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