The moment of inertia of a flywheel having kinetic energy 360 J and angular speed of 20 rad/s is

Options

A. 18 kg `m^(2)`
B. 1.8 kg `m^2`
C. 2.5 kg `m^2`
D. 9 kg `m^2`

Answer

B

Explanation

Given kinetic energy , K = 360 J
Angular speed , `omega = 20` rad/s .
As `K = (1)/(2) I omega^(2)` (where I = moment of inertia)
`therefore I = (2K)/(omega^2) = (2 xx 360)/(20 xx 20) = 1.8 kg m^2`