# Biophysics Problem 46

A uniform, solid sphere of mass \(\mathrm{m}\) and radius \(\mathrm{r}\) rolls down a hill of height \(\mathrm{20 \;m.}\) Ignoring energy losses due to friction, what is the sphere's speed at the bottom? (Recall for a sphere, the moment of inertia \(\mathrm{I = 2/5 \;M\; R^2}\)).

Assume the sphere starts from rest so that initially it has no kinetic energy. At the bottom of the hill, all its initial potential energy will have been converted to kinetic energy.

This problem must be solved using the concepts of conservation of energy. Remember, the sphere will have both translational and rotational kinetic energies when it gets to the bottom.

Write out the equation for conservation of energy for this system.

Your equation should be:

\(\mathrm{m \;g\; h = \frac{1}{2}\; m\; v^2 + \frac{1}{2}\; I\; \omega^2}\)

There seems to be, at first glance, a lot of unknowns in this equation. Let's see if we can get rid of some of them. Don't lose sight of our objective, which is to find the final velocity \(\mathrm{v.}\)

Remember, we were given \(\mathrm{ h}\) and we also know that \(\mathrm{\omega = v/r.}\) Substitute these into our energy conservation equation.

You should have written down:

\(\mathrm{m \;g\; h = \frac{1}{2}\; m\; v^2 + \frac{1}{2}\; \biggl( \frac{2}{5}\; m\; r^2 \biggr) \biggl(\frac{v^2}{r^2} \biggr)}\)

Notice that a lot of things will cancel out now. Rewrite the equation after cancelling out the appropriate variables, but do not rearrange the equation.

Your equation should now look like this:

\(\mathrm{g\; h = \frac{1}{2}\; v^2 + \frac{1}{5}\; v^2}\)

\(\mathrm{gh = \frac{7}{10}\; v^2}\)

Now calculate \(\mathrm{v}\) in \(\mathrm{m\; s^{-1}.}\)

You should have had:

\(\mathrm{v = \sqrt{ \frac{10\; g \;h}{7} } = \sqrt{ \frac{ 10 (9.8\; m\; s^{-2}) (20\; m)} {7} } = 16.7 \;m\; s^{-1} = 17\; m \;s^{-1}.}\)