Motion of mass constrained to rail

[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.

 

[Dale]

I have been assuming the track is a monorail and the rod attached fore and aft.

That is a different scenario than the one in the OP, which explicitly assumes a thin rod meaning no difference between fore and aft.

The frictionless track, the thin rod, and the rigid rod are simplifying assumptions. Are you familiar with the term "simplifying assumption"?

 

[sophiecentaur]

That is a different scenario than the one in the OP, which explicitly assumes a thin rod meaning no difference between fore and aft.

The frictionless track, the thin rod, and the rigid rod are simplifying assumptions. Are you familiar with the term "simplifying assumption"?

That's just being cheeky!

Yes, I get it now.

 

[Fantasist]

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

Not if the force attaches at a right angle to the rod, and especially not if you have opposite force vectors on both sides of the rod.

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?

No, not for any generalized coordinate. It depends obviously on the scenario i.e. on the specific constraints. And in this case, if I am not mistaken, the constraint is such that it does (as I tried to explain through the graphical illustration above).

But if you think that your derivation wraps up all possible scenarios, why don't you answer on this basis my above question what happens if the moving rod collides with a stationary point mass (same mass as the rod) in the curved section of the rail? What are the velocities of the rod and the point mass after the collision? An answer to this question should help to clarify the issue.

 

[A.T.]

Consider a rod, with the entire mass concentrated in two equal point masses, at the same distance from the rod center. Make that distance equal to the curve radius, so that the inner mass stops when the rod goes around the curve, and transfers all KE to the outer mass. The KE of the outer mass will double, so its speed will be √2*v₀. The speed of the rod center in the cruve is half of that v₀/√2, which is less than v₀.

You can generalize this, because due to the squared velocity, you always need more KE to accelerate by Δv, than you gain by slowing down the same mass from the same v by the same Δv. You can consider the uniform rod as a collection of such symmetrical point mass pairs.

 

[Dale]

Not if the force attaches at a right angle to the rod, and especially not if you have opposite force vectors on both sides of the rod.

Who said the force attaches at a right angle to the rod and is on both sides of the rod? Those weren't any of your constraints. Your constraints were that the COM followed the track and that the thin rod was perpendicular to the track at each point.

Under the constraints in the OP you get the above results. The forces from the track are whatever are needed to achieve those constraints on the motion. You cannot constrain both the motion and also the forces. Once you fix one then the other is determined.

No, not for any generalized coordinate. It depends obviously on the scenario i.e. on the specific constraints. And in this case, if I am not mistaken, the constraint is such that it does (as I tried to explain through the graphical illustration above).

Yes, you are mistaken. Your graphical illustration is irrelevant. For any rigid object you can always easily find generalized coordinates where the object is stationary in those coordinates.

So the fact that the plot of the generalized coordinates matches your illustration is completely uninformative. Any rigid body in any scenario undergoing any possible motion can match your illustration for some suitable choice of generalized coordinates.

Therefore, you have to derive the expression for the KE in the generalized coordinates some other way. For this problem, I showed how in 5 and again in 33.

But if you think that your derivation wraps up all possible scenarios

I never made that claim. My derivation wraps up THIS scenario without doubt, but certainly not all possible scenarios.

why don't you answer on this basis my above question what happens if the moving rod collides with a stationary point mass (same mass as the rod) in the curved section of the rail? What are the velocities of the rod and the point mass after the collision? An answer to this question should help to clarify the issue.

I fail to see the relevance of this new scenario to the scenario of the OP. You keep on bringing in irrelevant side issues. I don't know why.

If you want to solve this new system (I don't) then all you need to do is introduce another generalized coordinate representing the position of the point mass along the track. I assume that you still want KE conserved (rigid, frictionless, etc.). The KE of the system is the KE for the rod (given above) plus the KE for the point mass (easy to derive). The only hard part is handling the inequality constraint that the point mass must always be in front of the rod, but there are standard methods for doing that in Lagrangian mechanics.

