Question

A rope of negligible mass is wound round a hollow cylinder of mass $3 \,kg$ and radius $40 \,cm$. If the rope is pulled with a force of $30 \,N$, then the angular acceleration produced in the cylinder is

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

Answer 1

The Correct Option is C

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