# Why do wheels not topple over while rolling?

#### RoyalCat

Well, I've recently been introduced to the wonderful world of angular momentum and precession motion, and a question popped into mind.

Why, exactly, does a wheel rolling across the ground, not topple over even if tipped to one side?

I've tried approaching the problem by looking at the angular momentum vector and looking at what torque is acting to change it, but I haven't had much luck in the matter. Maybe if I could draw in 3D I'd succeed.

An approach involving measuring the torque about the center of mass proved quite fruitless, since gravity doesn't have a torque about the center of mass. I tried finding the torque applied by the static friction, but that didn't get me anywhere (Isn't the static friction just 0 for rolling motion?)

Trying to measure the torque about the point of contact got me a bit more ahead, but nothing came to fruition.
If I tilt the wheel by an angle $$\theta$$ on its side, in the direction of its angular momentum vector, then gravity applies a torque relative to the point of contact with the ground of $$mgR\sin{\theta}$$
The direction of this vector is perpendicular to the angular momentum, but I'm having trouble seeing how it contributes to stabilizing the wheel, if my analysis is at all correct.

I've tried running a google search for it, but that didn't help out either.

If someone could point me to a link, or explain it himself, I'd be very appreciative. :)

EDIT:
I think I've hit upon something.
Analyzing what happens when an external force is applied to the stable rolling wheel shows that there is a torque vector about the center of mass preventing the change in the direction of the angular momentum by the toppling over the point of contact?
Heh, that made a lot more sense in my head. Assistance is still needed. ^^;

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#### diazona

Homework Helper
Trying to measure the torque about the point of contact got me a bit more ahead, but nothing came to fruition.
If I tilt the wheel by an angle $$\theta$$ on its side, in the direction of its angular momentum vector, then gravity applies a torque relative to the point of contact with the ground of $$mgR\sin{\theta}$$
The direction of this vector is perpendicular to the angular momentum, but I'm having trouble seeing how it contributes to stabilizing the wheel, if my analysis is at all correct.
OK, you're right about that - but remember that the direction of the torque is perpendicular to the radius vector and to the applied force. (As well as to the angular momentum, in this case) If you're familiar with the right-hand rule, that's how you figure out the direction of the torque. So the torque about the point of contact is actually horizontal, since it has to be perpendicular to the vertical force of gravity. That means that the angular momentum vector is going to move horizontally, but it's not going to drop - that is, the wheel will curve around, but it'll keep rolling.

#### RoyalCat

OK, you're right about that - but remember that the direction of the torque is perpendicular to the radius vector and to the applied force. (As well as to the angular momentum, in this case) If you're familiar with the right-hand rule, that's how you figure out the direction of the torque. So the torque about the point of contact is actually horizontal, since it has to be perpendicular to the vertical force of gravity. That means that the angular momentum vector is going to move horizontally, but it's not going to drop - that is, the wheel will curve around, but it'll keep rolling.
Thanks for explaining that, it was much more clear when you talked about it. ^^;

Well, this is an explanation on why leaning into a turn, makes you turn, isn't it?

I still don't see how that torque pushes the angular momentum vector back up so the wheel doesn't topple.

#### diazona

Homework Helper
Well, this is an explanation on why leaning into a turn, makes you turn, isn't it?
Well, yeah, it does that...
I still don't see how that torque pushes the angular momentum vector back up so the wheel doesn't topple.
The point is that there's nothing pushing the angular momentum vector down in the first place, so it will stay at the level it starts at - you don't need anything to push it back up. If the wheel were going to topple, you would have to have something pushing the angular momentum vector down to cause the toppling.

#### RoyalCat

Well, yeah, it does that...

The point is that there's nothing pushing the angular momentum vector down in the first place, so it will stay at the level it starts at - you don't need anything to push it back up. If the wheel were going to topple, you would have to have something pushing the angular momentum vector down to cause the toppling.
Ah, so the point isn't that there's something to restore it to the upright position, but rather something to divert the force of gravity from pulling the wheel down, to just acting as a torque changing the direction of the angular momentum?
So the wheel just tries to make a turn instead of toppling, is that correct?

#### Atropos

Ah, so the point isn't that there's something to restore it to the upright position, but rather something to divert the force of gravity from pulling the wheel down, to just acting as a torque changing the direction of the angular momentum?
So the wheel just tries to make a turn instead of toppling, is that correct?
There's no force that restores the wheel to its upright position, but there is a force that resists any change in the position. Before we can figure out what that force is, we need to make a few assumptions.
(1) Because the wheel is rolling, it is rotating about a fixed axis, so it has angular velocity.
(2) The wheel is rolling on a rough suface, and there is also angular acceleration.
(3)And finally, because friction is present, energy is dissipated in the form of heat.
If these assumptions hold, then a law of gyroscopic systems is applicable that says: In the presence of energy dissipation, a body will end up rotating about the axis with the largest moment of inerta. In the case of a circular hoop, I=mr$$^{2}$$. This gyroscopic stability is what makes rolling objects resistant to SMALL distubances. (Obviously, if you decide to kick the wheel like it's a soccerball, the gyroscopic stability can't keep the wheel from toppling over.) As the angular velocity diminishes, the wheel becomes more and more unstable, until it stops and falls over. The actual mathematical analysis of the motion is wildly complex but, as a general rule, the motion is stable so long as
V$$^{2}$$$$\geq$$$$\frac{AMgr^{3}}{C(C+Mr^{2}}$$. V is the velocity, A and C are the principal moments of inertia, M is the mass, r is the radius, and g is acceleration due to gravity. For a hoop, A=$$\frac{1}{2}$$Mr$$^{2}$$ and C=Mr$$^{2}$$, and the stability condition becomes V$$^{2}$$$$\geq$$$$\frac{gr}{4}$$.

