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

## Homework Statement:

Problem 24 in US Physics Olympiad FMA packet A

## Relevant Equations:

F=ma
Because this problem is easier to understand with a picture, I'll just copy paste the original problem. There is no question about the validity of the solution. My question is about the statement in the solution
"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."

I know in fact this is a correct statement, however, is there a simple logical argument to support this statement. #### Attachments

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For an object in uniform circular motion about an axis, the centripetal acceleration is orthogonal to the tangential velocity, so when the centripetal acceleration is perfectly horizontal, the tangential velocity is perfectly vertical.

haruspex
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For an object in uniform circular motion about an axis, the centripetal acceleration is orthogonal to the tangential velocity, so when the centripetal acceleration is perfectly horizontal, the tangential velocity is perfectly vertical.
I don’t think that answers the question. For one thing, will the motion be uniform? I doubt it. Secondly, centripetal acceleration is always orthogonal to velocity, by definition. Thirdly, centripetal acceleration is not the issue - we want to know about the tangential acceleration.

Homework Statement:: Problem 24 in US Physics Olympiad FMA packet A
Relevant Equations:: F=ma

Consider the instant when the mass is moving vertically upward. In this instant the mass’s acceleration is perfectly horizontal,
I see no simple argument to support that deduction. Why should there not also be a vertical (so tangential) acceleration? That does not conflict with a perfectly circular trajectory.
I also don't like the wording "so that the mass can swing in a vertical circle". It is far from obvious that it will, so it would be better worded as "assume the mass ...".

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• BvU and sysprog
haruspex said:
I don’t think that answers the question. For one thing, will the motion be uniform? I doubt it. Secondly, centripetal acceleration is always orthogonal to velocity, by definition. Thirdly, centripetal acceleration is not the issue - we want to know about the tangential acceleration.

I see no simple argument to support that deduction. Why should there not also be a vertical (so tangential) acceleration? That does not conflict with a perfectly circular trajectory.
I also don't like the wording "so that the mass can swing in a vertical circle". It is far from obvious that it will, so it would be better worded as "assume the mass ...".
I interpreted the request as seeking a simple explanation of the first 2 sentences in the solution: It seems to me that the solution as provided is simple enough, but I thought that perhaps the definitional fact of the centripetal force being orthogonal to the tangential velocity was not part of what the requestor already understood as going without saying. I didn't want to drag in unnecessary complexities; there's a circular motion article in Wikipedia that explains everything in the problem but the role of the spring.

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haruspex
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I interpreted the request as seeking a simple explanation of the first 2 sentences in the solution
As do I, but those two sentences read "Consider the instant when the mass is moving vertically upward. In this instant the mass’s acceleration is perfectly horizontal,". The second sentence does not follow as a logical deduction from the first.

I must be missing something, this seems trivial.
If the motion is circular, energy considerations require that the center of the circle won't shift as the size of the circle increases. The center in the limit R small is ##\frac {mg} k## .

• ehild
haruspex said:
As do I, but those two sentences read "Consider the instant when the mass is moving vertically upward. In this instant the mass’s acceleration is perfectly horizontal,". The second sentence does not follow as a logical deduction from the first.
I contend that [the proposition that [at the one instant per revolution at which the (centroid of the) mass is moving vertically, the centripetal acceleration of that mass is horizontal], given the other problem conditions, is entailed by the fact that the tangential velocity of the mass is orthogonal to its centripetal acceleration].

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haruspex
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this seems trivial.
The issue is not whether the individual equations and statements are true but whether the reasoning is valid. I.e., does the second of those two sentences follow from the first?
If the motion is circular, energy considerations require that the center of the circle won't shift as the size of the circle increases.
For one thing, you are assuming that it will be circular motion regardless of the radius. The information given only implies we are to assume it is circular motion in the specific case being discussed. If we are to take it as being circular at any energy then we need to prove it.
Even then, I am not seeing why it immediately implies that the centre of the circle doesn't vary as the energy of the system changes. I don't doubt you can prove the centre will be the same, but how is it obvious?

