# How torque and friction cause wheel to roll

I apologize if this question has been answered before, but I did not find the explanation that I needed.

If a torque is applied to a wheel situated on a frictional surface, what forces cause the wheel to roll?

I know that if P is the contact point, the static frictional force counteracts the torque, making P stationary with respect to the surface.

What are the forces that cause the translation motion and how can I compute them?

Thank you.

A.T.
...the static frictional force ...

...What are the forces that cause the translation motion ...
You have mentioned it yourself.

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But I should be able to find a force that can be considered to be applied at the center of mass in order for a pure translational acceleration to exist. Is there any source that explains the entire rolling motion?

A.T.
But I should be able to find a force that can be considered to be applied at the center of mass in order for a pure translational acceleration to exist. Is there any source that explains the entire rolling motion?
But you don't have a pure translational acceleration, when a wheel starts rolling.

I am aware, but what is the force that "helps" with the translational acceleration and how I can find both these accelerations?

A.T.
I am aware, but what is the force that "helps" with the translational acceleration
You have mentioned it yourself. There is only one external force, if you just apply a pure torque at the axis.

and how I can find both these accelerations?
You have to write down the equations for linear and angular acceleration and solve them.

maybe it would help draw a diagram and the forces at play; for the moment, go ahead and replace the "torque" by an equivalent pair of forces, one above the center of the wheel, one below, both at a radius half of that of the wheel...

...don't forget to draw the frictional static force tangent to wheel right at the point where the wheel touches the ground (you can skip weight, if you want)

What forces combine to produce what?

So a force applied at the base of the wheel, tangent to it, is responsible for both a linear acceleration and torque?
In other words, a force applied to a free body (not at the center of mass) causes a linear acceleration (a=F/m) and a torque (F*r/I) as well? Thank you.

woahhh...not so fast.

o.k., allow me another attempt at explaining this.

The friction force is static, it does not move; but, at least, it also keeps the bottom of the wheel from moving, i.e., it creates some kind of pivot, if you will.

Once you represent the central torque as two separate forces, one above the center of the wheel and one below, you could think of these two forces as producing moments about the point where the wheel touches the ground (momentary pivot). Then, you will see that the moment produced by the force above the center of the wheel is greater than the one produced by the one below the center of the wheel....and, so, the summation of these two moments is not zero and a rotation is produced about the point currently touching the ground...this is a momentary pivot that comes about thanks to the static frictional force...

Another mental exercise could be to imagine what would happen if instead of a wheel, you have a gear and then you pivot the gear from a tangent point and apply torque at the center of the gear...what happens?...this should help you "forget" about linear translation, for a moment, and see the rotation motion alone...but a rotation about a point different than the center of the gear

Then, start increasing the size of the pivot to be a small gear...what happens?

Then increase the "pivot" gear to a large gear...what happens?

The increase the "pivot" gear to the size of the earth...

I understand. So because the bottom of the wheel does not move and the lever arm of the "top part of the torque" is longer than that of the "bottom part of the torque" the wheel rotates around that point.

As for the gear analogy, it is as if the wheel is made of very small flat parts. I think this is what you were trying to say.

I know this is not to the point, but the part about the force being applied to a free body (off center) is correct?
Thank you.

Hhhhmmm...I guess it depends how quickly you apply the force...meaning, if the force is applied instantaneously at its final constant velocity, it may produce quite a few oscillations/rotations of the body before it settles in its final state...if the force is applied gradually, starting from a velocity of zero and gradually increasing, the body may not oscillate as much...

...at the end, though, I am thinking the "torque" should disappear with the applied force and center of mass co-linear...and pure translation...at least with drag and friction present...you can this, not with a standing wheel, but with a disc flat on a smooth surface.

In other words, the applied force off the center of mass and the inertia at the center of mass do produce a torque that produce rotation, but because the applied force is moving in a straight line and there is nothing holding the center of mass....the rotation produced will tend to bring the center of mass onto the line of action of the applied force, reducing the "torque" at the same time...when the center of passes the line of action, the torque is zero and then switches sign and starts acting opposite direction to the rotation...I am thinking the center of mass will behave like a pendulum about the point of the applied force.

nasu
Gold Member
So a force applied at the base of the wheel, tangent to it, is responsible for both a linear acceleration and torque?
In other words, a force applied to a free body (not at the center of mass) causes a linear acceleration (a=F/m) and a torque (F*r/I) as well? Thank you.

Yes. You've got it.

For a rigid body, you can write Newton's second law as
∑F=m*a_cm
where a_cm is the acceleration of the center of mass and ∑F is the sum of all forces acting on the body, no matter where their point of application is.