Hot Wheels on a treadmill? Why didn’t anyone think of that earlier? As you can see in the video, this guy put a whole bunch of tiny cars on an inclined treadmill. After that, he just slowly increased the track speed. And there you have it: an instant demolition derby. It’s kind of fun to watch as some cars collide and then get thrown off the back of the treadmill.

But of course there are questions here—physics questions. I’m going to answer them for you.

Let’s imagine that these cars are just blocks on a low-friction inclined plane. (It’s simpler that way.) With that, I can show the forces acting on each car (block).

We have three forces. First there is the downward-pulling gravitational force. This depends on both the mass of the object (*m*) and the gravitational field (g = 9.8 Newtons/kilogram). That’s the easy force.

The next one is *FN*. This is the normal force. It’s an interaction between the car and the surface. The whole point of this force is to prevent the car from moving *through* the surface. It’s a force of constraint—that means it has just the right value to keep the block on the inclined plane.

Finally, there is the kinetic frictional force (*FK*). This force depends on two things: the magnitude of the normal force and a coefficient of friction that is set for the two materials that are interacting. For the actual Hot Wheels car, the kinetic friction is not between the wheels and the track, but rather between the wheels and the axles.

As an equation, the kinetic friction force can be modeled as:

So, the more these two surfaces—the car and inclined plane—are pushed together, the greater the kinetic frictional force. Note: It’s called kinetic friction because the two surfaces are sliding with respect to each other. If there is no sliding, then it would be static friction (and it would be modeled a little bit differently).

But what does this have to do with the motion of the car down the track? Since the car is constrained to only move in the direction down the plane, let’s set that as the x-axis, with the y-axis perpendicular to that. The first step is to find the normal force. The car has to have a y-acceleration of 0 meters/second2 otherwise it would accelerate off the track. With that, the normal force must be equal to the y-component of the gravitational force. (But not all of the gravitational force, since it’s not just in the y-direction.)

For the x-direction, things are a little different since the car actually accelerates down the incline. We can use Newton’s second law, which says that the net force in that direction is equal to the mass of the car multiplied by the x-acceleration. There are two forces that are pushing in the x-direction: the frictional force and a component of the gravitational force. Putting this together, I get the following:

If I put in the model for the kinetic friction force along with the expression for the normal force (from the y-direction), I can solve for the coefficient of kinetic friction in terms of the acceleration down the incline.

But what is this good for? Well, how about I actually find the coefficient of kinetic friction for an actual Hot Wheels car? It’s not difficult. I just need to roll a car down an incline and then find the acceleration (and the incline angle). Check it out:

Now I can use my favorite video analysis program (Tracker Video Analysis) to mark the position of the car in each frame of the video. Since there is a ruler on the incline, I can get position and time data as measured down the incline. Here’s what that looks like:

Since the car has a constant acceleration, this data should match the following kinematic equation:

By getting a parabolic fit to this data, the term in front of the t2 must match up to the (1/2), a term in the kinematic equation. This means that the acceleration of this particular car will be 0.248 m/s2. I can also measure the angle of the incline—I get 3.7o. Now I can just plug into my equation above to find the coefficient of kinetic friction (for *this* particular car) with a value of 0.039. That’s fairly low—it’s almost as low as the coefficient for ice sliding on ice. (That’s a good thing.)

OK, now we have an answer to the question: Why do some cars go faster? Well, if they have a lower coefficient of kinetic friction, the car will have a greater acceleration and speed up more.

If all the cars just went completely straight, this would be a boring race. Luckily, they don’t. There are many things that could make a car turn, but it’s probably due to one of two causes. First, the axle could be crooked. This would be very similar to turning the steering wheel on a real car.

The other reason would be different coefficients of friction for a pair of wheels. Yes, a Hot Wheels car has two axles, each with two independently rotating wheels. Let’s say that the friction on one side of the car is different than on the other. Here is a diagram showing the forces on a car (as seen from above) just showing the frictional forces on the front wheels. The same would be true for the back.

If the force on the left wheel is greater than the right, this will produce a net torque that will rotate the car to the right. However, for some turning cars this isn’t a problem. Let’s say a car turned to the left and is moving down the track in a diagonal path (not straight down). Now there will be a sideways force on the wheels. This will push a wheel on one side of the car into the axle and pull the other wheel away from the axle. It’s possible that this pushing and pulling of wheels can change the effective coefficient of kinetic friction such that the differential friction forces cause it to turn the other way and head directly back down the incline. These are the lucky cars that are more likely to win.

Let’s say a car turns left and moves to the left side of the treadmill until it comes in contact with the side wall. It can’t keep moving to the left since there’s a barrier there. If it hits at a shallow angle, the wall can exert a sideways force to turn it back “downhill.” However, if it keeps pushing against the sidewall, there will be a friction force between the side of the car and the wall. This frictional force will push up the incline and decrease the net force down the incline. If this wall frictional force is just the right amount, the net force will be zero and the car won’t accelerate. It will just stay in the same position.

In the analysis above, none of the forces depend on the speed of the treadmill. And if a car is moving straight down the track, then the treadmill speed doesn’t matter. But what about a car moving down at an angle? Clearly, in a real-life race with cars that can move in any direction, the track speed does matter. OK, so just assume we have two cars with the same speed (*v*) moving on a track. What happens when a car turns?

What are those labels on the velocities? It turns out that velocities are relative to our frame of reference. The two cars have velocities relative to the track. So, A-T is the velocity of car A with respect to the track. What about the velocity of the track? That is measured with respect to the reference frame of the ground (T-G). But what we want is the velocity of the cars with respect to the ground. For that, we can use the following velocity transformation. (Here is a more detailed explanation.)

Since velocity is a vector, both the magnitude and direction are important. For car A, the velocity of the car with respect to the track and the velocity of the track have the same magnitude, but opposite directions. When these two are added together, the velocity of car A with respect to the ground is the zero vector. (The two velocities perfectly cancel.) However, for car B, the velocity of the car with respect to the track and the track with respect to the ground are in different directions. They don’t add up to the zero vector, but instead give a velocity that is sideways and backwards with respect to the ground. That means the velocity of car B down the track will be less than car A. It will lose the race.

So, in this case, turning is losing. But if all the cars “won,” that just wouldn’t be any fun—would it?

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