The Vertical Force of a Spring: A Logical Argument

[sysprog]

It did not "point out" any such thing! There was no mention of it.
On what basis did it claim that there was no vertical acceleration??

The solution said that at the point at which the mass is moving vertically, the vertical component of the force of the spring is equal and opposite to the force of gravity; it made no claim that there was no vertical acceleration.

 

[haruspex]

The solution said that at the point at which the mass is moving vertically, the vertical component of the force of the spring is equal and opposite to the force of gravity; it made no claim that there was no vertical acceleration.

You have it backwards. Read it again:
"Consider the instant when the mass is moving vertically upward. In this instant the mass’s acceleration is perfectly horizontal, which means that the vertical component of the force from the spring must be equal and opposite to the force of gravity."
It claims without proof that when the motion is vertical the acceleration is purely horizontal, i.e. no vertical acceleration. From that it deduces that the force from the spring must be equal and opposite to the force of gravity.

 

[sysprog]

Well, that the speed is constant has been established in the thread (post #22)

I think that constancy of the speed has not in this thread been established.

That's a matter between you and @TSny, for now at least.

You said that what @TSny said established that the speed is constant; I ask you how so.

 

[haruspex]

You said that what @TSny said established that the speed is constant; I ask you how so.

Please stop deflecting and answer my question first, then I'll take a closer look at post #22.

 

[sysprog]

Please stop deflecting and answer my question first, then I'll take a closer look at post #22.

Please while you're at it look at my post #25.

In my view, you were deflecting when, after you claimed that something had been established, you then, in response to my saying that I thought that it had not been established, said something to the effect that whether it had been established was a matter between me and the person whose post you had proclaimed to have established what you had said had been established.

The most recent question you posited in this thread is (from post #30):

On what basis did it claim that there was no vertical acceleration??

to which my response was (in post #31):

it made no claim that there was no vertical acceleration.

 

[sysprog]

Hey @haruspex, (aside) I started my participation in this thread with an off-the-cuff response; I've done much more thinking and investigating regarding the matter than I would have had it not been for your insightful posts (and those of @hutchphd and @TSny) $-$ so thanks for your (and their) as usual being thought-inspiring to others.

 

[haruspex]

to which my response was (in post #31):

To which I responded:

"Consider the instant when the mass is moving vertically upward. In this instant the mass’s acceleration is perfectly horizontal,

That is the same as saying there is no vertical acceleration. This is the key claim that led to the thread being created.

I don't know why you brought up, in post #25, whether constancy of speed had been established within the thread. That was nothing to do with the debate we were having, so it felt like deflection. What was relevant was whether the official solution had, and I think we agree it had not.
I replied as I did because it seemed to me TSny at least thought he had established that in post #22, and he has an unbeatable track record, but I'm happy to give it a closer look later.

 

[sysprog]

Do you now agree that the textbook answer is inadequate because it fails to demonstrate that a purely vertical velocity implies no vertical acceleration?

I think that the solution didn't say that; it did say, as you quoted,
"Consider the instant when the mass is moving vertically upward. In this instant the mass’s acceleration is perfectly horizontal,", and I recognize that "perfectly horizontal" means no vertical; however, in the context of the problem, "the mass's acceleration" refers to the conditions of the already postulated forces; not to some other unspecified source of tangential acceleration $-$ I agree that it arguably was remiss of the testers to not say anything about how an orbit would be incipiated to begin with.

 

[sysprog]

I don't know why you brought up, in post #25, whether constancy of speed had been established within the thread. That was nothing to do with the debate we were having, so it felt like deflection. What was relevant was whether the official solution had, and I think we agree it had not.
I replied as I did because it seemed to me TSny at least thought he had established that in post #22, and he has an unbeatable track record, but I'm happy to give it a closer look later.

You had stated that only with the uniform circular motion would the argument in the first two sentences of the solution be correct $-$ given that on the way down, $mg$ is added, while on the way up, it is subtracted, the question whether the interaction of the spring force with the gravitational force keeps the velocity constant seems germane to me.

 

[TSny]

All the possible motions of the particle P in the original setup are the same as all the possible motions of P in another setup where the same spring is attached to point C rather than point A and you switch off gravity. See comment 3 in @hutchphd 's post #16.

The two equivalent systems are shown below.

Start P at the same position relative to point A in both systems and pick the same initial velocity for both systems. Then the motion of P will be the same in both systems.

Since system II clearly allows uniform circular motion about C, system I also allows uniform circular motion about C.

I believe that the only trajectories that you can get in system II are ellipses with center at C, or special cases of an ellipse: circular motion and linear motion. So, the same can be said for system I.

 

[sysprog]

All the possible motions of the particle P in the original setup are the same as all the possible motions of P in another setup where the same spring is attached to point C rather than point A and if you switch off gravity. See comment 3 in @hutchphd 's post #16.

To me that looks like a bare unsupported assertion. Clearly, "switching off gravity", would remove any gravitationally-induced speed variation, but I don't ex ante see how it is removed in "system I".

The two equivalent systems are shown below.

