# Car Acceleration - Torque Balance

I am confused about a topic regarding torque and slippage.

I have been led to believe that for a car initially traveling at some constant speed: the torque at the axle of a wheel will be equal to the torque caused by static friction at the point of contact of the wheel. This is true so that the wheel doesn't angularly accelerate and so the point of contact does not slip with the road.

Now that's all well and good. However, before the axle provided a torque, the car was traveling at a speed v = w/r, which satisfied the no slip condition. However, afterwards the static friction force applied at the wheel speeds the car up (while maintaining angular velocity w). Now v is larger than the initial w/r, so the car must slip, right?

So didn't this initial process that was meant to stop the car from slipping, cause it to anyway?

I'm somewhat confused.

Dale
Mentor
2020 Award
(while maintaining angular velocity w).
This is the part that is getting you mixed up. The angular velocity increases as the car speeds up.

wrobel
When a wheel rolls with constant velocity and without slipping on horizontal plane then the friction force from the ground is equal to zero.

This is the part that is getting you mixed up. The angular velocity increases as the car speeds up.

Thanks for the response. I guess I don't get that. When I apply a torque at the axle, the static friction force is constrained to provide a torque (ideally) equal to that at the axle. So how does the angular velocity increase? Is this due to kinetic friction after the car slightly speeds up?

When a wheel rolls with constant velocity and without slipping on horizontal plane then the friction force from the ground is equal to zero.

Thanks to you too. However, I'm referring to the situation where a torque is applied to the axle, and not simply free rolling.

wrobel
When I apply a torque at the axle, the static friction force is constrained to provide a torque (ideally) equal to that at the axle.
it is not equal

it is not equal

Well why not. Doesn't the static friction apply a torque such that the the wheel doesn't rotationaly accelerate and therefore slip along the road? Doesn't this mean they have to be equal?

wrobel
I think that you should decide which statement of the problem you consider: with slipping or without slipping. And then to write equations of motion just for the wheel

Dale
Mentor
2020 Award
When I apply a torque at the axle, the static friction force is constrained to provide a torque (ideally) equal to that at the axle.
This is not generally true. There is nothing constraining those two torques to be equal. If they happen to be equal then the wheel will not be undergoing angular acceleration. Conversely, if the wheel is not undergoing angular acceleration then those two torques will be equal. But there is nothing constraining them to be equal and in the accelerating case they are not equal.

Is this due to kinetic friction after the car slightly speeds up?
In normal operation the friction is always static friction.

This is not generally true. There is nothing constraining those two torques to be equal. If they happen to be equal then the wheel will not be undergoing angular acceleration. Conversely, if the wheel is not undergoing angular acceleration then those two torques will be equal. But there is nothing constraining them to be equal and in the accelerating case they are not equal.

Well I suppose a better question would be: what decides what the force of static friction will be? I understand there is a maximum static friction, but how is it that when I accelerate my car the force of static friction and the unbalanced torque that creates an angular acceleration works out perfectly so that the velocity of the car matches the no slip condition of v = wr. (Since I notice I don't start slipping whenever I accelerate while driving)

I think I may have actually answered my own question. Is this explanation reasonable? When the car provides a torque at the wheel, to prevent the bottom of the wheel from slipping with the road, static friction provides a force, and therefore a linear acceleration that prevents this from happening. As a result the no slip condition continues to be met v = wr (and upon differentiating a = r (alpha) ). Correct?

I think my confusion stemmed from my robotics class last semester. For some reason we were told to find a maximum acceleration of robots by diving the maximum torque delivered to the axle from a motor by the radius to find the force of friction. This would mean the torque's were equal at the wheel and the axle, which as you said and I realize now to not be true. Any idea why they would have had us do this?

Dale
Mentor
2020 Award
what decides what the force of static friction will be?
The constraint is the no slip condition. The force of static friction takes whatever value is necessary to prevent slipping.

Dale
Mentor
2020 Award
to prevent the bottom of the wheel from slipping with the road, static friction provides a force,
Yes, this is correct.

I think my confusion stemmed from my robotics class last semester. For some reason we were told to find a maximum acceleration of robots by diving the maximum torque delivered to the axle from a motor by the radius to find the force of friction. This would mean the torque's were equal at the wheel and the axle, which as you said and I realize now to not be true. Any idea why they would have had us do this?
I don't know, it doesn't sound right to me either. I would have to work that out to see, but it would seem that the highest acceleration would be when there is no friction. Or do they mean linear acceleration rather than angular acceleration?

wrobel
Let us write the equations

We have a wheel of radius ##r## and ##S## stands for its center; ## M## is the torque applied to the axle; ##F## is a friction force, ##F\le F^*##
First assume that there is no slipping (##F<F^*##), then the equations of motion are
$$F=\frac{mrM}{J+mr^2}.$$ Thus the condition of non-slipping is $$\frac{mrM}{J+mr^2}<F^*.$$ If this condition is broken then we have equations for the case of slipping