Question

A rope of negligible mass is wound round a hollow cylinder of mass $3kg$ and radius $40cm$. If the rope is pulled with a force of $30N$, then the angular acceleration (in rad/s${}^2$) produced in the cylinder is

• A. $15rad{s}^{-2}$
• B. $20rad{s}^{-2}$
• C. $25rad{s}^{-2}$
• D. $30rad{s}^{-2}$

Answer 1

Answer: (C)

Here, $M=3kg$
$R=40cm=40\times 10^{-2}m$
Force applied, $F=30N$
Torque, $\tau =FR=(30N)(40\times 10^{-2}m)=12Nm$
Moment of inertia of hollow cylinder about its axis is
$I=M{R}^2=(3kg)(40\times 10^{-2}m{)}^2=0.48kg{m}^2$
Let a is the angular acceleration produced.
As $\tau =I\alpha$
$\therefore \alpha =\frac{\tau }{I}$
$=\frac{12Nm}{0.48kg{m}^2}$
$=25rad{s}^{-2}$