Motion of mass constrained to rail

[Fantasist]

I have a classical mechanics problem related to the constrained motion of a mass:

as per the attached graphics, assume a thin rigid rod of length L and mass m moving frictionless constrained to a rail such that is always locally perpendicular to the latter. The rod starts off with speed v=v0 in the straight path of the rail. My question is, what is the speed in the apex of the curved section (half a circle with radius R), and what is the speed after it is in the straight section again? There are no external forces acting.

Thanks.

 

[A.T.]

If it is frictionless and the rail is immobile then the velocity of the rod center is constant.

 

[siddharth5129]

There must be an external force on the rod. ( A central reaction force from the rail acting inward on the thin rigid rod. )

 

[mfb]

If it is frictionless and the rail is immobile then the velocity of the rod center is constant.

In the curve, the motion along the rail is not the only contribution to the motion - rotation will need some energy.
The velocity is lower, but goes back to the initial value once the rod leaves the curve - this is a result of energy conservation.

The force between rod and rail is internal.

 

[Dale]

You have $KE=\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$ during the curve. From that you can do a little algebra to get:
v=\frac{2\sqrt{3}R}{\sqrt{L^2+12 R^2}}v_0

 

[Fantasist]

In the curve, the motion along the rail is not the only contribution to the motion - rotation will need some energy.
The velocity is lower, but goes back to the initial value once the rod leaves the curve - this is a result of energy conservation.

The force between rod and rail is internal.

But the rod could only be slowed down and accelerated by a force along the rail (after all, the motion is essentially restricted to one degree of freedom). Where does this force come from?

Also, wouldn't the rod have to be slowed down and sped up instantaneously to the appropriate speed when it enters and leaves the curved section?

 

[Dale]

Where does this force come from?

It comes from whatever it is that accomplishes this:

constrained to a rail such that is always locally perpendicular

 

[mfb]

Also, wouldn't the rod have to be slowed down and sped up instantaneously to the appropriate speed when it enters and leaves the curved section?

Yes, in the same way your rail needs "infinite forces" to start the rotation. A more realistic setup would have a curvature that increases gradually, then the forces and accelerations stay finite.

 

[alva]

You have $KE=\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$ during the curve. From that you can do a little algebra to get:
v=\frac{2\sqrt{3}R}{\sqrt{L^2+12 R^2}}v_0

Wrong, I think.

$KE=\frac{1}{2} m v^2$

or

$KE=\frac{1}{2} I \omega^2$

whatever you choose. You can not apply both formulae to a spinning mass.

It's common sense that the velocity of the rod center is constant.

 

[mfb]

You can not apply both formulae to a spinning mass.

Wrong, you have to account for both contributions (or choose a frame where the mass is not moving, only rotating - the center of the curve).

 

[AlephZero]

But the rod could only be slowed down and accelerated by a force along the rail (after all, the motion is essentially restricted to one degree of freedom). Where does this force come from?

If you want the problem to be physically realistic (no "infinite forces" or "finite impulses applied instantaneously") then the track much have a continuously varying radius of curvature. Also, to apply a moment to the rod, there must be at least two points of contact, at different positions along the length of the track.

With those assumptions, the curvature at the different points of contact will be different, and the resultant forces on the rod can have a component that changes the "linear" speed of the rod along the track, as well as changing its angular velocity.

 

[sophiecentaur]

Yes, in the same way your rail needs "infinite forces" to start the rotation. A more realistic setup would have a curvature that increases gradually, then the forces and accelerations stay finite.

I don't think that would be necessary because the rod itself, by taking time to enter the curve, would not need an infinite impulse to get it turning. The front section would be deflected a bit, on entering the curve and the deflection would reach a maximum once the rear section was on the curve. (It's a low pass temporal filter, in effect). But, in any case, it's only a step change in acceleration, which doesn't imply an infinite force - just a step change in force.
I don't think the angular rotation is any more relevant than the lateral deflection of the path. With an infinitely rigid rail and infinite mass, no energy can be lost from the rod.

 

[AlephZero]

We can have different interpretations of exactly how to turn this into a physically realistic problem, but the bottom line is, the rod must be able to move along the track without any step changes in velocity (either linear or angular).

