Q. A string is wound round the rim of a mounted fly wheel of mass $20\ kg$ and radius $20\ cm$. A steady pull of $25\ N$ is applied on the cord. Neglecting friction and mass of the string, the angular acceleration of the wheel is
A) $50\ s^{- 2}$
B) $25\ s^{- 2}$
C) $12.5\ s^{- 2}$
D) $6.25\ s^{- 2}$
Explanation
(c) Here, $M = 20kg,\ R = 20cm = \frac{1}{5}m$
Moment of inertia of flywheel about its axis is
[image:7_image100.emf]
$I = \frac{1}{2}MR^{2} = \frac{1}{2} \times 20kg \times \left( \frac{1}{5}m \right)^{2}$
$= 0.4kgm^{2}$
As $\tau = I\alpha$
Where $\alpha$ is the angular acceleration
$\therefore\alpha = \frac{\tau}{I} = \frac{FR}{I} = \frac{25 \times \frac{1}{5}}{0.4} = \frac{5Nm}{0.4kgm^{2}} = 12.5s^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.