Energy conservation paradox for constrained mass

[A.T.]

To constrain the mass you must have torque at point P where the mass connects and therefore friction.

As DaleSpam has shown no friction is required.

 

[Fantasist]

@Fantasist

I really don't see where you are going with this.

Hopefully to a better understanding of the physics of constrained mechanical systems.

If there is a complete disc (massless etc) supporting the mass then there is no work done in rotating it.
If you use number of spokes, there is still no work done.
Two spokes, progressively closer and closer together, will produce no work. If you say there is something special about your single 'rigid arm' with 'non flexing' shoe on the rail then you would need to say what the limiting case is (i.e. when two spokes start behaving as one, only differently). If you are calculating "for small ϕ" and you are getting a non zero result, you have to conclude that you have got the sums wrong. I don't see that, in any way, you have discovered a 'paradox' - you have just produced an error in your derivation. We have all been there and usually look for our error, rather than cry 'paradox'.

I have indicated above already where the reason of the paradox may lie, namely because of assuming the work is done directly at the mass, when in fact it is done at the point of constraint.

 

[sophiecentaur]

Hopefully to a better understanding of the physics of constrained mechanical systems.



I have indicated above already where the reason of the paradox may lie, namely because of assuming the work is done directly at the mass, when in fact it is done at the point of constraint.

Are you saying that your method of calculation, incontrovertibly shows that work is done "at the point of constraint"? I suggest that the calculation will involve taking limits and that has to be done in a valid way or the answer can't be relied on.
Imo, the way to deal with situations like this is to start with the basic principle and not aim at showing it's wrong. As soon as we start getting into zeros and infinities we depart from the real world so spurious paradoxes start to suggest themselves. "A better understanding" doesn't seem likely when the system being studied is unreal.

 

[Fantasist]

The variables $F_A$ and $F_B$ in the equations of posts 20 and 38 and are the magnitudes of vectors which point it different directions. It doesn't make any sense to add them or subtract them this way. If you actually calculate the force vectors $\mathbf{F_A}$ and $\mathbf{F_B}$ and add them then you will see, as AlephZero said, that their sum is mg upwards regardless of the size of $\phi$, as it must be.

OK, my mistake. I apparently misunderstood AlephZero's statement. Obviously, if you multiply $\sin(\phi)$ to $F_A - F_B$ then this would be equal to $-mg \sin(\theta)$ (as this is what you put in as your assumption in the first place).

Newton's 2nd law is ∑F=ma, where ∑F is the sum of all of the forces acting on a body, m is the mass of the body, and a is the acceleration of the center of mass of the body. Because this is a vector equation it does indeed constrain the motion of the center of mass in both the horizontal and vertical direction or equivalently in both the direction parallel to and perpendicular to the arm.

However, the a in the equation is the acceleration of the center of mass. It does not constrain in any way the motion of the rest of the body around the center of mass. In particular, it does not prevent the object, including the massless arm, from rotating around the center of mass.

I fully agree with this in principle. Only this is not how you set up your equations, where the forces are taken at the constraint, not at the mass. I quote your earlier post:

For concreteness, let's assume that the structure consists of two pairs of frictionless ball bearings, each located an angle of $\phi$ away from the angle of the rod which is itself at an angle $\theta$ from the vertical. Then the forces at the bearings are:
$F_A=-mg \csc(2\phi) \sin(\theta-\phi)$
$F_B=mg \csc(2\phi) \sin(\theta+\phi)$

I already proved the contrary. This is simply false, so please don't repeat it again. Two radial forces together can indeed provide a net force which can compensate for the gravitational force. In general, a radial force at some point A and a radial force at some point B will usually produce a net force which is not radial at a third point C.

Sorry, but you haven't proved anything. You made the assumption that the system is static by postulating the force equations at the constraint as

$mg \cos(\theta) = F_A \cos(\phi) + F_B \cos(\phi)$
$mg \sin(\theta) = F_B \sin(\phi) - F_A \sin(\phi)$

You then calculate from this the forces $F_A$ and $F_B$ that satisfy this assumption. These forces are exactly the internal stress forces in the material when you have constrained not only the radial motion but also the tangential motion along the rail. But it would be a circular conclusion to say that the system will be static without a tangential constraint. I might just as well postulate the second of the equations as

$F_B \sin(\phi) - F_A \sin(\phi) = 0$

and then say I have proven that the system is not static in general.

I don't understand why you are bringing in the torque here. If you apply a force $\mathbf{F}$ to some point of a rigid body and move this point a distance $d\mathbf{s}$ in the process, then the work associated with this is $\mathbf{F}\cdot d\mathbf{s}$, irrespectively of how much of this results in a translation of the center of mass and how much in a rotation due the torque associated with the force. If you say the total work done in the process is zero, then $\mathbf{F}\cdot d\mathbf{s}$ must be zero. But this contradicts your equation $\mathbf{F}\cdot d\mathbf{s}=m\mathbf{g}\cdot d\mathbf{s} = -mg \sin(\theta) ds$.
Like I said already earlier, the important point here is that the work is done at the point of constraint, not directly at the mass.

You have specified two kinematical constraints: one constraint on the position and one constraint on the direction. A single constraint force is simply incapable of explaining the dynamics of your kinematical constraints. It leads to an underdetermined set of equations. If you disagree then I encourage you to set up the system of equations and see for yourself.

I don't know why you assume the system would be underdetermined. Consider the case of tightening a nut with a spanner (wrench): you exert a vertical force $\mathbf{F}$ at a distance r from the nut; when you move this through one rotation you'll do the work $W = 2\pi*r*\mathbf{F}$. There is no additional contribution from the torque (of course, alternatively you could calculate the work from the torque as well, which would give the same result). The present scenario is pretty much identical to this.P.S.: I make this my last post on this subject, as I don't think there is a lot to add anymore here.

 

[Dale]

I fully agree with this in principle. Only this is not how you set up your equations, where the forces are taken at the constraint, not at the mass. I quote your earlier post:

For concreteness, let's assume that the structure consists of two pairs of frictionless ball bearings, each located an angle of $\phi$ away from the angle of the rod which is itself at an angle $\theta$ from the vertical. Then the forces at the bearings are:
$F_A=-mg \csc(2\phi) \sin(\theta-\phi)$
$F_B=mg \csc(2\phi) \sin(\theta+\phi)$

And what is wrong with that? Obviously, the forces on the bearings act at the bearings! Where else would the forces on the bearings act other than at the bearings?

It is completely baffling to me that you would consider this a point of contention. If I push a door then the force of my finger on the door acts where I push on the door and not at the window near the door, and the force of the door on my finger acts on my finger and not on my nose. How can you possibly think otherwise? Why would you think that the forces on the bearings act elsewhere?

Furthermore, why do you think it matters? Look at Newtons second law. Where in the law does the point of application matter? Regardless of where on an object a force is exerted the acceleration of the center of mass of the object responds the same.

Sorry, but you haven't proved anything. You made the assumption that the system is static

Nonsense. YOU made that assumption when you specified the geometry of the problem and "a mass m" and "a rod (with negligible mass)". It is part of your OP that the center of mass did not accelerate, which you even recognize in the OP with "the mass m has always stayed in the same location".

My derivation is simply a consequence of applying Newton's laws to the system that YOU specified. YOU specified that ma=0, not me. As a consequence we know by Newton's laws that ∑f also equals 0, but that is not an assumption I imposed. That is a consequence of your assumption.

I don't know why you assume the system would be underdetermined.

Why don't you actually set up the system of equations and see for yourself.