Question

A uniform solid cylinder of mass 'm' can rotate freely about its axis which is kept horizontal. A particle of mass $m_0$ hangs from the end of a light string wound round the cylinder. When the system is allowed to move, the acceleration with which the particle descends is

• A. $\frac{2m_{0}g}{m_0+2m}$
• B. $\frac{2m_{0}g}{m+2m_0}$
• C. $\frac{m_{0}g}{m+m_0}$
• D. $\frac{2m_{0}g}{m+2m_0}$

Answer 1

The correct option is B $\frac{2m_{0}g}{m+2m_0}$

Suppose that the tension in the string is T.
Then, $=m_{0}g-T=m_{0}a$
$\therefore T=m_0(g-a)$
where a = acceleration
Further, T.r = I $\alpha$
$\alpha$ = angular acceleration and T.r = moment of force acting on cylinder
or $T=\frac{I\alpha }{r}$
$=\left(\frac{m{r}^2}{2}\right)\left(\frac{\alpha }{r}\right)=\frac{mr\alpha }{2}=\frac{ma}{2}$
$\therefore a=\frac{2T}{m}=\frac{2m_0(g-a)}{m}$
Solving for a, we get
$a=\left(\frac{2m_{0}g}{m+2m_0}\right).$