With an infinitely rigid rail and infinite mass, no energy can be lost from the rod.

I don't see why do you need "infinite mass" anywhere, unless you really meant "the track is fixed in an inertial coordinate system".

 

[Dale]

You can not apply both formulae to a spinning mass.

Not only can you apply both, you must apply both. This is common practice:

http://web.mit.edu/8.01t/www/materials/modules/chapter20.pdf eq 20.5.2
http://www2.cose.isu.edu/~hackmart/translation_rotation_work.pdf first two equations
http://physics.ucf.edu/~roldan/classes/phy2048-ch10_sp12.pdf section IX
Etc.

 

[Chestermiller]

I don't understand why there is so much shock over the concept that, in the problem as stated, the forces and moments at the instantaneous transition between the straight and curved sections have to be infinite. We deal with this all the time in problems involving impact. In this problem, we basically have impact forces and moments applied by the rail at the two transitions. Big deal.

Chet

 

[AlephZero]

I don't understand why there is so much shock over the concept that, in the problem as stated, the forces and moments at the instantaneous transition between the straight and curved sections have to be infinite. We deal with this all the time in problems involving impact. In this problem, we basically have impact forces and moments applied by the rail at the two transitions. Big deal.

I agree with that in principle, but the "big deal" is that the OP's question about the velocity of the rail is not answerable as an impact problem, unless you assume something about the energy loss during the impact.

And you still need to explain away how the the impact can change the rod's angular velocity, without friction forces acting.

But I suppose whoever invented the problem expected the "average student" to realize you could get an answer using conservation of energy and just do the algebra, instead of thinking too hard about whether that method was justifiable.

 

[sophiecentaur]

We can have different interpretations of exactly how to turn this into a physically realistic problem, but the bottom line is, the rod must be able to move along the track without any step changes in velocity (either linear or angular).


I don't see why do you need "infinite mass" anywhere, unless you really meant "the track is fixed in an inertial coordinate system".

I do. I was just putting it in a practical context.

 

[Chestermiller]

And you still need to explain away how the the impact can change the rod's angular velocity, without friction forces acting.

I think I can see how the impact can change the rod's angular velocity without friction acting. There is a "moment impulse" also. As you yourself indicated, there will be 2 points of contact in the curved section. Going into the curved section, there will first be one point of contact, and the, in rapid succession, a second point of contact. These normal contact forces will be very high at the transition, and produce the impulse moment. What is your opinion? Does this make any kind of sense?

Chet

 

[sophiecentaur]

Surely, as the rod goes round the curve, its tangential velocity must be less than when along the straight. This allows the total KE (Rotational + Translational) to be unchanged. There will be a compression force on the rod as it enters the curve (slowing it down (gently - as the whole transitions onto the curve; there needs to be no step function in force, I think) and a tension, speeding it up on exit.
We needn't have friction if we don't want it and neither do we need any instant changes of anything.

 

[Dale]

I don't understand why there is so much shock over the concept that, in the problem as stated, the forces and moments at the instantaneous transition between the straight and curved sections have to be infinite.

I agree completely. The problem as stated included a bunch of simplifying but unrealistic assumptions (thin, rigid, completely constrained). Those assumptions can be understood as approximations. Under those approximations you get an impulsive torque at either end. That impulsive torque is also unrealistic, but can similarly be understood as an approximation. If you want to get rid of it then you merely have to get rid of the unrealistic approximations that generated it.

 

[sophiecentaur]

I agree completely. The problem as stated included a bunch of simplifying but unrealistic assumptions (thin, rigid, completely constrained). Those assumptions can be understood as approximations. Under those approximations you get an impulsive torque at either end. That impulsive torque is also unrealistic, but can similarly be understood as an approximation. If you want to get rid of it then you merely have to get rid of the unrealistic approximations that generated it.

Can you help me with this, please? How is the sudden application of the centripetal acceleration on the front of the rod any more problematic than a lossless bounce, which is usually treated as an impulse?
I assumed that this rod is just held on the track at a point at both ends. Is that right?