#### RoyalCat

There's no force that restores the wheel to its upright position, but there is a force that resists any change in the position. Before we can figure out what that force is, we need to make a few assumptions.
(1) Because the wheel is rolling, it is rotating about a fixed axis, so it has angular velocity.
(2) The wheel is rolling on a rough suface, and there is also angular acceleration.
(3)And finally, because friction is present, energy is dissipated in the form of heat.
If these assumptions hold, then a law of gyroscopic systems is applicable that says: In the presence of energy dissipation, a body will end up rotating about the axis with the largest moment of inerta. In the case of a circular hoop, I=mr$$^{2}$$. This gyroscopic stability is what makes rolling objects resistant to SMALL distubances. (Obviously, if you decide to kick the wheel like it's a soccerball, the gyroscopic stability can't keep the wheel from toppling over.) As the angular velocity diminishes, the wheel becomes more and more unstable, until it stops and falls over. The actual mathematical analysis of the motion is wildly complex but, as a general rule, the motion is stable so long as
V$$^{2}$$$$\geq$$$$\frac{AMgr^{3}}{C(C+Mr^{2}}$$. V is the velocity, A and C are the principal moments of inertia, M is the mass, r is the radius, and g is acceleration due to gravity. For a hoop, A=$$\frac{1}{2}$$Mr$$^{2}$$ and C=Mr$$^{2}$$, and the stability condition becomes V$$^{2}$$$$\geq$$$$\frac{gr}{4}$$.
I would like to dispute those assumptions. Wouldn't any friction we're dealing with here be static? What angular acceleration is there for a wheel rolling at a constant angular velocity?

Or are you saying that once I apply a torque to the wheel, kinetic friction appears? If so, I'd be happy if you could explain why that happens.

I greatly appreciate the insight, but a look at why gyroscopic systems resist small disturbances is what I'm looking for. :) What they do once the energy starts dripping it is a different matter altogether! Not in the least bit less fascinating, I must say, but not what I was asking about.

I milled it over in my head a bit more today, let's consider a disk rolling without sliding at a constant angular velocity $$\omega$$ with its center of mass traveling at a constant velocity $$v=\omega R$$ where $$R$$ is the radius of the disk.
This disk has a mass $$m$$

This is a part I hand-waved away, so I'd love to be corrected if I'm wrong.
The angular momentum about the point of contact, $$P$$ is equal only to the contribution of the linear motion, as for every mass element rotating on the disk, there is one rotating with a velocity that is exactly its opposite, with the same lever arm.
So $$\vec L_p=mvR$$ in the direction perpendicular to the plane of the disk.

Now, suppose we apply some impulse to a point other than the point of contact (I'll assume the static friction reaction force can keep the wheel from sliding sideways, and for the sake of simplicity, that the force exerted in the impulse was constant).
This force will have a torque about that point equal to $$Fr'$$ where $$r'$$ is its lever arm.

First, looking at the case where the force is perpendicular to the plane of the disk, the change in angular momentum brought on by the torque, acting for a time $$\Delta t$$ is equal to $$Fr'\Delta t$$ and it is pointed along the plane of the disk. That is to say, it is perpendicular to the existing angular momentum vector.

The force was trying to tip the disk over, but the torque it exerted, forces the disk to turn away from the toppling force, is this the key to the stability of rolling wheels? Or is this approach flawed and I should keep looking?

#### RoyalCat

Sorry to double post, but I still need an answer. :s

#### Atropos

im having issues with LaTeX

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#### Atropos

$$\vec{}$$Forget what I wrote in my last post. I think I found a way to solve this using only algebra. Let me sum up the problem as I understand it. Let me know if it's correct or not.

**A circular disc of uniform density, mass M and radius R is rolling without slipping on the ground. Initially it is rolling in a straight line with constant angular velocity $$\omega$$ and constant translational velocity $$\vec{v}$$. At some time t a force is applied in a direction perpendicular to the plane of the disc. The force is constant over some time interval $$\Delta t$$. How is the motion altered by the impulse F?**

The setup: The center of mass C is located in the geometrical center of the disc. As a reference frame, imagine that you are standing at the origin facing the positive x axis and that the disc is rolling away from you. The angular momentum $$\vec{L}$$ caused by the rotation is pointing towards the positive y direction and originates at C. An impulse F$$\Delta t$$ in the negative y direction is applied that causes the disc to roll at an angle $$\theta$$ from its original direction. ($$\theta$$ is measured clockwise from the x axis) This force creates a torque $$\vec{\tau}$$pointed in the positive x axis, also originating from C.

Because the rate of change of the angular momentum $$\frac{d\vec{L}}{dt}$$ = $$\vec{\tau}$$, the disc precesses about a vertical axis (an axis through the center of mass and parallel to the z axis.) The angular momentum is "chasing the torque." In time $$\Delta t$$ the disc precesses through an angle $$\Delta\theta$$. So the angular velocity of the disc around the axis of precession $$\omega_p =\frac{\vec{\tau}}{\vecL}$$

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