Question

A plank is kept over a cylinder as shown in figure and there is no slipping at any contact. $15\text{N}$ is the horizontal force applied on the plank. Then, identify the correct statement(s):

• A. Acceleration of plank is $\frac{60}{7}{\text{m/s}}^2$
• B. Acceleration of centre of mass of the cylinder is $\frac{30}{7}{\text{m/s}}^2$
• C. Friction force between plank and cylinder is $\frac{45}{7}\text{N}$
• D. Friction force between cylinder and ground is $\frac{15}{7}\text{N}$

Answer 1

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

The correct options are
A Acceleration of plank is $\frac{60}{7}{\text{m/s}}^2$
B Acceleration of centre of mass of the cylinder is $\frac{30}{7}{\text{m/s}}^2$
C Friction force between plank and cylinder is $\frac{45}{7}\text{N}$
D Friction force between cylinder and ground is $\frac{15}{7}\text{N}$

Let the acceleration of the centre of mass $\left(a_{\text{com}}\right)$ of the cylinder be $a$.
Since the tangential acceleration $(a_t=r\alpha )$ at the top-most point and acceleration of the centre of mass of the cylinder $\left(a_{\text{com}}\right)$ are in the same direction,
Acceleration of topmost point is,
$a^{\prime }=a_{\text{com}}+a_t...\left(i\right)$
From condition of no slipping on ground,
$a_{\text{com}}=r\alpha =a$
From Eq.$\left(i\right)$,
$a^{\prime }=r\alpha +r\alpha =2r\alpha$
$\Rightarrow a^{\prime }=2a...\left(ii\right)$
Since, slipping is not taking place at any contact, therefore acceleration of plank $\left(a_P\right)$ will be equal to the acceleration of the point of contact with the cylinder i.e $a_P=2a$.
Friction$\left(f\right)$ will act in forward direction on cylinder and in backward direction on the plank to prevent slipping (at top-most position).


For plank, on applying Newton's second law along its direction of acceleration,
$15-f=1\times 2a=2a...\left(1\right)$
For cylinder, along its direction of acceleration of $\text{COM}$
$f+f^{\prime }=2a...\left(2\right)$

[For cylinder, the point of contact with the ground will tend to slip backwards on application of $15\text{N}$, hence friction $f^{\prime }$ will act along forward direction.]

On taking instantaneous axis of rotation ($I.A.R$) at the point of contact of cylinder and ground (because point of contact for cylinder with ground is always at rest)
Applying equation of torque for cylinder about $I.A.R$,
$\sum \tau =I\alpha$
$\because$ Normal reaction from ground, $f^{\prime },mg$ will pass through $I.A.R$, their torque will vanish.
$\Rightarrow {\tau }_f=I\alpha$
$\Rightarrow 2rf=\frac{3m{r}^2}{2}.\frac{a}{r}$
[$I=I_0+m{d}^2,I_0=\frac{m{r}^2}{2},d=r$ and $a=\alpha r$ for pure rolling]
$\Rightarrow f=\frac{3ma}{4}$
$\Rightarrow f=\frac{3\times 2\times a}{4}=\frac{3a}{2}...\left(3\right)$
On putting Eq.$\left(3\right)$ in Eq.$\left(1\right)$,
$15-\frac{3a}{2}=2a$
$\Rightarrow a=\frac{30}{7}{\text{m/s}}^2$ is the acceleration of COM of cylinder.
Acceleration of plank is,
$a_P=2a=\frac{60}{7}{\text{m/s}}^2$
On putting the value of $a$ in Eq.$\left(3\right)$, we get friction between plank and cylinder
$\Rightarrow f=\frac{45}{7}\text{N}$
On putting $f$ in Eq. $\left(2\right)$, we get friction between cylinder and ground
$\Rightarrow f^{\prime }=\frac{15}{7}\text{N}$
Hence options $\left(a\right),\left(b\right),\left(c\right),\left(d\right)$ are correct.