The moment of inetia of a flywheel having kinetic energy 360 J and angular speed of `"20 rad s"^(-1)` is

Options

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

Answer

B

Explanation

Rotatational kinetic energy of flywheel, K = 360 J
Angular speed of flywheel, `omega="20 rad s"^(-1)`
Rotational kinetic energy, `K=(1)/(2)lomega^(2)`
`therefore" Moment of inertia, l"=(2K)/(omega^(2))=(2xx360)/((20^(2))=1.8kg-m^(2)`