 

[Fantasist]

Who said the force attaches at a right angle to the rod and is on both sides of the rod? Those weren't any of your constraints. Your constraints were that the COM followed the track and that the thin rod was perpendicular to the track at each point.

Under the constraints in the OP you get the above results. The forces from the track are whatever are needed to achieve those constraints on the motion. You cannot constrain both the motion and also the forces. Once you fix one then the other is determined.

Your suggestion was that the force of constraint changes the dynamics of the mass moving on the rail. This is impossible according to d'Alembert's principle. The force of constraint can't do any work. It just constrains. It would only be noticeable in the form of internal stress forces in the material.

Your graphical illustration is irrelevant. For any rigid object you can always easily find generalized coordinates where the object is stationary in those coordinates.

So the fact that the plot of the generalized coordinates matches your illustration is completely uninformative. Any rigid body in any scenario undergoing any possible motion can match your illustration for some suitable choice of generalized coordinates.

You make it sound as if I had plucked the general coordinate out of thin air. It strictly relates to the physical constraint here. The rail and rigidly connected rod nail down the path of any mass element of the rod to be a curve parallel to the rail. So taking the general coordinate as the path along the rail is not only the natural thing to do here but even what you should do unless you want to get into difficulties.

Therefore, you have to derive the expression for the KE in the generalized coordinates some other way. For this problem, I showed how in 5 and again in 33.

You have assumed a rotational degree of freedom that simply isn't there. 'Degree of freedom' means just that: the object is free to move in this dimension. But in this case it isn't because of the constraint. You can't rotate the rod without translating it. The system has only 1 degree of freedom. The rod may get rotated but it is not rotating. If you assume it is rotating you count something twice here.

I fail to see the relevance of this new scenario to the scenario of the OP. You keep on bringing in irrelevant side issues. I don't know why.

It is not irrelevant. An elastic collision can conveniently separate the linear momentum from the rotational momentum. If you assume the rod with linear momentum p hits elastically a point mass with the same mass at rest on the rail, the conservation laws for the linear momentum and kinetic energy require that after the collision

1) the point mass has linear momentum p
2) the rod has linear momentum 0

But because the orientation of the rod is constrained to its linear motion, 2) implies also

3) the rod has intrinsic angular momentum 0

So since the total energy is p^2/2m after the collision, energy conservation requires that it is also p^2/2m before the collision, i.e. there was no intrinsic rotational energy of the rod present in the first place.

 

[Dale]

Your suggestion was that the force of constraint changes the dynamics of the mass moving on the rail. This is impossible according to d'Alembert's principle. The force of constraint can't do any work. It just constrains.

I agree fully with d'Alembert's principle. The KE is conserved, therefore the constraint force does no work.

Your suggestion is the one which violates d'Alembert's principle since you would have the constraint do work to increase the KE around the bend so that the COM velocity is unchanged.

You make it sound as if I had plucked the general coordinate out of thin air. It strictly relates to the physical constraint here. The rail and rigidly connected rod nail down the path of any mass element of the rod to be a curve parallel to the rail. So taking the general coordinate as the path along the rail is not only the natural thing to do here but even what you should do unless you want to get into difficulties.

I agree that your s is a very natural coordinate, but that is not relevant to the argument. The naturalness of the coordinate is not in discussion, only the expression for the KE in terms of the coordinate.

You have assumed a rotational degree of freedom that simply isn't there. 'Degree of freedom' means just that: the object is free to move in this dimension. But in this case it isn't because of the constraint. You can't rotate the rod without translating it. The system has only 1 degree of freedom. The rod may get rotated but it is not rotating. If you assume it is rotating you count something twice here.

Again, completely irrelevant. It doesn't matter if it is free to rotate or constrained to rotate. Either way it undergoes rotational motion wrt any inertial coordinate system and therefore has rotational KE. Remember, s is a generalized coordinate not an inertial coordinate. So you cannot simply assume that pure translation wrt s implies an absence of rotation wrt inertial coordinates.

