Total KE of a Rolling Body

The formula

A body rolling without slipping has two types of kinetic energy:

$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Using $v = R\omega$ (rolling condition) and $I = mk^2$ (where $k$ is radius of gyration):

$KE = \frac{1}{2}mv^2 + \frac{1}{2}mk^2 \cdot \frac{v^2}{R^2} = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$

The factor $\left(1 + \frac{k^2}{R^2}\right)$

This factor tells you what fraction of the total KE goes to rotation vs translation:

• $\frac{k^2/R^2}{1+k^2/R^2}$ goes to rotation
• $\frac{1}{1+k^2/R^2}$ goes to translation

For the same speed $v$, the hollow cylinder has the most total KE — more energy stored in rotation. The solid sphere has the least.

Application: rolling down an incline

Using energy conservation (height $h$, starting from rest):

$mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$

$v = \sqrt{\frac{2gh}{1+k^2/R^2}}$

The body with the smallest $k^2/R^2$ reaches the bottom fastest — the solid sphere wins any rolling race.

The rolling race order

From fastest to slowest:

1. Solid sphere ($k^2/R^2 = 2/5$): $v = \sqrt{10gh/7}$
2. Solid cylinder/disc ($k^2/R^2 = 1/2$): $v = \sqrt{4gh/3}$
3. Hollow sphere ($k^2/R^2 = 2/3$): $v = \sqrt{6gh/5}$
4. Hollow cylinder/ring ($k^2/R^2 = 1$): $v = \sqrt{gh}$

Note that this order is independent of mass, radius, or the angle of incline — it depends only on $k^2/R^2$.