Obtain an expression for total kinetic energy of a rolling body in the form `1/2MV^(2)[ 1 +(K^(2))/(R^(2))]`.

Explanation

Let M and R be the mass and radius of the body, v is the translation speed, `omega` is the angular speed and I is the moment of inertia of the body about an axis passing through the centre of mass.
Kinetic energy of rotation ,
`E_(R) = 1/2 Iomega^(2)`
Kinetic energy of translation,
`E_(T) = 1/2 Mv^(2)`
Thus, the total kinetic energy E of the rolling body is
`E = E_(T) +E_(R)`
`= 1/2 Mv^(2) + 1/2 I omega^(2)`
`= 1/2 Mv^(2) + 1/2MK^(2) omega^(2)`
( `:' I = MK^(2)` and K is the radius of gyration)
`= 1/2 MR^(2) omega^(2) + 1/2 MK^(-2) omega^(2)`
`( :' v = Romega)`
`:. E = 1/2 Momega^(2) (R^(2) + K^(2))`
`:.E = 1/2 M (v^(2))/(R^(2)) (R^(2) + K^(2))`
`:. E = 1/2Mv^(2) (1+(K^(2))/(R^(2)))`