# Seeking a simple logical argument to an interesting statement (spring-mass motion)

haruspex
Homework Helper
Gold Member
Whether the compensatory action of the spring suffices not only to keep the orbit circular and its center fixed, but also suffices to keep the velocity constant, is yet to be fully established in this discussion.
Well, that the speed is constant has been established in the thread (post #22), but I agree it was not established by the textbook answer. And this is why the textbook reasoning is wrong: merely knowing it is moving in a circle is not sufficient basis for saying that when the velocity is vertical the acceleration is horizontal. (So are you agreeing with me now?)

haruspex said:
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 clearly established.

haruspex
Homework Helper
Gold Member
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.
Do you now agree that the textbook answer is inadequate because it fails to demonstrate that a purely vertical velocity implies no vertical acceleration?

That's a matter between you and @TSny, for now at least.
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 provided answer did not say anything to the effect that a purely vertical velocity implies no vertical acceleration; rather, I think that it sought to address the gravitational force by pointing out that it was offset at the equilibrium point by the spring force.

hutchphd
haruspex
Homework Helper
Gold Member
it sought to address the gravitational force by pointing out that it was offset at the equilibrium point by the spring force.
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??

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
Homework Helper
Gold Member
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.

haruspex said:
Well, that the speed is constant has been established in the thread (post #22)
sysprog said:
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
Homework Helper
Gold Member
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.

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):
haruspex said:
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.

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
Homework Helper
Gold Member
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.

haruspex said:
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.

haruspex said:
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.

Last edited:
TSny
Homework Helper
Gold Member
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.

Last edited:
TSny said:
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?

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??????

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.

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

TSny
Homework Helper
Gold Member
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.

Last edited:
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.

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

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?