Question

A cylinder of mass 'm; is kept on the edge of a plank of mass '2m' and length 12 metre, which in turn is kept on smooth ground. Coefficient of friction between the plank and the cylinder is 0.1. The cylinder is given an impulse, which imparts it a velocity 7 m/s but no angular velocity. Find the time after which the cylinder falls off the plank.

• A. 2 sec
• B. 3 sec
• C. 2.5 sec
• D. 2.25 sec

Answer 1

The correct option is C 2.5 sec

You should infer by now that there will be kinetic friction on the cylinder in the backward direction till it starts pure rolling and then no frictional force at all.

Let's assume that radius of cylinder be R. let us investigate the frictional force upon the cylinder by the plank & vice versa by this simple free body diagram of both as shown.

Now the acceleration of cylinder = $-\mu g$ and acceleration of plank =$\frac{\mu g}{2}$

Let the time spent till pure rolling starts bet. Now from torque equation of cylinder about its centre O,

$\mu \text{/}mg\text{/}R=\frac{\text{/}M{R}^{\text{/}2}}{2}\alpha$

$\frac{2\mu g}{R}=\alpha$

The velocity of cyclinder of and plank till time 't' will vary as,

$V_c\left(t\right)=7-\mu gt$

$V_p\left(t\right)=\frac{\mu gt}{2}$

and angular velocity will vary as,

$\omega ={\omega }_0+\alpha t$

$\omega =\frac{2\mu g}{R}t$

But at the instant the cylinder starts rolling purely, the point of contact should be at rest with respect to the plank.

$\Rightarrow \frac{\mu gt}{2}=(7-\mu gt)-\left(\frac{2\mu gt}{R}\right)\times R$

Taking $\mu =0.1andg=10m/s^2$

$\frac{t}{2}=7-t-2t$

$3t+\frac{t}{2}=7\Rightarrow t=2s$

Now the distance covered by cylinder & plank can be obtained simply by using $s=ut+\frac{1}{2}a{t}^2$

$U_{rel}=7-0=7m/s$

T = 2

$a_{rel}=-\mu gt-\frac{\mu gt}{2}$

$s=ut+\frac{1}{2}a{t}^2$

$s_c-s_p=11m$

And after pure rolling starts,

$V_c=u+at$

$=7-\mu gt$

$V_c=5m/s$

$v_p=\frac{\mu gt}{2}$

$V_p=1m/s$

$V_c-V_p=4m/s$

$S_{\frac{c}{p}}=1m$

t=0.25s

so,total time=2.25 s