Question

A cylindrical rod of mass $m$, length $L$ and radius $R$ has two light strings wound over the rod 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 and the cylinder is rolling vertically down. The total tension in two strings is:

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

Answer 1

The correct option is B $\frac{mg}{3}$

Let the tension in each string be T.
The equation of motion: $mg-2T=ma$ .....(1)
Writing the torque equation: $2TR=I\alpha =\frac{m{R}^2}{2}\alpha =\frac{m{R}^2}{2}\frac{a}{R}$
$\therefore T=\frac{ma}{4}$......(2)
Substituting $T$ from (2) in (1), we get, $a=\frac{2g}{3}$
$\Rightarrow T=\frac{mg}{6}$
Total tension : $2T=\frac{mg}{3}$