My derivation in 33 starts with the known expression for KE in an inertial coordinate system and then calculates the expression in the generalized coordinates. If you disagree with my conclusion then please point out the exact step where I made my mistake.

It is not irrelevant. An elastic collision can conveniently separate the linear momentum from the rotational momentum.

Momentum isn't conserved in this problem, just energy. Again, I fail to see the relevance, but if you wish to work the problem, please be my guest. I gave instructions for doing so previously.

 

[Fantasist]

f you disagree with my conclusion then please point out the exact step where I made my mistake.

The mistake is in your last step

$$KE = \frac{1}{2} m R^2 \omega^2 + \frac{1}{24} L^2 m\omega^2$$
$$KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$

where effectively you dropped the constraint.
Note that in the first equation the same angular frequency $\omega$ appears in both terms, which means that if one term gets larger (due to an increase of $\omega$), the other would have to get larger as well, so it would violate energy conservation. Your final equation just hides this fact and pretends that you can compensate the increase/decrease of one term by correspondingly adjusting the other. This would be justified if you had actually two degrees of freedom (e.g. if the rod would only accidentally have a synchronous rotation) but not if the orientation is rigidly fixed relatively to the path.

 

[mfb]

where effectively you dropped the constraint.

This does not change the validity of the equations.

2x=2x remains true even if I define y=x and replace it to get 2x=x+y as modified equation.

Note that in the first equation the same angular frequency ω appears in both terms, which means that if one term gets larger (due to an increase of ω), the other would have to get larger as well, so it would violate energy conservation.

This just shows ω is constant for constant curvature. A reduced curvature radius corresponds to an increased angular velocity - but not in a linear relation due to the second term, which is exactly the effect we are looking at.

 

[Dale]

The mistake is in your last step
where effectively you dropped the constraint.

Where is the mistake?
$v=R\omega$
and
$I=\frac{1}{12}L^2 m$

So the last equation follows from the previous by simple algebraic substitution. This is very basic and valid algebra. The math doesn't care about whether two variables are constrained by some separate equation. The substitution remains valid regardless.

Note that in the first equation the same angular frequency $\omega$ appears in both terms, which means that if one term gets larger (due to an increase of $\omega$), the other would have to get larger as well, so it would violate energy conservation.

Obviously, if one gets larger then the other would indeed have to get larger as well. I don't know why you think this is a mistake.

This does not in any way imply a violation of energy conservation. The only way that one could get larger is by some external force doing work on the system. In such a circumstance the external force would have to increase both terms, as shown by the equation.

The only way that you would get a violation of conservation of energy here is if ω were to change without any external work. If that were to happen then the conservation of energy would still be violated even if we used only the translational KE term. So the expression does not indicate anything amiss, and I don't know why you think it does.

 

[Fantasist]

Where is the mistake?
$v=R\omega$
and
$I=\frac{1}{12}L^2 m$

So the last equation follows from the previous by simple algebraic substitution. This is very basic and valid algebra. The math doesn't care about whether two variables are constrained by some separate equation. The substitution remains valid regardless.

The substitution is valid, but not your conclusions from it. Your argument was that if in

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

the second (rotational) term on the right hand side gets larger due to an increase in $\omega$, the translational velocity v has to get smaller to keep the total energy KE constant. But from the original equation

$$KE = \frac{1}{2} m R^2 \omega^2 + \frac{1}{24} L^2 m\omega^2$$

you can see that this is not possible as the same angular frequency $\omega$ appears in both terms (and R does not change at the moment when the rod hits the curved section).

From this you can conclude that it was a mistake in the first place to assume a separate rotational energy term when there is no rotational degree of freedom (because of the constraint).

 

[Dale]

The substitution is valid

If the substitution is valid then the derivation is correct and that is the correct equation for the KE during the curve. You cannot have it both ways.

