Q.Obtain an expression for total kinetic energy of a rolling body in the form $\dfrac{1}{2}MV^2\left(1+\dfrac{K^2}{R^2}\right)$.

A solid sphere rolling without slipping has both translational and rotational kinetic energy; total KE $= \tfrac{7}{10}mv^2$.

For a solid sphere, moment of inertia about its diameter is $I = \dfrac{2}{5}mR^2$. Rolling without slipping means $v = \omega R$.

Total kinetic energy = translational KE + rotational KE:

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2$

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Add the translational KE of the centre of mass to the rotational KE about the centre of mass, using $v=\omega R$ for rolling without slipping.

A body of mass $M$, radius $R$ and radius of gyration $K$ (so its moment of inertia about its own axis through the centre of mass is $I=MK^2$) rolls without slipping with the velocity of its centre of mass equal to $V$.

Since it rolls without slipping, its angular velocity about the centre of mass is

$\omega=\frac{V}{R}$

Translational KE (motion of the centre of mass):

$KE_{trans}=\frac12 MV^2$

Rotational KE (spin about the centre of mass):

$KE_{rot}=\frac12 I\omega^2=\frac12 (MK^2)\left(\frac{V}{R}\right)^2=\frac12 MV^2\frac{K^2}{R^2}$

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