Question

A string is wrapped over the curved surface of a uniform solid cylinder and the free end of the string is fixed with rigid support. Find the speed of the centre of mass of the cylinder as it falls through a vertical distance $l$.

• A. $\sqrt{\frac{2gl}{3}}$
• B. $\sqrt{4gl}$
• C. $\sqrt{3gl}$
• D. $\sqrt{\frac{4gl}{3}}$

Answer 1

The correct option is D $\sqrt{\frac{4gl}{3}}$

Let mass of the solid cylinder be $m$;
Tension in the string be $T$
$a=$ acceleration of the centre of mass of cylinder.


Applying Newton's $2^{\text{nd}}$ law in the direction of acceleration of the cylinder,
$mg-T=ma...\left(i\right)$
For rotational motion of the cylinder, equation of torque
${\tau }_{\text{com}}=I_{\text{com}}\alpha$
$\Rightarrow TR=\left(\frac{m{R}^2}{2}\right)\alpha ...\left(ii\right)$
$\because {\tau }_{mg}=0$ 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. $\left(ii\right)$,
$\Rightarrow T=\frac{ma}{2}$
Substituting in Eq. $\left(i\right)$,
$mg-\frac{ma}{2}=ma$
$\frac{3ma}{2}=mg$
$\therefore a=\frac{2g}{3}$

Initially system was at rest i.e $u=0$ Applying equation of motion $v^2-u^2=2as$
$\Rightarrow v^2=2al$
$\Rightarrow v^2=\frac{4gl}{3}$
$\therefore v=\sqrt{\frac{4gl}{3}}$