Either the math is valid and the derivation is correct or there is a mistake in the math. You cannot admit that the math is valid and yet logically claim that the result is wrong.

and R does not change at the moment when the rod hits the curved section

Yes it does. It is infinite before the curve and finite after.

 

[A.T.]

From this you can conclude that it was a mistake in the first place to assume a separate rotational energy term when there is no rotational degree of freedom (because of the constraint).

See my post #54. It uses only linear KE, and arrives at the same conclusion (center must slow down in the curve).

 

[sophiecentaur]

Just lost a half made-up post. Try again.

Can one of you chaps help me with your disagreement, please?

Is it, basically, that momentum conservation and energy conservation are not both followed in your model? Can you not just equate KE before and after and see how the speed along the rail is changed due to the change of linear KE into some rotational KE? The point of transition onto the curve should not be a fundamental problem because any change of curvature would introduce the same problem - however great the radius.

Take time off to give me a heads up on the story so far. (I'll hand round orange segments to suck)

 

[AlephZero]

The substitution is valid, but not your conclusions from it...

... you can see that this is not possible as the same angular frequency $\omega$ appears in both terms (and R does not change at the moment when the rod hits the curved section).

From this you can conclude that it was a mistake in the first place to assume a separate rotational energy term when there is no rotational degree of freedom (because of the constraint).

No, your mistake is trying to describe the KE using variables that don't make sense. On the straight part of the track, $\omega = 0$ and $R$ is "infinite". You are trying to multiply "0 by infinity" to get the finite value of the KE.

The problem goes away if you write the KE in terms of $R$ and $v$, not $R$ and $\omega$.

$$\begin{aligned}KE &= \frac{1}{2} m v^2 + \frac{1}{24} mL^2 \frac{v^2}{R^2}\\
&= \frac{1}{2}mv^2\left(1 + \frac{L^2}{12R^2}\right)\end{aligned}$$

Now it is clear that on the straight track, the "infinite" value $R$ gives the correct KE of $\frac{1}{2}mv^2$, and that if the KE is constant (because there are no forces acting that do work), then if $R$ decreases, so does $v$.

 

[Dale]

Can one of you chaps help me with your disagreement, please?

Fantastist believes that the correct formula for KE is $$KE = \frac{1}{2} m v^2$$ even though multiple references and a derivation have been provided which confirm that the correct formula is $$KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$

 

[sophiecentaur]

Fantastist believes that the correct formula for KE is $$KE = \frac{1}{2} m v^2$$ even though multiple references and a derivation have been provided which confirm that the correct formula is $$KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$

What is there to disagree with about that? Except, perhaps that ω can be stated as v/R, which could be useful in this case because the v term would be in both parts. I can see that there could be a problem at the transition but isn't that a common situation in Physics? That's the beauty of the Impulse.

 

[A.T.]

This seems to be the root of the confusion:

The force of constraint can't do any work.

The force of constraint does no net work, so the total KE (linear + angular) doesn't change.

But the force of constraint can do negative linear work, and the same amount of positive angular work, so linear KE is converted to angular KE.

 

[Dale]

I think his confusion is deeper than that. I think that his confusion is that he believes that the mere presence of a constraint changes the KE.

In other words, he believes given two identical rigid bodies following identical motions wrt an inertial frame, if one is following the motion due to constraints and the other is following the identical motion freely or due to a potential then the KE of the two objects will be different.

It is clearly wrong, completely unjustified, contradicted by references, and disproven mathematically. I don't know what else can be done for Fantasist.

 

[Fantasist]

and R does not change at the moment when the rod hits the curved section

Yes it does. It is infinite before the curve and finite after.

$R$ should be infinite before the curve? $R$ is the radius vector from the origin of the coordinate system to the center of mass of the rod. And this can not change from something finite to something infinite (or vice versa) within an infinitesimal path length $ds$.
Also, you would then have an infinite angular momentum $l=m R\times v$ of the center of mass before the curve, which clearly can not be correct (whether you assume it changes or not).