What force causes precession?

[test123]

Homework Statement

Take a rope attached to the ceiling, with a rod perpendicular to it. On the end of the rod there is a spinning bicycle wheel that can be treated as a disk. What force causes the wheel to precess, if any?

Relevant Equations

T = r x F
L = I*w
dL/dt = T

In class, the prof. gave the following explanation for why the wheel precesses:
Whenever the wheel is spinning, it has a angular momentum perpendicular to its face. The torque due to gravity changes the direction of the angular momentum, causing precession.
However, this does not seem like a good enough explanation for this problem, because:
1. There is not an explanation as to why the angular momentum must be perpendicular to the wheel.
2. It does not explain where the force that causes precession comes from.

Below is my attempt to link her explanation to a force:

Diagram:
View attachment 371335

Explanation:
1. I draw my FBD. The dL represents the small change in L.
2. I combine dL and L to get L'. Experimentally, we know that the wheel will try to flatten itself at with the plane of rotation. In order for this to occur, a force, F must apply.
3. This is more evidence for the force. We know that the precession causes a z component of L. We don't start with any z component, and the torque due to gravity has no z component, so we must have some force in the x-y plane causing it.

Attempt at identifying force:
Based on my diagram I am 90% sure that there is a force, however I can not identify it. At first I thought the bearing of the wheel would give the force, but I don't know how it would know what direction the wheel is spinning. My next idea was that the wheel would become unstable when not rotating like it normally would, allowing external forces to push it in the direction of L'. Once the chaotic forces eventually pushed it toward L', it would stabilize. However, I don't like this solution either because it isn't very scientific sounding and would require some degree of luck.

 

[A.T.]

..., so we must have some force in the x-y plane causing it.

Not really, except for the eventual small radial centripetal force to make the center of mass go in circles slowly. But you can make gyros that precess with a static center of mass.

In general, to change L in the xy-plane, you need a torque in the xy-plane, not forces in the xy-plane.

Maybe this will help you:

 

[Lnewqban]

Welcome!
Your attached diagram can't be seen for some reason.
Try Control+P to insert the image, please.

Regardless explanations or understandings, the phenomena is real.
It happens to airplanes with big propellers and to electric hand grinders.
There is resistance from any rotating body to move from its plane of rotation.

The fastest it spins and the greater its moment of inertia is, the greater the resistance to that.
Precession is just a manifestation of that resistance in the form of a reactive torque to the one applied.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/rotv2.html#rvec4

 

[kuruman]

However, this does not seem like a good enough explanation for this problem, because:
1. There is not an explanation as to why the angular momentum must be perpendicular to the wheel.

The angular momentum of the wheel is given by
$$\mathbf L = I\boldsymbol {\omega}$$ which says that the angular momentum vector is in the same direction as the angular velocity vector. What direction is this?

2. It does not explain where the force that causes precession comes from.

If you put the spinning wheel away from the Earth in free space (zero external forces), it will keep on spinning without precession. What force, do you think, causes the precession?

Spoiler

The Earth's gravity.

 

[test123]

Homework Statement: Take a rope attached to the ceiling, with a rod perpendicular to it. On the end of the rod there is a spinning bicycle wheel that can be treated as a disk. What force causes the wheel to precess, if any?
Relevant Equations: T = r x F
L = I*w
dL/dt = T

In class, the prof. gave the following explanation for why the wheel precesses:
Whenever the wheel is spinning, it has a angular momentum perpendicular to its face. The torque due to gravity changes the direction of the angular momentum, causing precession.
However, this does not seem like a good enough explanation for this problem, because:
1. There is not an explanation as to why the angular momentum must be perpendicular to the wheel.
2. It does not explain where the force that causes precession comes from.

Below is my attempt to link her explanation to a force:

Diagram:
View attachment 371335

Explanation:
1. I draw my FBD. The dL represents the small change in L.
2. I combine dL and L to get L'. Experimentally, we know that the wheel will try to flatten itself at with the plane of rotation. In order for this to occur, a force, F must apply.
3. This is more evidence for the force. We know that the precession causes a z component of L. We don't start with any z component, and the torque due to gravity has no z component, so we must have some force in the x-y plane causing it.

Attempt at identifying force:
Based on my diagram I am 90% sure that there is a force, however I can not identify it. At first I thought the bearing of the wheel would give the force, but I don't know how it would know what direction the wheel is spinning. My next idea was that the wheel would become unstable when not rotating like it normally would, allowing external forces to push it in the direction of L'. Once the chaotic forces eventually pushed it toward L', it would stabilize. However, I don't like this solution either because it isn't very scientific sounding and would require some degree of luck.

