Problem: A disc ($K^2/R^2 = 1/2$) and a ring ($K^2/R^2 = 1$), each of mass 1 kg and radius 0.2 m, roll down from height h = 2 m. Find the speed of each at the bottom. ($g = 10$ m/$s^{2}$)

Method: Energy conservation (rolling without slipping, so friction does no work).

$mgh = \frac{1}{2}mv_{cm}^2\!\left(1 + \frac{K^2}{R^2}\right)$

Solving for $v_{cm}$: $v_{cm} = \sqrt{\frac{2gh}{1 + K^2/R^2}}$

For the disc: $v_{disc} = \sqrt{\frac{2 \times 10\,\text{m/s}^2 \times 2\,\text{m}}{1 + 1/2}} = \sqrt{\frac{40\,\text{m}^2/\text{s}^2}{3/2}} = \sqrt{\frac{80}{3}\,\text{m}^2/\text{s}^2} \approx 5.16\,\text{m/s}$

For the ring: $v_{ring} = \sqrt{\frac{2 \times 10\,\text{m/s}^2 \times 2\,\text{m}}{1 + 1}} = \sqrt{\frac{40\,\text{m}^2/\text{s}^2}{2}} = \sqrt{20\,\text{m}^2/\text{s}^2} \approx 4.47\,\text{m/s}$

Result: The disc (5.16 m/s) arrives before the ring (4.47 m/s).

Note on mass and radius: Both bodies had mass 1 kg and radius 0.2 m, but these values cancelled in the formula. A 10 kg disc of radius 1 m would give the same speed — only shape ($K^2/R^2$) matters.