Rotational Kinetic Energy

wIn the following scene from the Princess Bride (1987) our hero and heroine both tumble down the side of a mountain.

If we assume they roll without slipping how fast will they be going when they get to the bottom of the hill?

Let’s take the hill to be 100m high and the mass of the people to be 80kg each.  We can approximate them as cylinders with a radius of 0.5m.  If there is no friction then energy is conserved as they roll so

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2

Since they roll without slipping v=Rω. The moment of inertia of a cylinder is I =½mR²,  so we can write the above equation as

mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)v^2/R^2 = \frac{3}{4}mv^2

which means v² = 4/3 gh and v = \sqrt{4/3(9.8m/s^2)(100m)} = 36m/s.  Of course you can see from in the film that the end up with zero velocity at the end because there is friction which converts their kinetic energy into work.