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) Iomega^(2)`
the moment of inertin of the ring is `I=MR^(2)`
`I=9xx3^(2)=9xx9=81 kg m^(2)`
the angular speed of the ring is,
`omega=240 pm =(240xx2pi)/(60) "rad" s^(-1)`
`KE=(1)/(2)xx81xx((240xx2pi)/(60))=(1)/(2)xx81xx(8pi)^(2)`
`KE=(1)/(2)xx81xx64xx(pi)^(2)=2592xx(pi)^(2)`
`KE ~~25920 J " " :. (pi)^(2)=10`
`KE~~25.9250 KJ`