Force on a rotating wheel/disc

[Buckethat_fisherlad]

TL;DR Summary: A force which stops a wheel/disc from rotating

Hi! I made a problem myself inspired by a typical mechanics problem, where you have two equal forces acting on each end of a wheel/disc and a third one that is suppose to stop the wheel/dic from rotating. So let me get into the details,

Let's say, we have a bicycle that is upside-down and you start to rotate the frontwheel with two equal forces on the distance R from the center of the wheel on opposite sides of the wheel (the wheel is also fixated to its center). It starts to rotate and the net-force is equal to zero (if we set the positive direction to one of the forces direction and the negative direction to the other force), whilst the torque is FR + FR = 2FR.
Now, we add a third force which is located on the arc of the wheel between the two equal forces, and call it G. The force, G, makes a 180 degree angle between itself and the radius of the wheel, and the force points away from the wheel (similar to if you had a rope attached to the arc of the wheel and you pull it to stop the wheel). Now comes the part where I'm stuck,

The net-force is now, F + G - F = 0 => G = 0 (?)
The torque is, FR + FR - G x R = 0 => lim v->180, 2F/(sin(v)) = G = infinity (?)

I know there's a detail I've missed but I really can't see it. Can any of you?

 

[Buckethat_fisherlad]

Also, ignore friction and other typical forces that can stop the wheel on its own

 

[Ibix]

Your analysis looks correct. The force $\vec G$ must be zero if the wheel is not to move and, since it is radial it cannot affect the angular momentum of the wheel.

I suspect you do not intend the force $\vec G$ to be radial. That is certainly not how a rope attached to the wheel would behave; it would be always horizontal (or that would be good enough for a first attempt anyway).

Also, you need a fourth force to stop the bicycle accelerating. This would normally be friction from the ground transmitted to the wheel through the forks, so would counter the acceleration of $\vec G$ without affecting its torque.

 

[Baluncore]

Welcome to PF.

I know there's a detail I've missed but I really can't see it.

The missing detail is the Free Body Diagram.
The direction of G is confused.

 

[Buckethat_fisherlad]

Here's a free body diagram of the problem. I had trouble uploading the image before...

 

[jbriggs444]

View attachment 346037
Here's a free body diagram of the problem. I had trouble uploading the image before...

Per the problem description you have four forces.

(the wheel is also fixated to its center)

 

[Buckethat_fisherlad]

Per the problem description you have four forces.

Alright, that is missleading. I tried to say "the wheel/disc rotates around it's center" but I wanted it to be clear that the body can't leave its place. It (can) spin freely (in the air or around its axis) but the force G stops it from spinning.

 

[Ibix]

Alright, that is missleading. I tried to say "the wheel/disc rotates around it's center" but I wanted it to be clear that the body can't leave its place. It (can) spin freely (in the air or around its axis) but the force G stops it from spinning.

Yes, so it needs a fourth force through its axis pointing to the left in order to stop it from moving.

You also need to think about how $\vec G$ is applied. If it's a rope attached to a point on the rim of the wheel, what will your diagram look like a moment later when the rope attachment point is (e.g.) 15° above where it's currently shown?

 

[Buckethat_fisherlad]

what will your diagram look like a moment later when the rope attachment point is (e.g.) 15° above where it's currently shown?

The force G becomes; G = 2F/sin(180-15), which is a very big force

 

[Ibix]

The force G becomes; G = 2F/sin(180-15), which is a very big force

So you are assuming that the force G is always applied horizontally. Fine.

But the point is that as the wheel rotates, the plce where the rope is attached moves with the wheel. So the force needed to counteract the torque of the two Fs varies with time. It is infinite in your initial situation, but it is finite at other instants.

 

[Baluncore]

Here's a free body diagram of the problem.

The wheel will rotate until it is stopped by the force G being applied to a fixed point on the wheel, in a direction other than radial.
You must redraw the FBD with the G at some other angle.

 

[Buckethat_fisherlad]

It is infinite in your initial situation, but it is finite at other instants.

