Obtain an expression for total kinetic energy of a rolling body in the form

`1/2 (MV^2)[1+K^2/R^2]`

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#### Solution

Let M and R be the mass and radius of the body, V is the translation speed, ω 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=1/2MV^2`

Kinetic energy of translation, `E_r=1/2Iomega^2`

Thus, the total kinetic energy 'E' of the rolling body is

E = E_{R} + E_{r}

`=1/2MV^2+1/2Iomega^2`

`=1/2MV^2+1/2MK^2omega^2 ......(I=MK^2 " and K is the radius of gyration")`

`=1/2MR^2omega^2+1/2MK^2omega^2 ........(V=Romega)`

`:.E=1/2Momega^2(R^2+K^2)`

`:.E=1/2MV^2/R^2(R^2+K^2)`

`E=1/2MV^2(1+K^2/R^2)`

Hence proved.

Concept: Definition of M.I., K.E. of Rotating Body

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