# Newtonian Mechanics - How does something like a car wheel roll?

Ok so please bear with me here and what is almost certainly a really stupid question, partly because I don't quite know how to ask it.

When you have a wheel such as one attached to a car, a torque is applied to it from the engine. It's my understanding that the wheel would slide over the road and the car would not move if it were not for static friction which is of course the concept of rolling without slipping.

The problems I am having is when I consider what the magnitude of the force exerted on the road from the wheel would be, or specifically the equal and opposite force exerted back onto the wheel from the road via static friction. The reason for this is because I want to say that it is just the torque exerted on the wheel by the engine divided by the radius of the wheel but that seems to lead to a ridiculous conclusion. If I follow that logic, then the equal and opposite force exerted on the wheel by the road, multiplied by the radius of the wheel to get the counter-torque, is equal to the torque exerted on the wheel by the engine but in the opposite direction. That of course would mean that there is no net torque and the wheels would never move (unless the force were greater than that available from static friction but then it would just slip) which means that the car would never be able to move! So simply put, what am I missing that is almost certainly staring me in the face (again)? Clearly things roll and cars move so I know i'm going horribly and embarrassingly wrong somewhere.

Also, I often hear people say that static friction is the force responsible for propelling a car forward. How is that the case when it acts in the direction opposite to the direction that the wheels are trying to roll? That's sort of the same question but I think it may help highlight my misunderstandings of rotational motion in this area.

I never knew wheels could be so complicated! At least for me anyway.

The static friction will be pointing forwards not backwards. It comes into existence to counteract sliding when the torque pulls the wheel around - the wheel tries to push the ground backwards and static friction is the ground's equal and opposite response.

That static friction is then the only force on the wheel. Newton's 2nd law says that the wheel will then accelerate forward.

Update

Since this means that there is translational acceleration $a$, there must be angular acceleration $\alpha$ of the wheel as well. The well spins faster and faster while the car goes faster and faster. This is because of the so-called geometric bond between wheel and ground:

$$a=R\alpha$$

You can't have $a$ without $\alpha$ in this case.

This means that the torques do no balance out, which seemed to be the assumption that caused your whole confusion. The applied torque from the engine will not equal the torque from friction. It will be a bit bigger in order to also cause some angular acceleration. You must use $$\sum \tau =I\alpha$$ And not $$\sum\tau=0\;.$$

This equation along with the geometric bond and Newton's 2nd law should be enough equations for you to solve for the static friction.

• Right so the static friction will point forwards on the bottom of the wheel, but that part of the wheel is trying to move backwards due to the torque from the engine. That's part of the problem I am trying to understand. – An_African_Ape Jun 17 '17 at 16:39
• @An_African_Ape If you force the wheel to turn but the bottom point is stuck, then the centre will "fall" forward. Moving the wheel Centre means moving the car. This is essentially how the forward motion happens. – Steeven Jun 17 '17 at 16:53
• Your phrase "if you force the wheel to turn" implies a net torque and that is the main problem I am having. How can there be a net torque if the force exerted on the bottom of the wheel by static friction (equal and opposite to the force exerted onto the road by static friction) is equal to the torque from the engine on the wheel divided by its radius? It seems as though the static friction force from the road to the bottom of the wheel multiplied by the wheels radius is equal to the torque from the engine but in the opposite direction so they cancel out. Clearly that is not the case but how? – An_African_Ape Jun 17 '17 at 17:29
• @An_African_Ape Aha, I see your point now. You think that the applied torque will be cancelled out by the torque from static friction. This is not the case. When your car speeds up, it accelerates translationally and the wheel rotation accelerates as well (angular acceleration). The sum of torques is not equal to zero but to $\sum\tau=I\alpha$ in the same way that the sum of forces is not zero but equal to $\sum F=ma$. The reason we know for sure that the angular acceleration must be present is because of the connection $a=\alpha R$. – Steeven Jun 17 '17 at 19:30
• @An_African_Ape I have added an update to the answer explaining this. – Steeven Jun 17 '17 at 19:42

Torque due to frictional force is equal to applied torque on wheel only when the vehicle is moving at a constant velocity. That means the traction force completely balances all the resistance forces and hence there is no linear or angular acceleration.

Net force = Total Traction force-Total Resistance Force

Acceleration = 0 means Total Traction Force = Total Resistance Force

In normal case (neglecting resistance forces): $$F_tR=T-I\alpha$$ where $F_t$ is the traction force, $R$ is the radius of the wheel, $T$ is the applied torque on the wheel, $I$ is the moment of inertia of the wheel, and $\alpha$ is the angular acceleration of the wheel.

• You can format mathematical expression much more cleanly using mathjax. If you click edit you can see a simple example applied to your answer. – Kyle Oman Jan 8 '18 at 10:34

With problems like this, a diagram always helps:

In this front wheel drive car, the green arrow represents the direction in which the engine is turning the wheel.

The red arrow is the reaction force of the road onto the wheel - and that is a net external force on the car-plus-wheels. If there are no other forces of friction, this red force is what accelerates the car.

If you are interested in analyzing the torque more carefully, you have to account for the fact that if the car is accelerating, the inertial reaction of the center of mass (which is higher than the surface of the road) will tend to shift some of the weight to the rear wheels of the car; this ensures that there is once again no net torque, so the car doesn't flip over (if there is a net torque the angular momentum of the car will change; in fact if the car is accelerating there will of course be a change in the angular momentum of the wheels, so there is some small net torque needed).

I hope I didn't end up confusing you more...

It appears as though you're taking the origin to be the center of the wheel in your analysis. In this case, you have a torque both from the axle (applied very close to but not at the center, so nonzero) and from static friction at point of contact between the tire and road.

Instead, take as the origin the point of contact between the tire and the road. Then, the torque from static friction is zero, since $\vec r = \vec 0$, and the only torque on the tire is that supplied by the axle, which gives a nonzero net torque in the direction one would expect.

If you were to analyze it in your original frame, it's important to realize that the friction opposes the motion of the tire (consider the case of a crate sitting in a truck bed), not the torque, therefore $\tau_{axle} \neq \tau_{friction}$.