You're saying the only case where the force G (which is always horizontal) can be finite is when the force [G] is at instances where the angle between G and the radius is non-zero/non-180-degree? Mathematically, I get it, but if I would do the experiment at home, I wouldn't need an infinite big force to stop the wheel from spinning.

 

[A.T.]

but if I would do the experiment at home,

What is stopping you from doing it?

 

[Buckethat_fisherlad]

The wheel will rotate until it is stopped by the force G being applied to a fixed point on the wheel, in a direction other than radial.
You must redraw the FBD with the G at some other angle.

I can do a variant of the problem,

We have the wheel spinning when the two forces F is applied to the wheel (shown in figure). Now we have a rod that is attached to the right wall and stops the wheel from spinning. Due to Newton's third law, the force G that the rod applies to the wheel should equal a horizontal "force-rope" as previously discussed. There's no friction between the rod and the wheel, so it goes for the center of the wheel and the stand (dashed line).

 

[Buckethat_fisherlad]

I can do a variant of the problem,
View attachment 346046
We have the wheel spinning when the two forces F is applied to the wheel (shown in figure). Now we have a rod that is attached to the right wall and stops the wheel from spinning. Due to Newton's third law, the force G that the rod applies to the wheel should equal a horizontal "force-rope" as previously discussed. There's no friction between the rod and the wheel, so it goes for the center of the wheel and the stand (dashed line).

In this scenario, the rod has to stop the centripetal force that the two F's contribute to. Then we get,
G = 2mw^(2)*R (where m is the mass of the wheel)
G in this case is not zero, and the angle between the rod and the radius is constant (180 degrees). So shouldn't this principle also apply to the case when there's a rope attached to the rim and pulling the wheel in a horizontal direction (and the placement is the same as previously discussed)?

 

[A.T.]

I can do a variant of the problem,
View attachment 346046
... a rod that is attached to the right wall and stops the wheel from spinning. ....There's no friction between the rod and the wheel...

If there is no friction, how would the rod stop the wheel from spinning?

the centripetal force that the two F's contribute to.

What are you talking about? The two F's are tangential, not centripetal.

 

[Buckethat_fisherlad]

If there is no friction, how would the rod stop the wheel from spinning?

Maybe I'm just yapping at this point. Sorry!

What are you talking about? The two F's are tangential, not centripetal.

I'm pretty sure you can apply the centripetal force to the system. Sure, omega is kind of misleading but I'm very sure that the velocity of the wheel is tangential along the rim.
So F = mv^(2)/R. And if I continue

G = 2mw^(2)/R

we would get G = 2mv^(2)/R. Is it still wrong?

 

[jbriggs444]

Maybe I'm just yapping at this point. Sorry!

Then stop yapping?

I'm pretty sure you can apply the centripetal force to the system.

Yes. You can apply any force you like. But if you apply a centripetal force, it will supply zero torque.

Sure, omega is kind of misleading but I'm very sure that the velocity of the wheel is tangential along the rim.
So F = mv^(2)/R.

The centripetal force required to keep an object in uniform circular motion is $F_c=\frac{mv^2}{R}$, yes.

Presumably any such force is supplied by the spokes of the bicycle wheel. What has this to do with a hypothetical braking mechanism?

we would get G = 2mv^(2)/R. Is it still wrong?

Yes. It is quite wrong.

 

[Buckethat_fisherlad]

Yes. It is quite wrong.

Then what is correct? Is the force G infinite as previously discussed or just a VERY big force?
It just feels absurd that the force has to be infinite to stop the wheel from rotating...

 

[haruspex]

the body can't leave its place.

because it is "fixed at its centre", which means the bicycle forks are exerting a horizontal force -G on the wheel.
While G is where you have drawn it, aligned with the centre of the wheel, there is also no net torque from the G and -G forces. At first, the wheel continues to turn at the same speed. But as it does so, the line of action of G changes height. (I assume it remains horizontal.) G and -G are no longer aligned, so now there is an increasing torque opposing that from F, -F. When the "G" torque magnitude exceeds that of the of the "F" torque the wheel will start to slow down.

