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,$\omega$ 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,$E_R=\frac{1}{2}M{V}^2$

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

Thus, the total kinetic energy $E$ of the rolling body is $E=E_R+E_r$

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

$=\frac{1}{2}M{V}^2+\frac{1}{2}M{K}^2{\omega }^2$ where $I=M{K}^2,K$ is the radius of gyration.

$=\frac{1}{2}M{R}^2{\omega }^2+\frac{1}{2}M{K}^2{\omega }^2$ where $V=R\omega$

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

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

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

Hence proved.