Question

Find the acceleration of a system consisting of a solid cylinder of mass $m$ and radius $R$ and a plank of mass $M$ placed on a smooth surface if it is pulled with a force $F$ as shown in the figure. Given that sufficient friction is present between the cylinder and the plank surface to prevent sliding of the cylinder.

• A. $\frac{F}{m+M}$
• B. $\frac{F}{2m+M}$
• C. $\frac{F}{m+2M}$
• D. $\frac{F}{3m+M}$

Answer 1

The correct option is D $\frac{F}{3m+M}$

Let the acceleration of the system be $a$. Free body diagram of the given system is


From FBD,
$F-T-f=Ma----\left(i\right)$
$T-f=ma---\left(ii\right)$
Torque experienced by the cylinder is,
$f\times R=I\alpha$
$\Rightarrow f\times R=\frac{1}{2}m{R}^2\alpha --\left(iii\right)$
Also, $\alpha R-a=a$ (condition for pure rolling)
$\Rightarrow 2a=\alpha R$
$\Rightarrow a=\frac{\alpha R}{2}---\left(iv\right)$
Now, $f=\frac{mR\alpha }{2}$ (from equation $\left(iii\right)$)
$\Rightarrow \alpha R=\frac{2f}{m}$
On adding equation $\left(i\right)$ and $\left(ii\right)$
$F-2f=a(m+M)$
$\Rightarrow F-a(m+M)=2f$
$\Rightarrow f=\frac{F-a(m+M)}{2}$
Substituting $f$ and $\alpha$ in the equation $\left(iii\right)$, we get,
$\frac{F-a(m+M)}{2}\times R=\frac{1}{2}m{R}^2\times \frac{2a}{R}$
$F-a(m+M)=2ma$
$\Rightarrow a=\frac{F}{3m+M}$