Question

A plank of mass $M$ is placed on a smooth surface over which a cylinder of mass $m$ $(=M)$ and radius $R=1m$ is placed as shown in figure. Now the plank is pulled towards the right with an external force $F(=2MG)$. If the cylinder does not slip over the surface of the plank, then :
(Take $g=10$ $m/s^2$)

• A. linear acceleration of the plank is $5m/s^2$
• B. linear acceleration of the cylinder is $10m/s^2$
• C. angular acceleration of the cylinder is $10rad/s^2$
• D. angular acceleration of the cylinder is $5rad/s^2$

Answer 1

Answer: (A), (C)

The correct options are
A linear acceleration of the plank is $5m/s^2$
C angular acceleration of the cylinder is $10rad/s^2$

Here we are given that cylinder does not slip over the plank surface, it is the case of pure rolling, we can use friction on cylinder in any direction. Here we choose towards the right. As friction is actin on the cylinder towards the right, it must be towards the left on the plank as shown in the force diagram of figure, Let the plank move towards the right with an acceleration $a_1$, the cylinder will experience a psuedoforce $m{a}_1$ in the left directionm due to which it will roll towards the left with respect to the plank with an acceleration $a_2$. As we have used pseudoforce, $a_2$ must be with
respect to the plank. Let its angular acceleration during rolling be $\alpha$, we have $a_2=R\alpha$.
For translational motion of the plank, we have
$F-f=M{a}_1..........\left(i\right)$
For translational motion of the cylinder with respect to the plank, we have
$m{a}_1-f=m{a}_2..........\left(ii\right)$
For rotational motion of the cylinder with respect to the plank, we have
$fR=I\alpha$
$fR=\left(\frac{1}{2}m{R}^2\right)\left(\frac{a_2}{R}\right)$
or $f=\frac{1}{2}m{a}_2.........\left(iii\right)$
From eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$m{a}_1-\frac{1}{2}m{a}_2$
$a_1=\frac{3}{2}{a}_2........\left(iv\right)$
Using Eqs. $\left(i\right),\left(iii\right),\left(iv\right)$, we get
$F-\frac{1}{2}m{a}_2=\frac{3}{2}M{a}_2$
$a_2=\frac{2F}{3M+m}=10m/s^2$
From Eqs.$\left(iv\right)$, $a_1=\frac{3F}{3M+m}=15m/s^2$
As we have already discussed that the value of $a_2$ is relative to the plank, the net acceleration of the cylinder will be given as $a_1-a_2$.
Hence, the acceleration of the cylinder is
$a_{cylinder}=a_1-a_2=15-10=5m/s^2$
The angular acceleration of the cylinder is
$\alpha =a_2/R=10rad/s^2$