 

[russ_watters]

Maybe I'm missing something, but I see no impact and not much complexity in identifying the energy loss:

If the rod is held onto the track with pins at front and back, then when it enters the curve, one pin is on the curve and the other is not. The rotational acceleration is finite and the force applied by the curve is not perpendicular.

 

[rcgldr]

The force applied by the curve is not perpendicular.

Since the curve is frictionless, the force at any point on the curve is perpendicular to the curve, but not perpendicular to the velocity of the center of mass of the rod.

 

[sophiecentaur]

Since the curve is frictionless, the force at any point on the curve is perpendicular to the curve, but not perpendicular to the velocity of the center of mass of the rod.

So that's ok, isn't it? I couldn't find anything wrong with this simple model and your point about the direction of the force could be what helps to make it all right.
I hate it when there seems to be no problem as far as I'm concerned and then someone with clout says there is a problem. haha

 

[russ_watters]

Since the curve is frictionless, the force at any point on the curve is perpendicular to the curve, but not perpendicular to the velocity of the center of mass of the rod.

Yes, that's what I was getting at; not perpendicular to the rod.

 

[Dale]

How is the sudden application of the centripetal acceleration on the front of the rod any more problematic than a lossless bounce, which is usually treated as an impulse?

I don't think that it is any more problematic. In fact, you can think of a lossless bounce of a thin rod which in certain circumstances can convert linear KE to or from rotational KE. Think of a thin rod dropped with high angular velocity, depending on how it hits the ground it can bounce sideways with greatly increased linear KE and reduced rotational KE. It is a very similar concept.

 

[sophiecentaur]

Yes, that's what I was getting at; not perpendicular to the rod.

I wouldn't expect it to be perpendicular to the rod. The rod has to slow down to provide rotational KE. I've lost count about who is holding which side of the argument, (Poor old sod). It all seems to fit together with no paradox nor any infinite forces.

 

[sophiecentaur]

I don't think that it is any more problematic. In fact, you can think of a lossless bounce of a thin rod which in certain circumstances can convert linear KE to or from rotational KE. Think of a thin rod dropped with high angular velocity, depending on how it hits the ground it can bounce sideways with greatly increased linear KE and reduced rotational KE. It is a very similar concept.

You could possibly replace the curve with two reflectors at 45° to the straight track and get the same effect - with the above argument. But I think you would need to get the spacings right.

 

[alva]

Not only can you apply both, you must apply both. This is common practice:

http://web.mit.edu/8.01t/www/materials/modules/chapter20.pdf eq 20.5.2
http://www2.cose.isu.edu/~hackmart/translation_rotation_work.pdf first two equations
http://physics.ucf.edu/~roldan/classes/phy2048-ch10_sp12.pdf section IX
Etc.

I looked at your 2 first references. They talk about other problem: a spinning body that has also transversal movement. This is not the case.

Perhaps we should do this experiment.

 

[Fantasist]

Yes, in the same way your rail needs "infinite forces" to start the rotation. A more realistic setup would have a curvature that increases gradually, then the forces and accelerations stay finite.

But there are no infinite forces here: in the straight section there is no force at all, in the curved section there is a (finite) centripetal force that keeps the rod on the rail. A force that changes from 0 to something finite shouldn't cause any problems, even if it changes instantaneously.

 

[Fantasist]

Not only can you apply both, you must apply both. This is common practice:

http://web.mit.edu/8.01t/www/materials/modules/chapter20.pdf eq 20.5.2
http://www2.cose.isu.edu/~hackmart/translation_rotation_work.pdf first two equations
http://physics.ucf.edu/~roldan/classes/phy2048-ch10_sp12.pdf section IX
Etc.

The case of a rolling cylinder is very different from the present one: as the third of your references makes it very clear (p.17), the constraint for the rolling cylinder is only caused by the friction force. You can actually consider an equivalent setup to the present here: assume a cylinder that is first sliding (not rotating) over a smooth surface, but suddenly the surface changes to a rough one. Now there will be a friction force that a) slows down the velocity of the center of mass and b) sets the cylinder in rotation due the torque involved. A steady state is only reached when the velocity at the contact point reaches zero (i.e. when the friction forces reaches zero and the cylinder is properly rolling). The crucial point is that for the sliding/rolling cylinder the friction force is directed parallel to the initial velocity i.e. parallel to the constrained path. In our case however there is, by assumption, no friction force. The only force is the centripetal force that keeps the rod on the rail in the shown configuration, but that force is strictly radial and has no components parallel to the rail i.e. can not do any work on the rod and thus not change its speed.

