The moment of inertia of a flywheel having kinetic energy 360 J and angular speed of 20 $\mathrm{rad

Question

The moment of inertia of a flywheel having kinetic energy 360 J and angular speed of 20 $\mathrm{rad} / \mathrm{s}$ is\

Professor PC Thomas Classes · Answer

Given: Kinetic energy $\mathrm{K}=360 \mathrm{~J}$ Angular speed $\omega=20 \mathrm{rad} / \mathrm{s}$\ $	herefore \mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}$\ where $\mathrm{I}=$ moment of inertia\ or , $\mathrm{I}=\frac{2 \mathrm{~K}}{\omega^{2}}=\frac{2 	imes 360}{20 	imes 20}=1.8 \mathrm{~kg} \mathrm{~m}^{2}$.\