A ring of mass `m` and radius `r` rotates about an axis passing through its centre and perpendicular to its plane with angular velocity `omega`. Its kinetic energy is

Explanation

Suppose the disc is placed in xy-plane and is rotated about the z-axis. Consider an annular ring of radius r and of thickness dr, the charge on this ring
`dq=sigma(2pirdr)`
As the ring rotates with angular velocity `omega`, so the current
`i=(dq)/(dt)=(sigma(2pidr))/((2pi)/(omega))=sigmaomegardr`
The torque on the current loop
`vectau=ivecAxxvecB`.

Hence the torque on this annular ring `dvectau=i(dvecAxxvecB)`
`=sigmaomegardr(pir^(2)Bsin90^(@))=pisigmaomegar^(3)Bdr`
and `tau=pisigmaomegaBint_(0)^(a)r^(3dr)=(pisigmaomegaBa^(4))/4`