Question

A cylinder of mass m is kept on the edge of a plank of mass $2m$ and length $12m$ 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 $7m/s$ but no angular velocity. Find the time after which the cylinder falls off the plank. $(g=10m/s^2)$

Answer 1

Initially the cylinder will slip on the plank, therefore kinetic friction will act between the cylinder and the plank
$a_c=\frac{\mu mg}{m}=-\mu g$
$a_p=+\frac{\mu mg}{2m}=+\frac{\mu g}{2}$
${\alpha }_c=+\frac{\left(\mu mg\right)\left(R\right)}{\left(m{R}^2/2\right)}=+\frac{2\mu g}{R}$
For pure rolling,
$v_p=v_c-R{\omega }_c$
$\therefore \frac{\mu g}{2}t=v_0-\mu gt-\left(R\right)\left(\frac{2\mu g}{R}\right)\left(t\right)$
$\therefore t=\frac{v_0}{3.5\mu g}=\frac{7}{3.5\times 0.1\times 10}=2s$
$\therefore s_c-s_p=v_{0}t-\frac{1}{2}\times \left(\mu g\right)\left(t^2\right)-\frac{1}{2}\left(\frac{\mu g}{2}\right)\left(t^2\right)$
$=(7\times 2)-\frac{1}{2}\left(0.1\right)\left(10\right)\left(4\right)$
$-\frac{1}{2}\left(\frac{0.1\times 10}{2}\right)\left(4\right)=11m$
Also, $v_c-v_p=(v_0-\mu gt)-\left(\frac{\mu g}{2}\right)\left(t\right)$
$=7-0.1\times 10\times 2-\frac{0.1\times 10\times 2}{2}=4m/s$
Hence, the remaining distance $(12-11=1m)$ is travelled in a time,
$t^{\prime }=\frac{1.0}{4}=0.25s$
$\therefore Totaltime=2+0.25=2.25s$