Question

A cylinder of mass $m=1\text{kg}$ is suspended through two strings wrapped around it as shown in figure. Find the acceleration of $COM$ of the cylinder.

• A. $\frac{2g}{5}$
• B. $\frac{2g}{3}$
• C. $g$
• D. $\frac{g}{4}$

Answer 1

The correct option is B $\frac{2g}{3}$

The portion of the strings between the ceiling and the cylinder is at rest. Hence the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings.
Suppose the centre of the cylinder falls with an acceleration $a$. The angular acceleration of the cylinder about its axis is $\alpha =a/R$, as the cylinder does not slip over the strings.
The equation of motion for the centre of mass of the cylinder is
$mg-2T=ma$ ... (i)
For rotational motion about axis passing through C.O.M,
$R\times 2T=I.\alpha$
where $R\to$ radius of cylinder
$\Rightarrow 2TR=\frac{m{R}^2}{2}\times \left(\frac{a}{R}\right)$
$\Rightarrow T=\frac{ma}{4}$ ... (ii)
From eqn. (i) and (ii)
$mg-2\left(\frac{ma}{4}\right)=ma$
$\Rightarrow a=\frac{2g}{3}$