Here is my diagram, it doesn't let me edit my question anymore.

 

[test123]

Not really, except for the eventual small radial centripetal force to make the center of mass go in circles slowly. But you can make gyros that precess with a static center of mass.

In general, to change L in the xy-plane, you need a torque in the xy-plane, not forces in the xy-plane.

Maybe this will help you:

I understand why L changes, I just don't understand why the wheel attempts to be flush with the plane of rotation. In other words, why does the wheel rotate so that L' must be normal to it? The wheel could technically rotate at a slight angle right? Also I attached my diagram above to help explain my question. Essentially, if the gyro precesses, it must have some L in the z direction. This can only be caused by a force in the x-y plane because right hand rule will make the torque go in the z direction. That was essentially what I was asking.

 

[Janus]

Maybe this image taken from an old Physics text may help.

 

[test123]

The angular momentum of the wheel is given by
$$\mathbf L = I\boldsymbol {\omega}$$ which says that the angular momentum vector is in the same direction as the angular velocity vector. What direction is this?


If you put the spinning wheel away from the Earth in free space (zero external forces), it will keep on spinning without precession. What force, do you think, causes the precession?

Spoiler

The Earth's gravity.

The direction of the angular velocity would point at the next point in the precession. I understand that if the wheel rotated to the next point in precession, its angular velocity would be the predicted one. However there are many scenarios in which the wheel would have that angular velocity dictated by L'. Take my expected behavior for example:

The orbit would be strange, but it would still have the same omega value. What in this scenario would make the wheel correct to become normal to L'?

Edit: I realize the silhouette should intersect with the rod and be wider, assume that it is.

 

[Lnewqban]

Copied from:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-3-precession-of-a-gyroscope/

 

[test123]

Copied from:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-3-precession-of-a-gyroscope/

View attachment 371344
View attachment 371345

I understand the whole L changing thing, and the T dt = dL thing. What I don't understand is why the spindle of the top has to be aligned with L + dL. Take the first picture for example. Notice how the L vector is always through the spindle. What makes this true? What about physics makes it so that we can only rotate through that spindle?

 

[Lnewqban]

What I don't understand is why the spindle of the top has to be aligned with L + dL. Take the first picture for example. Notice how the L vector is always through the spindle. What makes this true? What about physics makes it so that we can only rotate through that spindle?

Could you please explain "the spindle of the top" a little more in detail?
Which other way do you believe it could rotate and why?

 

[A.T.]

If you put the spinning wheel away from the Earth in free space (zero external forces), it will keep on spinning without precession. What force, do you think, causes the precession?

Spoiler

The Earth's gravity.

If you put the spinning wheel in uniform gravity, but don't apply any other force (so it falls free), it will not precess.

If you put the spinning wheel away from gravity sources, but an apply a force, that doesn't go through the center of mass, it will precess.

So the 'cause of precession' you give makes no sense.

 

[test123]

Could you please explain "the spindle of the top" a little more in detail?
Which other way do you believe it could rotate and why?

By the spindle I meant the axis of the top that passes through the little beam at the top. I know the top could theoretically rotate around a crooked axis that is not through that beam. In my reply to kuruman I drew an example of this with the bicycle wheel rotating crooked.

 

[kuruman]

The direction of the angular velocity would point at the next point in the precession.

Please study and understand the diagram on the right that uses the right hand rule to find the direction of the angular velocity vector $\boldsymbol {\omega}$ and hence the direction of the angular momentum vector $\mathbf L.$ In the absence of the torque generated by gravity, the wheel will spin with constant angular momentum $\mathbf L$ directed along its axis and perpendicular to the plane of the wheel.

Now consider the spinning wheel in the gravitational field of the Earth which exerts force $\mathbf {W}$ (the weight) at the center of mass of the wheel which is located at position $\mathbf R$ from the point of support. The torque about point O is $\boldsymbol {\tau}=\mathbf R\times \mathbf W$ in a direction that is in the plane of the wheel and perpendicular to both $\mathbf R$ and $\mathbf W$ as shown in the figure on the right.

