Question

A solid cylinder of mass $m$ is kept on the edge of a plank of mass $3m$ and length $15\text{m}$ which in turn is kept on smooth ground. The coefficient of friction between the plank and the cylinder is $0.1$. The cylinder is given an impulse which imparts it a velocity of $6\text{m/s}$ but no angular velocity. Find the time after which the cylinder will start pure rolling.

• A. $1.8\text{s}$
• B. $3.8\text{s}$
• C. $4.8\text{s}$
• D. $2.8\text{s}$

Answer 1

The correct option is A $1.8\text{s}$

The cylinder will slip on the plank and after some time, it will start pure rolling. The necessary torque will be provided by the friction from the plank.
FBD of cylinder and plank-


For cylinder,
$N=mg$ and
$f=\mu N=\mu mg=ma$
$\Rightarrow \mu mg=ma$
$\Rightarrow a=\mu g$
$\Rightarrow a=0.1\times 10$
$\Rightarrow a=1{\text{m/s}}^2$
Let us suppose the time after which the cylinder starts pure rolling is $t$.
For the bottom most point (point of contact) of the cylinder, velocity due to pure translation,
$v_1=u_1+at$
$\Rightarrow v_1=6-t$ ..............(1)
For the cylinder,
$fR=I\alpha$
$\Rightarrow \alpha =\frac{fR}{I}$
$\Rightarrow \alpha =\frac{\mu mgR}{m{R}^2/2}$
$\Rightarrow \alpha =\frac{2\mu g}{R}=\frac{2\times 0.1\times 10}{R}=\frac{2}{R}$
Now, ${\omega }_f={\omega }_i+\alpha t$
${\omega }_f=0+\frac{2t}{R}=\frac{2t}{R}$
For the bottom most point (point of contact) of the cylinder, velocity due to pure rotation,
$v_2={\omega }_{f}R=2t$ ...........(2)
So, net velocity of bottom most point,
$v=v_2-v_1=6-t-2t=6-3t$ .....(3)
[from (1) and (2)]
For plank, $f=3m{a}^{{}^{\prime }}$
$\Rightarrow \mu mg=3m{a}^{{}^{\prime }}$
$\Rightarrow a^{{}^{\prime }}=\frac{\mu g}{3}=\frac{0.1\times 10}{3}=\frac{1}{3}{\text{m/s}}^2$
So, $v=u+a^{{}^{\prime }}t$
$\Rightarrow v=\frac{t}{3}$.........(4)

For pure rolling, the plank should have the same velocity as the velocity of the point of contact.
From (3) and (4),
$6-3t=\frac{t}{3}$
$\Rightarrow t=1.8\text{s}$