 

[Fantasist]

Surely, as the rod goes round the curve, its tangential velocity must be less than when along the straight. This allows the total KE (Rotational + Translational) to be unchanged.

Maybe one should take the potential energy into account as well (after all there is no energy conservation law for kinetic energies as such). When the rod enters the curved section it is subject to a centripetal force that corresponds to a negative potential energy (equal and opposite to the rotational kinetic energy). After all, it is commonplace that an object temporarily 'borrows' kinetic energy from potential energy and then gives it back again.

 

[Dale]

I looked at your 2 first references. They talk about other problem: a spinning body that has also transversal movement. This is not the case.

This is the same case. The center of mass is translating and the object is rotating about the center of mass. I don't know what you think makes it different. You must be focusing on some irrelevant aspect to conclude that it isn't the same. However, the point is that all three (and every other reference on the subject) clearly agree that you have to include both the translational and rotational KE.

In any case, it is a rather easy exercise to check. You can simply evaluate the KE of each differential element of the rod and see what you get.

KE = \int_M \frac{1}{2} v^2 \; dm
KE = \int^{R+L/2}_{R-L/2} \frac{1}{2} v^2 \frac{m}{L} \; dr
KE = \int^{R+L/2}_{R-L/2} \frac{1}{2} r^2 \omega^2 \frac{m}{L} \; dr
KE = \frac{1}{2} m R^2 \omega^2 + \frac{1}{24} L^2 m\omega^2
KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2

Just as I said above and as confirmed by all of the references I provided. You must include both the KE of the translation of the center of mass as well as the KE of the rotation about the center of mass.

 

[cjl]

But there are no infinite forces here: in the straight section there is no force at all, in the curved section there is a (finite) centripetal force that keeps the rod on the rail. A force that changes from 0 to something finite shouldn't cause any problems, even if it changes instantaneously.

An infinite torque is required though, since the object goes from 0 angular velocity to a constant positive angular velocity in zero time when it hits the curved section (since it is constrained to be always perpendicular to the rail).

 

[Dale]

The case of a rolling cylinder is very different from the present one: as the third of your references makes it very clear (p.17), the constraint for the rolling cylinder is only caused by the friction force. You can actually consider an equivalent setup to the present here: assume a cylinder that is first sliding (not rotating) over a smooth surface, but suddenly the surface changes to a rough one. Now there will be a friction force that a) slows down the velocity of the center of mass and b) sets the cylinder in rotation due the torque involved. A steady state is only reached when the velocity at the contact point reaches zero (i.e. when the friction forces reaches zero and the cylinder is properly rolling). The crucial point is that for the sliding/rolling cylinder the friction force is directed parallel to the initial velocity i.e. parallel to the constrained path. In our case however there is, by assumption, no friction force.

So what? The presence or absence of any external force in no way alters the correct formula for the KE. There is no doubt whatsoever that the formula I provided above for the KE is the correct formula for the problem as stated.

The presence or absence of any other force only determines whether or not KE is conserved, or if KE is converted to some other form of energy. It does not alter the correct expression for the KE.

The only force is the centripetal force that keeps the rod on the rail in the shown configuration, but that force is strictly radial and has no components parallel to the rail i.e. can not do any work on the rod and thus not change its speed.

No, that is not the only force. There is also a force which generates the torque to keep the rod oriented perpendicular to the rail. Every constraint that you introduce introduces another (generalized) force. The constraint along the path is one force, the constraint on the orientation is another.

 

[Dale]

Maybe one should take the potential energy into account

What potential energy? You specified that there are no external forces acting on the system, so there is no external potential energy. You also specified that it is rigid, so there is no elastic potential energy. If there is any available place to put any potential energy it is certainly not apparent from your description.

