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 center and perpendicular to its plane. (rpm is a unit of speed of rotation which means revolutions per minute)

Explanation

The rotational kinetic energy is, `KE=(1)/(2)Iomega^(2)`
The moment of inertia of the ring is, `I=MR^(2)`
`I=9xx3^(2)=9xx9=81kgm^(2)`
The angular speed of the ring is,
`omega=240"rpm"=(240xx2pi)/(60)"rad s"^(-1)`
`KE=(1)/(2)xx81xx((240xx2pi)/(60))^(2)=(1)/(2)xx81xx(8pi)^(2)`
`KE=(1)/(2)xx81xx64xx(pi)^(2)=2592xx(pi)^(2)`
`KE~~25920J because(pi)^(2)~~10`
KE = 25.920 kJ