A circular disc of mass m and radius r is rolling over a horizontal table top with angular velocity `omega`. Prove that the total energy of the disc , `K = (3)/(4) m omega ^(2) r^(2)`.

Explanation

Total kinetic energy of the disc,
K= translational kinetic energy+ rotational kinetic energy
`=(1)/(2) mv^(2)+(1)/(2) Iomega^(2)`
Here v = linear velocity of the disc = `omegar`
`I`= moment of inertia of the disc about the perpendicular axis passing through its centre
`= (1)/(2)mr^(2)`
`:. " " K = (1)/(2) m (omegar)^(2)+(1)/(2).(1)/(2) mr^(2).omega^(2)`
=`(1)/(2) momega^(2)r^(2) +(1)/(4)momega^(2) r^(2) = (3)/(4) m omega^(2)r^(2).`