 

[sophiecentaur]

Maybe one should take the potential energy into account as well (after all there is no energy conservation law for kinetic energies as such). When the rod enters the curved section it is subject to a centripetal force that corresponds to a negative potential energy (equal and opposite to the rotational kinetic energy). After all, it is commonplace that an object temporarily 'borrows' kinetic energy from potential energy and then gives it back again.

How can there be an PE if nothing changes height or shape (everything is rigid)? If the situation has only KE then it has to be just KE that's conserved. As with many problems, you could allow for some finite (elastic) flexing and what you suggest could apply but would it help the analysis at all?

There have been a few 'assertions' about infinite forces and impossible conditions here but I have yet to read anything to convince me that this is not a simple situation that can be treated as ideal.

 

[A.T.]

In the curve, the motion along the rail is not the only contribution to the motion - rotation will need some energy.
The velocity is lower, but goes back to the initial value once the rod leaves the curve - this is a result of energy conservation.

Right. I overlooked that.

 

[sophiecentaur]

Why doesn't someone work out the answer with actual sums, then?

 

[mfb]

But there are no infinite forces here: in the straight section there is no force at all, in the curved section there is a (finite) centripetal force that keeps the rod on the rail. A force that changes from 0 to something finite shouldn't cause any problems, even if it changes instantaneously.

The "infinite force" would be needed exactly at the transition between the two parts.

Why doesn't someone work out the answer with actual sums, then?

DaleSpam did this in post 5.

 

[russ_watters]

The "infinite force" would be needed exactly at the transition between the two parts.

An infinite torque is required though, since the object goes from 0 angular velocity to a constant positive angular velocity in zero time when it hits the curved section (since it is constrained to be always perpendicular to the rail).

I disagree with you guys, but I don't think one of us is "wrong", I think that the problem is just ill-defined.

The rod has length and can't be touching the whole curve at once, so there is no one answer to what it means when it "hits the curve" or is "exactly at the transition". Your positions assume that the center of mass is the point that matters; when the center of mass hits the curve, the rod starts to rotate at the rotation rate determined by the speed around the curve. The rod stays on a tangent to the curve.

That's a valid possibility, but in reality a difficult one. There is no easy way to get a rod to be constrained to move in that way. Most similar situations such as trucks and trains do not move that way. They move in secants to the curve, not tangents. They start to rotate when the front part hits the curve and the rotation rate slowly increases until the back part hits the curve. There is no infinite force/acceleration.

 

[Dale]

I don't think one of us is "wrong", I think that the problem is just ill-defined.
...
That's a valid possibility, but in reality a difficult one. There is no easy way to get a rod to be constrained to move in that way. Most similar situations such as trucks and trains do not move that way. They move in secants to the curve, not tangents. They start to rotate when the front part hits the curve and the rotation rate slowly increases until the back part hits the curve. There is no infinite force/acceleration.

I wouldn't say "ill-defined", it is just a highly idealized and simplified approximation. It is well-defined in the sense that there is sufficient information to set up the equations and get a definite answer, but no real system could be built to exactly match the system.

As the front and the back get closer, the torque is increased. It is never infinite for any finite secant, but it can be made arbitrarily high for small secants. So the infinite torque for a thin rod is just seen as an approximation, which is valid or invalid to the extent that the original assumptions are valid or invalid.

 

[alva]

This is the same case. The center of mass is translating and the object is rotating about the center of mass. I don't know what you think makes it different. You must be focusing on some irrelevant aspect to conclude that it isn't the same. However, the point is that all three (and every other reference on the subject) clearly agree that you have to include both the translational and rotational KE.

In any case, it is a rather easy exercise to check. You can simply evaluate the KE of each differential element of the rod and see what you get.


Just as I said above and as confirmed by all of the references I provided. You must include both the KE of the translation of the center of mass as well as the KE of the rotation about the center of mass.

You are right and I am wrong. After thinking for a quite long time I realized that the formula
KE=1/2*I*ω2 refers to the spinning movement of the rod around its center of mass (and not to the spinning movement of the whole rod around the center of the rail, as I first thought).

