Question

A cylindrical rod of mass $m$, length $L$ and radius $R$ has two light strings wound over it and two upper ends of strings are attached to the ceiling. The rod is horizontal and two strings are vertical when the rod is released. The strings unwound while the cylinder is rolling vertically down. The acceleration of centre of mass of rod is:

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

Answer 1

Answer: (C)

$\frac{2g}{3}$

Moment of inertia of the cylindrical rod :
$I=\frac{m{R}^2}{2}$
From force balance we get:
$mg-2T=ma$........(i)
and from torque balance we have:
$I\alpha =\left(2T\right)\left(R\right)$
$\Rightarrow \frac{m{R}^2}{2}\times \frac{a}{R}=2TR$
$\therefore \frac{ma}{4}=T$........(ii)

From (i)

$T=m(g-a)/2$

Substituting in (ii), we get:
$a=\frac{2g}{3}$