Question

A plank of mass $M$, whose top surface is rough with coefficient of friction $\mu$ is placed on a smooth ground. Now a disc of mass $(m=\frac{M}{2})$ and radius $R$ is placed on the plank. The disc is now given a velocity $v_0$ in the horizontal direction at $t=0$. Then, choose the correct option(s):

• A. Time when disc starts rolling without slipping is $\frac{2v_0}{7\mu g}$.
• B. Velocity of plank when rolling without slipping starts is $\frac{v_0}{7}$.
• C. Velocity of disc when rolling without slipping starts is $\frac{5}{7}{v}_0$.
• D. Angular velocity of disc when rolling without slipping starts is $\frac{4v_0}{7R}$.

Answer 1

Answer: (A), (B), (C), (D)

Angular velocity of disc when rolling without slipping starts is $\frac{4v_0}{7R}$.

Here, disc is moving rightward, hence friction will act leftward to oppose relative motion (at contact point).


For plank, $\sum F_x=Ma$
$f=Ma$
$\Rightarrow \mu mg=Ma$ $(\because f=\mu mg)$
$\Rightarrow \mu \frac{M}{2}g=Ma$
$\Rightarrow a=\frac{\mu g}{2}$

For disc, $\sum F_x=-m{a}_1$
$-f=-m{a}_1$ [Both friction and acceleration are acting leftward]
$\Rightarrow -\mu mg=-m{a}_1$
$\Rightarrow a_1=\mu g$

From $\tau =I\alpha$ (for disc about its centre)
$-f.R=-\left(\frac{m{R}^2}{2}\right)\alpha$ (clockwise)
$\Rightarrow \left(\mu mg\right)R=\left(\frac{m{R}^2}{2}\right)\alpha$
$\Rightarrow \alpha =\frac{2\mu g}{R}$

For pure rolling, velocity of contact point will be same as velocity of plank.
i.e $v=v^{\prime }-\omega R$ ... (1)
where $v^{\prime }=$ velocity of disc
$v=$ velocity of plank when disc starts pure rolling
From equation of motion,
Velocity of plank $v=u+at=at$ ... (2)
[initial velocity of plank = 0]
Velocity of disc, $v^{\prime }=v_0-a_{1}t$ ... (3)
(-ve because $a_1$ is in opposite direction of motion)
Angular velocity of disc $\omega ={\omega }_0+\alpha t=0+\alpha t$ ... (4)
[initial angular velocity of disc = 0]
Put in eq. (1)
$v=v^{\prime }-\omega R$
$\Rightarrow \left(at\right)=(v_0-a_{1}t)-\left(\alpha t\right)R$
$\Rightarrow \frac{\mu gt}{2}=(v_0-\mu gt)-\left(\frac{2\mu gt}{R}\right)R$
$\Rightarrow t=\frac{2v_0}{7\mu g}$
From eq. (2):
Velocity of plank when disc starts pure rolling is
$v=at=\frac{\mu g}{2}\times \frac{2v_0}{7\mu g}=\frac{v_0}{7}$
Velocity of disc when it starts pure rolling is
$v^{\prime }=v_0-a_{1}t$
$v^{\prime }=v_0-\left(\mu g\right)\times \left(\frac{2v_0}{7\mu g}\right)=\frac{5v_0}{7}$
Angular velocity of disc when it starts pure rolling is
$\omega =\alpha t$
$=\left(\frac{2\mu g}{R}\right)\times \left(\frac{2v_0}{7\mu g}\right)=\left(\frac{4v_0}{7R}\right)$
Hence, all options are correct.