Start P at the same position relative to point A in both systems and pick the same initial velocity for both systems. Then the motion of P will be the same in both systems.

Since system II clearly allows uniform circular motion about C, system I also allows uniform circular motion about C.

I believe that the only trajectories that you can get in system II are ellipses with center at C, or special cases of an ellipse: circular motion and linear motion. So, the same can be said for system I.

It seems to me that your assertion that "Since system II clearly allows uniform circular motion about C, system I also allows uniform circular motion about C." is a non sequitur, and that your calling of "system I" and "system II", the "two equivalent systems", is a petitio principii $-$

I think that it still here has yet to be shown how, in "system I", the spring compensation not only keeps the trajectory circular about $\text C$, but also keeps the velocity of $\text P$ from varying with its direction relative to the gravitational force.

You said "I believe that the only trajectories that you can get in system II are ellipses with center at C, or special cases of an ellipse: circular motion and linear motion.", but you didn't say which or why.

On what basis can we be sure that in "system I", the velocity is invariant?

 

[hutchphd]

It seems to me that your assertion that "Since system II clearly allows uniform circular motion about C, system I also allows uniform circular motion about C." is a non sequitur, and that your calling of "system I and "system II", "The two equivalent systems", is a petitio principii ,

The equations of motion, with that simple change in origin, are identical.
What more can you possibly need?

 

[sysprog]

The equations of motion, with that simple change in origin, are identical.
What more can you possibly need?

If we replace the spring with an inelastic tether, with gravity present, we get $-mg$ on the upstroke part of the rotation, and $+mg$ on the downstroke part, and that means that something else must correspondingly vary periodically in the system.

 

[hutchphd]

Sorry but I have no idea what you are trying to communicate here.

 

[sysprog]

Sorry but I have no idea what you are trying to communicate here.

Perhaps this communicates it better (no springs attached): https://www.khanacademy.org/science...tal-forces/v/yo-yo-in-vertical-circle-example

 

[TSny]

To me that looks like a bare unsupported assertion. Clearly, "switching off gravity", would remove any gravitationally-induced speed variation, but I don't ex ante see how it is removed in "system I".

It seems to me that your assertion that "Since system II clearly allows uniform circular motion about C, system I also allows uniform circular motion about C." is a non sequitur, and that your calling of "system I" and "system II", the "two equivalent systems", is a petitio principii −

The proof was presented in post #22. The net force acting on P in system I is equal to the net force acting on P in system II when P is at the same location in both systems. The particle "doesn't know the difference" between being in system I or being in system II.

You said "I believe that the only trajectories that you can get in system II are ellipses with center at C, or special cases of an ellipse: circular motion and linear motion.", but you didn't say which or why.

Assuming motion in a vertical plane, let $x(t)$ and $y(t)$ be the horizontal and vertical positions of P relative to point C in system II. It is easy to see that the equations of motion are $\ddot x = -\frac k m x$ and $\ddot y = -\frac k m y$. These have the general solutions

$x(t)=A \cos \omega t + B \sin \omega t$
$y(t)=C \cos \omega t + D \sin \omega t$

where $\omega = \sqrt{k/m}$ and $A,B,C, D$ are arbitrary constants corresponding to arbitrary initial conditions. For any choice of these constants, the above equations give a trajectory that is either an ellipse centered at point C, a circle centered at point C, or a straight line segment through point C.

On what basis can we be sure that in "system I", the velocity is invariant?

I'm not sure what you are asking here. The velocity will vary with time for elliptical, circular, or linear motion of the system. For the case of circular motion, the speed will remain constant.

 

[hutchphd]

Perhaps this communicates it better (no springs attached): https://www.khanacademy.org/science...tal-forces/v/yo-yo-in-vertical-circle-example

What am I supposed to see? I truly do not understand what point you are trying to make...sorry.

 

[sysprog]

What am I supposed to see? I truly do not understand what point you are trying to make...sorry.

I think that it has not in this thread been established that the speed of mass $\text P$, after initial tangential impulse to start it in orbit, would be constant. The direction of $\text P$ oscillates between toward and away from gravity. The spring keeping the radius of the orbit constant and its center at $\text C$ does not, in my view, ipso facto show that the rotational velocity of $\text P$ is constant.

 

[hutchphd]

OK. Let me try this. Do you believe that the constant speed is true if there were no gravity (g=0)?

 

[sysprog]

The proof was presented in post #22. The net force acting on P in system I is equal to the net force acting on P in system II when P is at the same location in both systems. The particle "doesn't know the difference" between being in system I or being in system II.
Assuming motion in a vertical plane, let $x(t)$ and $y(t)$ be the horizontal and vertical positions of P relative to point C in system II. It is easy to see that the equations of motion are $\ddot x = -\frac k m x$ and $\ddot y = -\frac k m y$. These have the general solutions

$x(t)=A \cos \omega t + B \sin \omega t$
$y(t)=C \cos \omega t + D \sin \omega t$

where $\omega = \sqrt{k/m}$ and $A,B,C, D$ are arbitrary constants corresponding to arbitrary initial conditions. For any choice of these constants, the above equations give a trajectory that is either an ellipse centered at point C, a circle centered at point C, or a straight line segment through point C.