 

[Dale]

I realized that the formula
KE=1/2*I*ω2 refers to the spinning movement of the rod around its center of mass

Yes, that is exactly correct. My apologies that I didn't make that clear to begin with and so caused unnecessary disagreement.

 

[Fantasist]

An infinite torque is required though, since the object goes from 0 angular velocity to a constant positive angular velocity in zero time when it hits the curved section (since it is constrained to be always perpendicular to the rail).

It's as much of a problem as the 'instantaneous' reversal of the momentum when you throw a ball against a wall. An infinite force/torque acting for a zero duration is theoretically perfectly well treatable (of course, in reality we dealing still with finite durations anyway, and then the force/torque is not infinite).
The rail (like the wall) will take up any changes in momentum (sudden or gradual), and in this case also any change in angular momentum. Assuming its mass is sufficiently large, it wouldn't be noticeable.

 

[Fantasist]

There is no doubt whatsoever that the formula I provided above for the KE is the correct formula for the problem as stated.

Maybe for the problem of the rolling cylinder, but the present case is distinctly different. I have created another graphic to illustrate this: if you take the path along the rail as a generalized coordinate s and plot the vertical distance z of both ends of the rod from this path against this, you get parallel lines i.e. there is no motion of any part of the rod in a dimension orthogonal to the path s, so there is no kinetic energy apart from the translational kinetic energy along the path s. The system has only 1 degree of freedom because of the constraint.
In contrast, for the rolling cylinder, there is an oscillatory motion in the z-direction, so there a 2 degrees of freedom (even though they are coupled).

No, that is not the only force. There is also a force which generates the torque to keep the rod oriented perpendicular to the rail.

A torque around the center of mass doesn't change the linear momentum of the latter.

To shed more light on the problem, one could consider an extension of the present thought experiment: assume in the curved section there is a point mass (same mass as the rod) at rest on the rail also constrained to move along the rail only. Assuming a totally elastic central collision of the two objects, what will happen if the rod hits the mass with velocity v? What will the velocities of the point mass and the rod along the rail be after the collision?

 

[Dale]

Maybe for the problem of the rolling cylinder, but the present case is distinctly different.

It isn't different. This isn't an assumption I am making. This is an obvious fact that I clearly derived in post 33.

I have created another graphic to illustrate this: if you take the path along the rail as a generalized coordinate s and plot the vertical distance z of both ends of the rod from this path against this, you get parallel lines i.e. there is no motion of any part of the rod in a dimension orthogonal to the path s, so there is no kinetic energy apart from the translational kinetic energy along the path s. The system has only 1 degree of freedom because of the constraint.
In contrast, for the rolling cylinder, there is an oscillatory motion in the z-direction, so there a 2 degrees of freedom (even though they are coupled).

So what? What has any of that got to do with the expression for the KE? Do you believe that the expression for KE is automatically $\frac{1}{2}m \dot q^2$ for any generalized coordinate, q?

A torque around the center of mass doesn't change the linear momentum of the latter.

No, but the force which provides the torque can. I.e. a single force can provide both torque and acceleration.

 

[sophiecentaur]

The "infinite force" would be needed exactly at the transition between the two parts.

DaleSpam did this in post 5.

DaleSpam wrote down the situation after the whole rod is on the track. That isn't the situation at the instant the front of the rod hits the curve. The rotation does not change instantaneously to its maximum, surely.

 

[Dale]

DaleSpam wrote down the situation after the whole rod is on the track. That isn't the situation at the instant the front of the rod hits the curve. The rotation does not change instantaneously to its maximum, surely.

For the thin rod approximation it does. A thin rod means that approximately the instant the front hits the curve so does the back, so it is rotating at its maximum approximately instantaneously. Again, it is just an approximation.

 

[sophiecentaur]

For the thin rod approximation it does. A thin rod means that approximately the instant the front hits the curve so does the back, so it is rotating at its maximum approximately instantaneously. Again, it is just an approximation.

I have been assuming the track is a monorail and the rod attached fore and aft.. Hence my ( and others ?) confusion. I don't remember seeing a diagram to help with this. Pictures can help a lot.
I now see where the question is coming from and that there is instantly a tension in the rod on the curve which I think needs some flexing to allow the problem to be solved.