Question

A cord is wound over the rim of a flywheel of mass $ 20\, kg $ and radius $ 25 \,cm $ . A mass $ 2.5\, kg $ attached to the cord is allowed to fall under gravity. Calculate the angular acceleration of the flywheel.

• A. $ 25\, rad/s^2 $
• B. $ 20\, rad/s^2 $
• C. $ 10\, rad/s^2 $
• D. $ 5\, rad/s^2 $

Answer 1

Answer: (C)

Given $M = 20 \,kg$
$R = 0.25 \,cm$
$R = \frac{1}{4} m$
$F = mg$
$ = 2.5 \times 10 = 25\,N$
Torque $= F \times R$
$ = 25 \times \frac{1}{4}$
$ = \frac{25}{4}$
Moment of inertia $= \frac{1}{2} MR^2$
$ = \frac{1}{2}\times 20 \times (\frac{1}{4})^2$
$ =\frac{5}{8}$
We have $\tau = I\alpha$
$\therefore $ Angular acceleration of flywheel
$\alpha = \frac{\tau}{I}$
$ = \frac{25/4}{5/8}$
$ \alpha = 10\,rad/s^2$