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

Explanation

The rotational kinetic energy is `KE = 1/2 I omega^2`
The momentum of inertia of the ring si `I = MR^2`
` I = 9 xx 3^2 = 9 xx 9 = 81 kg m^2`
The angular speed of the ring is
` omega = 240 "rpm" = (240 xx 2pi)/(60) rad s^(-2)`
`KE = 1/2 xx 81 xx ((240 xx 2pi)/(60) )^2 = 1/2 81 xx (8pi)^2`
`KE = 1/2 xx 81 xx 64 xx pi^2 = 2592 xx pi^2`
`KE = 25920 J therefore pi^2 = 10`
`KE = 25.9250 kJ`