Question

A rope is wound around a hollow cylinder of mass $3$ $kg$ and radius $40$ $cm$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30$ $N$?

• A. $25$ $rad{s}^{-2}$
• B. $20$ $rad{s}^{-2}$
• C. $25$ $m{s}^{-2}$
• D. $20$ $m{s}^{-2}$

Answer 1

The correct option is A $25$ $rad{s}^{-2}$

Given, $m=3kg$, $r=40cm$, $F=30N$

Moment of inertia of hollow cylinder about its axis
$=m{r}^2=3\times {\left(0.4\right)}^2=0.48kg{m}^2$

The torque is given by,
$\tau =Ia$
Where, $I$= moment of inertia and $a$ = angular accleration

In the given case
$\tau =rF,$as the force is acting perpendicularly to the radial vector.
$\therefore a=\frac{\tau }{I}=\frac{Fr}{m{r}^2}=\frac{F}{mr}=\frac{30}{3\times 40\times {10}^{-2}}a=25rad{s}^{-2}$