The change in angular momentum $\Delta \mathbf L$ over time interval $\Delta t$ is $\Delta \mathbf L=\boldsymbol {\tau}\Delta t.$ This means that one has to add a small element of angular momentum $\Delta \mathbf L$ to the existing $\mathbf L$ in the same direction as the torque, i.e. perpendicular to $\mathbf L$. Doing so, changes the common direction of the angular moment and the angular velocity along the spindle but not the magnitudes of $\mathbf L$ and $\boldsymbol {\omega}.$ This change of the common direction is illustrated in post #9 by @Lnewqban, see figure labeled (b).

To summarize, the force of gravity generates a torque about the point of support. This torque changes the direction of the angular momentum vector but not its magnitude.

 

[kuruman]

If you put the spinning wheel in uniform gravity, but don't apply any other force (so it falls free), it will not precess.

Oops, I forgot to specify that an off-center point of support is also needed. Thanks for pointing that out.

 

[Herman Trivilino]

Homework Statement: Take a rope attached to the ceiling, with a rod perpendicular to it. On the end of the rod there is a spinning bicycle wheel that can be treated as a disk. What force causes the wheel to precess, if any?
Relevant Equations: T = r x F
L = I*w
dL/dt = T

In class, the prof. gave the following explanation for why the wheel precesses:
Whenever the wheel is spinning, it has a angular momentum perpendicular to its face. The torque due to gravity changes the direction of the angular momentum, causing precession.
However, this does not seem like a good enough explanation for this problem, because:
1. There is not an explanation as to why the angular momentum must be perpendicular to the wheel.

That follows from the definition of angular momentum.

2. It does not explain where the force that causes precession comes from.

He states "torque due to gravity". So you don't see that a torque by definition requires a force? Or you don't see where the force of gravity comes from?

 

[test123]

View attachment 371347Please study and understand the diagram on the right that uses the right hand rule to find the direction of the angular velocity vector $\boldsymbol {\omega}$ and hence the direction of the angular momentum vector $\mathbf L.$ In the absence of the torque generated by gravity, the wheel will spin with constant angular momentum $\mathbf L$ directed along its axis and perpendicular to the plane of the wheel.






View attachment 371349Now consider the spinning wheel in the gravitational field of the Earth which exerts force $\mathbf {W}$ (the weight) at the center of mass of the wheel which is located at position $\mathbf R$ from the point of support. The torque about point O is $\boldsymbol {\tau}=\mathbf R\times \mathbf W$ in a direction that is in the plane of the wheel and perpendicular to both $\mathbf R$ and $\mathbf W$ as shown in the figure on the right.

The change in angular momentum $\Delta \mathbf L$ over time interval $\Delta t$ is $\Delta \mathbf L=\boldsymbol {\tau}\Delta t.$ This means that one has to add a small element of angular momentum $\Delta \mathbf L$ to the existing $\mathbf L$ in the same direction as the torque, i.e. perpendicular to $\mathbf L$. Doing so, changes the common direction of the angular moment and the angular velocity along the spindle but not the magnitudes of $\mathbf L$ and $\boldsymbol {\omega}.$ This change of the common direction is illustrated in post #9 by @Lnewqban, see figure labeled (b).

To summarize, the force of gravity generates a torque about the point of support. This torque changes the direction of the angular momentum vector but not its magnitude.

First let me clarify, I completely understand the concepts of angular momentum and that its time derivative is torque. I completely understand why the angular momentum vector must change direction. Both of these concepts were highlighted in my initial diagram, as L and the infintesimal dL, formed L'. I understand that the torque due to gravity will change the direction of L. What I am asking, is why the spindle has to point toward the angular momentum vector. Say you have a ruler which is rotating along itself (along the axis of its length). By the right hand rule it will have an angular momentum vector pointing along itself. However, if a torque was applied that changed the direction of L, the ruler would no longer rotate along itself. It would rotate like this (assume no external forces after the L has been changed):

Notice how after the angular momentum is changed, the ruler does not change its orientation to become inline with L. In the case of the question, however, we know that the rod attached to the wheel will change its orientation to become aligned with L. In my previous question I showed how I expected the rod to rotate, which experimentally is wrong. This is because the rod would trace a separate circle, whos plane is tangential to the precessional rotation, and whos center follows the precessional orbit. This would show that the rod vibrates up and down as it precesses, which is not the case. So I guess a better question would be, what thing (could be a force or something else) causes the rod to shift its orientation to the new angular momentum vector? As another example of this assume a football that the QB puts a heavy spiral on. This ball would have a large angular momentum vector, correct? Lets say the wind exerts a torque on this ball, shifting the angular momentum vector slightly to the side. The ball will then begin to point in this direction correct? It will not assume some wobbly rotation. What causes this effect?

