The Vertical Force of a Spring: A Logical Argument

[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.

 

[ehild]

let be $\vec r$ the position vector of the particle. A the spring has zero relaxed length, the spring force is -k $\vec r$. the total force also includes gryvity. WE set the coordinate system with horizontal x-axis and vertical down y axis. The equation of motion in coordinates is
$m\ddot x=-kx$
$m\ddot y=-ky +mg$
This is a linear , constant coeffient, inhomogeneous DE. the solution of the homogeneous part is a two-dimensional vibration with angular frequancy
$\omega= \sqrt{k/m}$ A particular solution of the inhomogeneous equation is $h= mg/k$ that corresponds to the steady state. For the new variables X=x and Y= y-h, the differential equation becomes $m\ddot X = - kX$ and $m\ddot Y=-kY$. The solution can be a circular motion with angular frequeny $\omega= \sqrt{k/m}$. and arbitrary radius R. Then the speed is constant, $\omega R$..
The problem maker might have took that straightforward, when assuming constant speed.

 

[haruspex]

The problem maker might have took that straightforward, when assuming constant speed.

I don't buy that. It's far too big a jump to be made knowingly in a "textbook solution".

 

[ehild]

I don't buy that. It's far too big a jump to be made knowingly in a "textbook solution".

I agree. Assuming constant speed was not at all straightforward.

 

[sysprog]

Is there a specific part of the argument of post #22 that you do not accept?

That post included:

The net force on $P$ is $\vec F_{net} = m \vec g -k \vec R =- k(- \vec h + \vec R) = -k \vec r$.

It seems to that it should be $+m \vec g$ on the way down, and $-m \vec g$ on the way up.

let be $\vec r$ the position vector of the particle. A the spring has zero relaxed length, the spring force is -k $\vec r$. the total force also includes gryvity. WE set the coordinate system with horizontal x-axis and vertical down y axis. The equation of motion in coordinates is
$m\ddot x=-kx$
$m\ddot y=-ky +mg$

It seems to that it should be $+mg$ on the way down, and $-mg$ on the way up.

 

[sysprog]

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

Yes, but then the rotation would be circular about $\text A$, rather than about $\text C$.

 

[hutchphd]

But gravity can be eliminated from the equation of motion by simply moving the origin : y→ y+$\frac {mg} k$ This constant translation cannot change any velocities. It does change the origin to C

I am frankly mystified that there is an argument here.

 

[TSny]

It seems to that it should be $+m \vec g$ on the way down, and $-m \vec g$ on the way up.

The force of gravity acting on an object is the same whether the object is standing still, moving upwards, downwards, sideways, or in any other direction. I wrote the force as $m \vec g$ where $\vec g$ is a vector pointing downwards with magnitude equal to the acceleration due to gravity.

 

[hutchphd]

The OP was answered completely with a three line explanation before post #10.
Perhaps it would serve us all well to start a new post with a precise statement of the outstanding issues.
I have no idea what they are, and so look forward to the new post.

 

[haruspex]

Yes, but then the rotation would be circular about $\text A$, rather than about $\text C$.

No, in the absence of gravity A and C would be the same point.
And as @TSny reminds you, gravity doesn't suddenly change sign according to which way an object moves.

 

[haruspex]

The OP was answered completely with a three line explanation before post #10.

Well, there does seem to have been some confusion even there.

What the OP asked was whether the official solution was valid, and I think we all agree it was 'inadequate', i.e. either the solver had made a huge unjustified jump or had used flawed reasoning.

Maybe there was also an implied question of what the simplest solution would be.
The crucial step, I would say, is to show that because of the relaxed zero length the horizontal and vertical components of the tension are in (the same) proportion to the horizontal and vertical components, respectively, of displacement from the tethered point. That makes the two motions SHM at the same frequency.
The circularity then fixes them as having the same amplitude and being 90 degrees out of phase.

But I know you argued that it suffices to use the info that the motion is circular. I can see that in principle that may do it, but I am afraid I still haven’t understood how your argument works. Maybe take that into a private discussion.