

How fast is a rolling ball moving when it reaches the bottom of a hill? Just as in the case of a block sliding down a frictionless hill, the energy of a rolling ball is conserved. We can use energy conservation to determine what kind of rolling objects will reach the bottom first!

A ball and block at the same height start with the same gravitational potential energy: E_{p} = mgh. At the bottom of the hill their final energies will also be equal. Whereas the sliding block has only kinetic energy, the rolling ball has both linear kinetic energy (½mv^{2}) and rotational kinetic energy (½Iω^{2}). The rolling ball has a lower velocity because it has less linear kinetic energy than the sliding block. The difference is the rolling ball’s rotational kinetic energy.

In the picture above, four different objects with the same mass and diameter roll down identical ramps. Which one reaches the bottom first? To answer the question, consider your results from the investigation in which you rolled differently shaped objects down a hill. What did you find?

If you start with the conservation of energy equation, substitute for the angular velocity using ω = v/r, and rearrange the terms, you get
$$mgh={\scriptscriptstyle \frac{1}{2}}m{v}^{2}+{\scriptscriptstyle \frac{1}{2}}I{\omega}^{2}={\scriptscriptstyle \frac{1}{2}}m{v}^{2}+{\scriptscriptstyle \frac{1}{2}}I{\left(\frac{v}{r}\right)}^{2}={\scriptscriptstyle \frac{1}{2}}{v}^{2}\left(m+\frac{I}{{r}^{2}}\right)$$

Simplifying the expression further gives
$$mgh={\scriptscriptstyle \frac{1}{2}}{v}^{2}\left(m+\frac{mI}{m{r}^{2}}\right)={\scriptscriptstyle \frac{1}{2}}m{v}^{2}\left(1+\frac{I}{m{r}^{2}}\right)$$

From this equation you can solve for the velocity of the ball or cylinder when it reaches the bottom of the inclined plane:
$$v=\sqrt{\frac{2gh}{1+\left(I/m{r}^{2}\right)}}$$

As the moment of inertia I increases, the velocity of a rolling object will decrease. This makes sense because a higher moment of inertia means more energy is in rotational kinetic energy—leaving less for linear kinetic energy. The solid sphere reaches the bottom first because it has the smallest moment of inertia and therefore the highest translational kinetic energy!

Look at the equation above: The velocity at the bottom of the hill is a function of (I/mr^{2}). Since the moment of inertia I is proportional to mass, the ratio (I/mr^{2}) is independent of mass. This is a general result: The speed of a rolling object at the bottom of a hill is independent of its mass, in the same way as the speed of a dropped ball does not depend on its mass. For rolling, the speed depends on how the mass is distributed in the object, which is its moment of inertia, not on the mass itself.

Two 250 g, solid metal spheres are rolling down the same ramp. One sphere has a radius of 2 cm, while the other has a radius of 3 cm. Which one reaches the bottom first?

They reach the bottom at the same time. Both are solid metal spheres, so they each have a moment of inertia of I = ⅖mr^{2}. If you plug this moment of inertia into the equation for velocity, you see that you get, for both spheres, the same velocity:
$$v=\sqrt{\frac{2gh}{1+\left(I/m{r}^{2}\right)}}=\sqrt{\frac{2gh}{1+\left({\scriptscriptstyle \frac{2}{5}}m{r}^{2}/m{r}^{2}\right)}}=\sqrt{\frac{2gh}{1+{\scriptscriptstyle \frac{2}{5}}}}=\sqrt{\frac{10gh}{7}}$$
The distribution of the mass within the geometrical object matters, not the exact value of the radius or mass!
