Question

A solid cylinder of mass $m=4\text{kg}$ and radius $R=10\text{cm}$ has a string wrapped around it as shown in figure.The cylinder is held horizontal by fixing the two free ends of the string to the hook on the ceiling such that both the strings are exactly vertical. The cylinder is released to fall under gravity. Find the linear acceleration $\left(a\right)$ of the cylinder.

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

Answer 1

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

Applying Newton's $2^{\text{nd}}$ law in direction of acceleration on the cylinder,


$mg-2T=ma...\left(i\right)$
For rotational motion of the cylinder, equation of torque
${\tau }_{\text{com}}=I_{\text{com}}\alpha$
$\Rightarrow 2TR=\left(\frac{m{R}^2}{2}\right)\alpha ...\left(ii\right)$
$\because {\tau }_{mg}=0$ about centre.

For string to not slip, tangential acceleration should be equal to acceleration of centre of mass $a$
$a_t=\alpha R$
$\therefore a=\alpha R...\left(iii\right)$
Substituting in Eq. $\left(ii\right)$,
$\Rightarrow T=\frac{ma}{4}$
Substituting in Eq. $\left(i\right)$,
$mg-\frac{ma}{2}=ma$
or $\frac{3ma}{2}=mg$
$\therefore a=\frac{2g}{3}$
Here, $a$ is the linear acceleration of the centre of mass of the cylinder.