Velocity of Rolling Body at Bottom of Incline

The formula

A body starting from rest at the top of an incline of height $h$ rolls down without slipping:

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

Derivation using energy conservation

Since static friction does no work (contact point is at rest), mechanical energy is conserved.

Initial state (top, at rest): $KE = 0$, $PE = mgh$

Final state (bottom): $KE = \frac{1}{2}mv^2(1+k^2/R^2)$, $PE = 0$

Energy conservation:

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

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

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

Derivation using kinematics

Alternatively, using $v^2 = 2as$ where $s = h/\sin\theta$:

$v^2 = 2 \cdot \frac{g\sin\theta}{1+k^2/R^2} \cdot \frac{h}{\sin\theta} = \frac{2gh}{1+k^2/R^2}$

Same result — consistent.

Comparison with free sliding

Sliding (no friction): $v = \sqrt{2gh}$

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

Rolling is always slower than sliding from the same height. Some energy goes into rotation, leaving less for translation.

The rolling race — which reaches bottom first?

From fastest to slowest (smallest $k^2/R^2$ wins):

The solid sphere always wins. The result is independent of mass, radius, and angle of incline — it depends only on $k^2/R^2$.

Why mass and radius don't matter

Both $m$ and $R$ cancel out of the energy equation. The velocity at the bottom depends only on the geometry ($k^2/R^2$ ratio) and the height $h$. This is why you can predict the winner of a rolling race from the shape alone, without knowing the mass or size.