Find the rotational kinetic energy of a ring of mass `9kg` and radius `3m` rotating with `240` rpm about an axis passing through its centre and perpendicular to its plane.

Explanation

`KE=(1)/(2)Iomega^(2)`
`I=MR^(2)`
`I=9xx3^(2)=9xx9=81kgm^(2)`
`omega=240rpm=(240xx2pi)/(60)rads^(-1)`
`KE=(1)/(2)xx81xx((240xx2pi)/(60))^(2)=(1)/(2)xx81xx(8pi)^(2)`
`KE=(1)/(2)xx81xx64(pi)^(2)xx=2592xx(pi)^(2)`
`KE~~25920J :'(pi)^(2)~~10`
`KE=25.920kJ`