Question

Obtain an expression for total kinetic energy of a rolling body in the form $\frac{1}{2}M{V}^2[1+\frac{K^2}{R^2}]$.

Answer 1

Let $M$ and $R$ be the mass and radius of the body, $V$ is the translation speed, to is the angular speed and $I$ is the moment of inertia of the body about an axis passing through the center of mass.

Kinetic energy of rotation, $E_R=\frac{1}{2}M{V}^2$

Kinetic energy of translation, $E_r=\frac{1}{2}I{\omega }^2$

$E=E_R+E_r$

$=\frac{1}{2}M{V}^2+\frac{1}{2}I{\omega }^2$

$\frac{1}{2}M{V}^2+\frac{1}{2}M{K}^2{\omega }^2$ ($I=M{K}^2$ and K is the radius of gyration)

$=\frac{1}{2}M{R}^2{\omega }^2+\frac{1}{2}M{K}^2{\omega }^2$ $(\because V=R\omega \omega )$

$\therefore E=\frac{1}{2}M{\omega }^2(R^2+K^2)$

$\therefore E=\frac{1}{2}M\frac{V^2}{R^2}(R^2+K^2)$

$\therefore E=\frac{1}{2}M{V}^2(1+\frac{K^2}{R^2})$ Hence proved.