Why do wheels not topple over while rolling?

In summary, the conversation discusses the stability of a wheel rolling on the ground and why it does not topple over when tilted to one side. The participants considered different factors such as torque, gravity, and friction to explain this phenomenon. One explanation is that there is no force pushing the angular momentum vector down, but rather a force diverting the force of gravity from pulling the wheel down. The wheel then tries to make a turn instead of toppling. The conversation also mentions assumptions about the wheel's rotation and friction.
  • #1
RoyalCat
671
2
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 [tex]\theta[/tex] 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 [tex]mgR\sin{\theta}[/tex]
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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  • #2
RoyalCat said:
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 [tex]\theta[/tex] 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 [tex]mgR\sin{\theta}[/tex]
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.
 
  • #3
diazona said:
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.
 
  • #4
RoyalCat said:
Well, this is an explanation on why leaning into a turn, makes you turn, isn't it?
Well, yeah, it does that...
RoyalCat said:
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.
 
  • #5
diazona said:
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?
 
  • #6
RoyalCat said:
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[tex]^{2}[/tex]. 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[tex]^{2}[/tex][tex]\geq[/tex][tex]\frac{AMgr^{3}}{C(C+Mr^{2}}[/tex]. 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=[tex]\frac{1}{2}[/tex]Mr[tex]^{2}[/tex] and C=Mr[tex]^{2}[/tex], and the stability condition becomes V[tex]^{2}[/tex][tex]\geq[/tex][tex]\frac{gr}{4}[/tex].
 
  • #7
Atropos said:
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[tex]^{2}[/tex]. 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[tex]^{2}[/tex][tex]\geq[/tex][tex]\frac{AMgr^{3}}{C(C+Mr^{2}}[/tex]. 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=[tex]\frac{1}{2}[/tex]Mr[tex]^{2}[/tex] and C=Mr[tex]^{2}[/tex], and the stability condition becomes V[tex]^{2}[/tex][tex]\geq[/tex][tex]\frac{gr}{4}[/tex].

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 [tex]\omega[/tex] with its center of mass traveling at a constant velocity [tex]v=\omega R[/tex] where [tex]R[/tex] is the radius of the disk.
This disk has a mass [tex]m[/tex]

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, [tex]P[/tex] 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 [tex]\vec L_p=mvR[/tex] 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 [tex]Fr'[/tex] where [tex]r'[/tex] 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 [tex]\Delta t[/tex] is equal to [tex]Fr'\Delta t[/tex] 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?
 
  • #8
Sorry to double post, but I still need an answer. :s
 
  • #9
im having issues with LaTeX
 
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  • #10
[tex]\vec{}[/tex]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 [tex]\omega[/tex] and constant translational velocity [tex]\vec{v}[/tex]. 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 [tex]\Delta t[/tex]. 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 [tex]\vec{L}[/tex] caused by the rotation is pointing towards the positive y direction and originates at C. An impulse F[tex]\Delta t[/tex] in the negative y direction is applied that causes the disc to roll at an angle [tex]\theta[/tex] from its original direction. ([tex]\theta[/tex] is measured clockwise from the x axis) This force creates a torque [tex]\vec{\tau}[/tex]pointed in the positive x axis, also originating from C.

Because the rate of change of the angular momentum [tex]\frac{d\vec{L}}{dt}[/tex] = [tex]\vec{\tau}[/tex], 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 [tex]\Delta t[/tex] the disc precesses through an angle [tex]\Delta\theta[/tex]. So the angular velocity of the disc around the axis of precession [tex]\omega_p =\frac{\vec{\tau}}{\vecL}[/tex]

?
 

1. Why do wheels not topple over while rolling?

Wheels do not topple over while rolling due to the concept of angular momentum. When a wheel is rolling, it has both linear and angular momentum. The angular momentum helps to keep the wheel in a stable position and prevents it from toppling over.

2. How does the shape of the wheel affect its stability?

The shape of the wheel plays a crucial role in its stability while rolling. A circular or cylindrical shape allows the wheel to distribute its weight evenly, making it less likely to topple over. This shape also helps to maintain a constant center of gravity, providing stability during the rolling motion.

3. Can a wheel topple over while rolling on an uneven surface?

Yes, a wheel can topple over while rolling on an uneven surface. In this case, the center of gravity of the wheel may shift, causing it to lose its stability and topple over. This is why it is essential to have a well-balanced and sturdy wheel for smooth rolling on uneven surfaces.

4. How does the weight of the load affect the stability of a rolling wheel?

The weight of the load can significantly impact the stability of a rolling wheel. If the load is too heavy, it can shift the center of gravity of the wheel, making it more prone to toppling over. It is important to distribute the load evenly on the wheel and choose a wheel with the appropriate weight capacity for the load.

5. Why do some wheels have treads or grooves on them?

The treads or grooves on wheels serve multiple purposes, one of which is to improve stability while rolling. These treads or grooves provide better traction and grip on the surface, preventing the wheel from slipping or toppling over. They also help to distribute the weight of the load more evenly, improving the overall stability of the wheel.

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