R$

$mR^2 \alpha = \mu mg R$

$\alpha = \frac{\mu g}{R}$

now we have

$\omega = \frac{\mu g}{R} t$

also for plank we will have

$\mu mg = Ma_1$

$a_1 = \frac{\mu mg}{M}$

$v_1 = \frac{\mu mg}{M} t$

now from above equation we have

$(v_o - \mu g t ) - R(\frac{\mu g}{R} t) = \frac{\mu mg}{M} t$

$v_o - 2\mu g t = \frac{\mu mg}{M} t$

$v_o = \mu g(2 + \frac{m}{M}) t$

$t = \frac{v_o}{\mu g(2 + \frac{m}{M})}$

#Learn

Topic : Rolling Motion

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