Q. The moment of inertia of a flywheel having kinetic energy $360\ J$ and angular speed of $20\ rad/s$ is
A) $18kgm^{2}$
B) $1.8kgm^{2}$
C) $2.5kgm^{2}$
D) $9kgm^{2}$
Explanation
(b) Given : kinetic energy $K = 360J$
Angular speed $\omega = 20\ rad/s$
$\therefore K = \frac{1}{2}I\omega^{2}$
Where $I =$ moment of inertia
$\Rightarrow I = \frac{2K}{\omega^{2}} = \frac{2 \times 360}{20 \times 20} = 1.8A\ kg\ m^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.