Write an expression for the KE of a body rolling without slipping with centre of mass as reference.

Explanation

The total kinetic energy (KE) as the sum of kinetic energy due to translation motion `(KE_("TRANS"))` and kinetic energy due to rotational motion `(KE_("ROT"))`
`KE=KE_("TRANS")+KE_("ROT")`
if the mass of the rolling object is `M`, the velocity of centre of mass is `v_(CM)`, its moment of inertia about centre of mass is `I_(CM)` and angular velocity is `omega` then
`KE=1/2 Mv_(CM)^(2)+1/2 I_(CM)omega^(2)`
With centre of mass as reference: The moment ofienrtia `(I_(CM))` of a rolling object about the centre of mass is `I_(CM)=MK^(2)` and `v_(CM)=R omega`. Here K is radius of gyration.
`KE=1/2 Mv_(CM)^(2)+1/2(MK^(2))(v_(CM)^(2))/(R^(2))`
`KE=1/2 Mv_(CM)^(2)+1/2Mv_(CM)^(2)((K^(2))/(R^(2)))`
`=1/2 Mv(CM)^(2)(1+(K^(2))/(R^(2)))`