 

[A.T.]

Oops, I forgot to specify that an off-center point of support is also needed

A force that produces a torque around the center of mass is all that is needed, regardless if there is gravity or not. So I wouldn't even say 'also' here.

 

[Lnewqban]

Say you have a ruler which is rotating along itself (along the axis of its length). By the right hand rule it will have an angular momentum vector pointing along itself. However, if a torque was applied that changed the direction of L, the ruler would no longer rotate along itself. It would rotate like this (assume no external forces after the L has been changed):
View attachment 371352

Could we compare your example to a bicycle wheel with its hub or bearings and natural axis of rotation not being perpendicular to the plane formed by the tire?
If so, nothing changes respect to the original experiment: the thing will still rotate around a verical axis.

 

[kuruman]

What I am asking, is why the spindle has to point toward the angular momentum vector.

You have it backwards. It is the angular momentum vector that points along the spindle. Consider an element of mass $dm$ on the circumference of the wheel at radius $\mathbf r$ having tangential velocity $\mathbf v$. Its angular momentum relative to the spindle is $$d\mathbf L = (dm)\mathbf r \times \mathbf v.$$ The right hand rule says that the angular momentum is perpendicular to both the radius and the velocity which puts it along the spindle. All other elements $dm$ likewise contribute angular momentum along the spindle because they describe circles perpendicular to the spindle. Therefore the total angular momentum is along the spindle.

 

[A.T.]

First let me clarify, I completely understand the concepts of angular momentum and that its time derivative is torque.

Then why did you ask about forces in the plane of angular momentum change?

What I am asking, is why the spindle has to point toward the angular momentum vector.

That's a more subtile issue. In general, there can indeed be different motions of a body that result in the same angular momentum.

However, when angular momentum changes smoothly, because a constant torque is applied, there is somtimes only one motion that transitions to the new angular momentum value. For example when it rotates around the axis of maximal moment of inertia, and angular momentum changes in direction only, which is the case for gyroscopes.

For a counter example see:
https://en.wikipedia.org/wiki/Tennis_racket_theorem

 

[test123]

You have it backwards. It is the angular momentum vector that points along the spindle. Consider an element of mass $dm$ on the circumference of the wheel at radius $\mathbf r$ having tangential velocity $\mathbf v$. Its angular momentum relative to the spindle is $$d\mathbf L = (dm)\mathbf r \times \mathbf v.$$ The right hand rule says that the angular momentum is perpendicular to both the radius and the velocity which puts it along the spindle. All other elements $dm$ likewise contribute angular momentum along the spindle because they describe circles perpendicular to the spindle. Therefore the total angular momentum is along the spindle.

Ok, I think I finally get it. Because L by definition is along the spindle, and we know L shifts because of torque, we know the spindle changes? So we are really just using what we know about L to to predict the location of the spindle? I'm sorry about the misleading question about force, I was trying to paraphrase a question proposed by my TA badly.

And would that make it so that the tension force from the bar provides the centripetal force for the z component of L?

 

[kuruman]

So we are really just using what we know about L to to predict the location of the spindle?

Yes, you can say that.

And would that make it so that the tension force from the bar provides the centripetal force for the z component of L?

You are not expressing yourself clearly. What bar? Do you mean the spindle? If so, you can say that it can exert centripetal tension on the center of mass of the wheel and cause it to rotate about the point of suspension with precession angular velocity $\boldsymbol {\omega}_p$ (not to be confused with the spinning angular velocity $\boldsymbol {\omega}.$) The z component of $\mathbf L$ is an abstract mathematical description and, as such, it cannot be affected by forces.

 

[A.T.]

Ok, I think I finally get it. Because L by definition is along the spindle,

$L_{total}$ is not necessarily exactly along the gyro axle (maximal moment of inertia axis) when the gyro is precessing. But if the $L_{axle}$ from the fast spin around the axle dominates $L_{total}$, then the axle follows the change of $L_{total}$, because that is the only motion that satisfies angular momentum conservation (taking into account all torque inputs) and kinetic energy conservation (assuming low dissipation).

But once the gyro spin rate gets slower, you have other possible motions that satisfy the above, so the gyro gets unstable, wobbles and drops at some point.

 

[Herman Trivilino]

I think I finally get it. Because L by definition is along the spindle, and we know L shifts because of torque, we know the spindle changes? So we are really just using what we know about L to to predict the location of the spindle?

By spindle "changing" you mean $\vec{\omega}$ of the rotating object changing direction? Then, yes.