I'm pretty sure you can apply the centripetal force to the system.

Then I am pretty sure you do not understand the meaning of centripetal.
The centripetal force on a moving body is that component of the net force on it which acts at right angles to its velocity.
Since the wheel has no overall velocity, there cannot be a centripetal force on it. If you just look at the part of the wheel where F is applied, F is in the same direction as its velocity. The acceleration of that part of the wheel is towards the wheel's centre, so there is a centripetal force on it, but that force comes from the structure of the wheel. (Partly from the spokes, maybe, but mostly from the tension in neighbouring parts of the rim.) There is no contribution from F.

 

[Ibix]

It just feels absurd that the force has to be infinite to stop the wheel from rotating...

If you only apply a radial force it can never slow the disc. It's like pressing down on a toy car: all you do is squash it, it doesn't start to accelerate horizontally. Similarly a radial force will never change tangential speed.

If you apply the force at the top of the wheel you will only need $2F$ to hold the acceleration to zero. You will need more at other angles because you are "wasting" some of your force trying to distort the wheel.

 

[Buckethat_fisherlad]

While G is where you have drawn it, aligned with the centre of the wheel, there is also no net torque from the G and -G forces. At first, the wheel continues to turn at the same speed. But as it does so, the line of action of G changes height. (I assume it remains horizontal.) G and -G are no longer aligned, so now there is an increasing torque opposing that from F, -F. When the "G" torque magnitude exceeds that of the of the "F" torque the wheel will start to slow down.

Alright, I'm seeing the bigger picture now. Thanks!

Since the wheel has no overall velocity, there cannot be a centripetal force on it.

I did not take acount for this. I see that now.

 

[Buckethat_fisherlad]

If you only apply a radial force it can never slow the disc. It's like pressing down on a toy car: all you do is squash it, it doesn't start to accelerate horizontally. Similarly a radial force will never change tangential speed.

If you apply the force at the top of the wheel you will only need $2F$ to hold the acceleration to zero. You will need more at other angles because you are "wasting" some of your force trying to distort the wheel.

So what you described is the same as what I drew in the following figure? If so then I totally understand that the force $G$ can't contribute to any movement, as long as $G$ is perpendicular to the roof of the car.

(I know there should be two normal forces acting from the ground that adds up to a net force equal to $G$).

 

[jbriggs444]

Partly from the spokes, maybe, but mostly from the tension in neighbouring parts of the rim

In a typical bicycle wheel, the spokes are in radial tension while the rim is under circumferential compression. The spoke tension is significantly larger than the centripetal force that would be required to keep the bits of rim in uniform circular motion at the rotation rates encountered in ordinary operation. You want to keep the spokes tight enough that they do not go slack when the wheel is deformed by the ground.

Obviously with only 24 to 36 spokes, the centripetal force acting on any small portion of rim away from a spoke will result from local internal stresses within the rim. These may include compression, tension and shear. The spokes are not purely radial, but are arranged at angles in groups of four. Right and left so that the wheel is rigidly [off-]centered relative to the hub. Spinward and anti-spinward so that the wheel rotates rigidly with the hub.

The situation with the tire is reversed. The air in the tire is under compression, obviously. Steel or Kevlar cords in the beads are under circumferential tension. From there, tension within the tire body keeps the rest of the tire body and tread attached.

If anyone cared enough to measure it, there would be (on average) a radial pressure gradient in the air, accounting for the centripetal force that keeps it circling within the tire.

 

[A.T.]

It just feels absurd that the force has to be infinite to stop the wheel from rotating...

Well, it's an absurd problem statement. You should complain to the one who posed it.

 

[Baluncore]

Then what is correct? Is the force G infinite as previously discussed or just a VERY big force?

As the direction of force G, approaches passing through the centre of the disc, the force needed to oppose rotation, increases exponentially, until you are prevented from dividing by zero. G is never infinite, it is simply undefined in your ideal model.

Radial and circumferential forces are mutually orthogonal. Neither shares a component of the other.