Derive an expression for kinetic energy in pure rolling.

Explanation

(i) The total kinetic energy (KE) can be written as the sum of kinetic energy due to translational motion `(KE_("TRANS"))` and kinetic energy due to rotational motion `(KE_("ROT"))`
`KE = KE_("TRANS") + KE_("ROT")`
(ii) If the mass of the rolling object is M, the velocity of center of mass is `v_(CM)`, its moment of inertia about center of mass is `I_(CM)` and angular velocity is `omega`, then
`KE = (1)/(2)Mv^(2)._(cm)+ (1)/(2) I_(CM)omega^(2)`
(iii) With center of mass as reference :
The moment of inertia `(I_(CM))` of a rolling object about the center of mass is, `I_(CM) = MK^(2) and v_(CM) = R omega`. Here, K is radius of gyration.
`{:(KE = (1)/(2) Mv^(2)._(CM) + (1)/(2) (MK^(2)) (v^(2)._(CM))/(R^(2))),(KE = (1)/(2)Mv^(2)._(CM) + (1)/(2) Mv^(2)._(CM) ((K^(2))/(R^(2)))),(KE = (1)/(2) Mv^(2)._(CM) (1+ (K^(2))/(R^(2)))):}`
(iv) With point of contact as reference :
We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). If we take the point of contact as O, then,
`KE = (1)/(2)I_(0) omega^(2)`
(v) Here, `I_(o)` is the moment of inertia of the object about the point of contact. By parallel axis theorem, `I_(o) = I_(CM) + MR^(2)`. Further we can write, `I_(o) = MK^(2) + MR^(2)`. With `v_(CM) = R omega or `omega = (v_(CM))/(R)`
`{:(KE = (1)/(2) (MK^(2)+MR^(2)) (v^(2)._(CM))/(R^(2))),(KE = (1)/(2) Mv^(2)._(CM) (1 + (K^(2))/(R^(2)))):}`
(vi) K.E. in pure rolling can be determined by any one of the following two cases.
(a) The combination of translational motion and rotational motion about the center of mass. (or)
(b) The momentary rotational motion about the point of contact.