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.

Find the rotational kinetic energy of a ring of mass 9 kg and radius 3

1.	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.
Answer» Solution :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`