If you live on flat terrain like I do, you might not get a chance to experiment with your car coasting down hills in neutral. It’s kind of dangerous even if you can. But let’s say you’re on the top of your driveway and beginning from a stop you coast down to the street below. If the total drop is 10 feet, that gravitational potential energy will be converted to kinetic energy and you’ll end up going something like 17 miles per hour. To do the calculation yourself, just set the potential energy at the top equal to the kinetic energy at the bottom, and solve for the velocity:

The mass cancels; it doesn’t matter how heavy the falling thing is, its speed is completely determined by how far it falls.

But your car *doesn’t* go that fast when it gets to the bottom of the hill. It probably isn’t even very close. I’ve done the experiment myself and my car isn’t anywhere near the predicted speed. Either physics is wrong or we’re doing something wrong. Well, I won’t keep you in suspense. It’s not the physics.

Not all the energy goes from potential to kinetic. Some of it will be lost to friction – both in the tires, axle, and air resistance. But especially at the low speeds of coasting from a standing start, friction should be very small. Where else might the energy have gone? Well, entirely apart from the forward motion of the car the wheels are also spinning. If a giant picked the coasting car straight up, the wheels would keep spinning. That spinning is mass in motion, so it has kinetic energy. So actually, our expression which turns potential energy into kinetic energy needs another term:

That last term is the rotational kinetic energy. The letter *I* denotes the moment of inertia, which is a way to quantify the way mass is distributed in different shapes – in this case a tire. The Greek letter omega is the angular velocity in radians per second. One radian per second is 1/(2π) times the number of revolutions per second. Now we want to figre out what the moment of inertia of our tires are. This isn’t possible yet because we haven’t related the linear velocity v to the angular velocity omega. Fortunately there’s an easy way to relate the two, straight out of what you learned in trig.

Where r is in this case just the radius of the tire. Plus this into the equation and you’ll get a complete formula that accurately relates the various energies involved in this situation:

Now solve this for I, and you can actually experimentally calculate the rotational inertia of your tires.

Assuming I haven’t screwed up the algebra, plugging in your starting height and your final velocity will give you the moment of inertia of your times. Sadly the equation is not linear, so if you don’t start from a standstill you’ll have to redo this derivation in a somewhat more tedious manner.

Now if we’re looking at a situation where we have very lightweight tires (or heck, we’re a stone falling from the sky), I will equal zero. This is only possible if the two terms in parentheses are equal – but that’s exactly the situation described in the very first equation. Things are perfectly consistent, as expected. Neat, huh?

If you like, try the experiment and calculate the moment of inertia of your tires. Given the informal nature of the setup I’ve used English units above, but you’ll definitely want to convert everything to metric to make the math easier if you actually want to try it in real life. If you can do so safely, give it a shot!