I'm not sure what you are asking here. The velocity will vary with time for elliptical, circular, or linear motion of the system. For the case of circular motion, the speed will remain constant.

Why does the periodic varying of the direction of $\text P$ between toward the ground and away from the ground not affect the speed of $\text P$, as it would if the spring were a rigid tether, or if $\text P$ were like the bob of a pendulum in that regard?

 

[haruspex]

however, in the context of the problem, "the mass's acceleration" refers to the conditions of the already postulated forces; not to some other unspecified source of tangential acceleration − I agree that it arguably was remiss of the testers to not say anything about how an orbit would be incipiated to begin with.

I have no idea what your point is.
The given solution claimed that when the velocity is vertical there is no vertical acceleration, it relied on this for completing the solution, yet it offered no reasoning to arrive at the claim. Yes, it can be shown to be true in the particular context of the problem, but that is a nontrivial step and is conspicuously absent from the given solution.
This has nothing to do with hypothesising other forces nor how the motion was initiated.

You had stated that only with the uniform circular motion would the argument in the first two sentences of the solution be correct $-$ given that on the way down, $mg$ is added, while on the way up, it is subtracted, the question whether the interaction of the spring force with the gravitational force keeps the velocity constant seems germane to me.

You are again confusing how to solve the original physics problem with whether the official solution is valid. If the official solution had first established that the motion is uniform, or if it had been given as a fact in the problem statement, I would have no complaint with those two sentences.

 

[haruspex]

Why does the periodic varying of the direction of $\text P$ between toward the ground and away from the ground not affect the speed of $\text P$, as it would if the spring were a rigid tether, or if $\text P$ were like the bob of a pendulum in that regard?

Because the string has zero relaxed length (which is not so in the other scenarios you mention) it happens that the vertical component of the tension is determined by the vertical height of the bob and the horizontal component by the horizontal displacement. The motions in the two directions become independent.
Therefore each motion is SHM and the two have the same frequency. By arranging that they also have the same amplitude and are 90 degrees out of phase, we have circular motion. It is clear that the peak velocities will also be the same magnitude, and that this will be the speed at all times.

 

[TSny]

Why does the periodic varying of the direction of $\text P$ between toward the ground and away from the ground not affect the speed of $\text P$, as it would if the spring were a rigid tether, or if $\text P$ were like the bob of a pendulum in that regard?

The spring in this problem is very different from a string (rigid tether). The important thing about the spring is that it has zero natural length.

In this problem, if you try to make the mass move as a pendulum, it just doesn't work. Suppose you let the mass hang at rest from point A. The mass will be located at point C. Now, by hand, move the mass along a circular arc (about A) to a new position as shown. Release the mass at rest from this point. Will it swing back down in a circular arc like a pendulum?

No, it won't move like that. P will move in SHM along the blue straight line segment shown, centered on C.

 

[sysprog]

I think that in your postulated experiment the mass would move in a circular arc.

 

[haruspex]

I think that in your postulated experiment the mass would move in a circular arc.

I showed in post #52 that the general motion for a zero length spring is an ellipse centred at the resting equilibrium position. So in general there is no point where it is instantaneously stationary.
The exception is for a degenerate ellipse, i.e. a straight line. So if it is released from rest it will oscillate in a straight line as @TSny depicts.

 

[sysprog]

I showed in post #52 that the general motion for a zero length spring is an ellipse centred at the resting equilibrium position. So in general there is no point where it is instantaneously stationary.
The exception is for a degenerate ellipse, i.e. a straight line. So if it is released from rest it will oscillate in a straight line as @TSny depicts.

I think that it would not move in a straight line.

 

[haruspex]

I think that it would not move in a straight line.

It is not enough to have intuition. Both @TSny and I have provided arguments why it is a straight line. If you believe otherwise either find the flaw in our arguments or present your own analysis. Or both.

 

[sysprog]

It is not enough to have intuition. Both @TSny and I have provided arguments why it is a straight line. If you believe otherwise either find the flaw in our arguments or present your own analysis. Or both.

Well, quoting TSny:

The spring in this problem is very different from a string (rigid tether). The important thing about the spring is that it has zero natural length.

In this problem, if you try to make the mass move as a pendulum, it just doesn't work. Suppose you let the mass hang at rest from point A. The mass will be located at point C. Now, by hand, move the mass along a circular arc (about A) to a new position as shown. Release the mass at rest from this point. Will it swing back down in a circular arc like a pendulum?

No, it won't move like that. P will move in SHM along the blue straight line segment shown, centered on C.

That is not a proof. It's not really even an argument. @TSny did not present it as such.

 

[TSny]

I think that in your postulated experiment the mass would move in a circular arc.

It is a simple consequence of what was proven in post #22 that if you release P from rest at any point other than point C, P will move in SHM along a straight line through C. Is there a specific part of the argument of post #22 that you do not accept?

 

[haruspex]

That is not a proof. It's not really even an argument.

You are deflecting again, ignoring the two posts that do present the argument (#22 and #52) and electing to respond to one that doesn’t.