1. There are clearly solutions where the motion is not circular. One chooses the correct initial conditions regardless.
2. In the absence of gravity it is certainly true
3. A simple coordinate transform removes gravity does it not?

haruspex
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1. There are clearly solutions where the motion is not circular. One chooses the correct initial conditions regardless.
2. In the absence of gravity it is certainly true
3. A simple coordinate transform removes gravity does it not?
None of which are relevant to the question being asked by the OP, namely, is the second of these two sentences a simple logical deduction from the first, when combined with the knowledge that the motion is circular:
"Consider the instant when the mass is moving vertically upward. In this instant the mass’s acceleration is perfectly horizontal"

Which would be at the equilibrium vertical position (no vertical force).

haruspex
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Which would be at the equilibrium vertical position (no vertical force).
First, what is meant by "equilibrium vertical position" in this context? Normally that refers to the position at which the mass will hang motionless. What does it mean for a mass undergoing this circular motion? Are you defining it to be that height? If so, how is it immediately obvious that there is no net vertical force wherever the mass is placed at that height? (I think it can be shown fairly readily, though.)

More seriously, why does being at that height coincide with the motion being purely vertical?

TSny
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I think the problem should have stated that the particle is given to move in uniform circular motion. This is what the solution apparently assumes.

I believe it's possible to show that the only possible circular motion is uniform. But I don't think that's something the student is expected to show. I could be wrong, of course.

• ehild
I don’t think that answers the question. For one thing, will the motion be uniform? I doubt it. Secondly, centripetal acceleration is always orthogonal to velocity, by definition. Thirdly, centripetal acceleration is not the issue - we want to know about the tangential acceleration.

I see no simple argument to support that deduction. Why should there not also be a vertical (so tangential) acceleration? That does not conflict with a perfectly circular trajectory.
I also don't like the wording "so that the mass can swing in a vertical circle". It is far from obvious that it will, so it would be better worded as "assume the mass ...".
Your understanding is correct. I originally posted this in advanced homework forum, the mod decided to move this here. I have done some lengthy calculations which does yield this peculiar result. I have thought about this statement for a while and thus far I have not found a simple way to show this must follow as effortlessly as the solution indicates.

and hi TSny.

haruspex
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I contend that [the proposition that [at the one instant per revolution at which the (centroid of the) mass is moving vertically, the centripetal acceleration of that mass is horizontal], given the other problem conditions, is entailed by the fact that the tangential velocity of the mass is orthogonal to its centripetal acceleration].
?
No one disputes that the centripetal acceleration is orthogonal to the velocity; it always is, by definition, as I wrote in post #3.
The questionable claim in the 'proof' is that when the velocity is vertical there is necessarily no tangential acceleration. It is easy to construct counterexamples to that in other contexts, such as a mass rotating in a vertical plane on an inelastic string. How does it become obviously true for an elastic string with zero relaxed length?

First, what is meant by "equilibrium vertical position" in this context? Normally that refers to the position at which the mass will hang motionless. What does it mean for a mass undergoing this circular motion? Are you defining it to be that height? If so, how is it immediately obvious that there is no net vertical force wherever the mass is placed at that height? (I think it can be shown fairly readily, though.)
1. I assume it means the "motionless" position as you say which in this case corresponds to the R→0 limit of the circular motion.
2. Yes that is the height
3. Seems obvious to me in light of the transformation of the origin in the usual way to obviate gravity. If it is not obvious, then it is not. But as you say it is easy to convince yourself it is true. The spring force is $$k(\vec x +\vec y)$$ and the problem separates. Does that help?

More seriously, why does being at that height coincide with the motion being purely vertical?
If the motion is circular (which it is chosen to be by the problem initial conditions) it must be the horizontal line through the "new" origin (i.e. the equilibrium height). Even if the motion is not circular the path will be symmetric about this line (the limit is 1D vertical oscillation) and hence vertical at the intersection.

• guv, sysprog and TSny
@hutchphd This is an excellent explanation using the vector nature of the spring tension force. It can also be seen now, the point mass must be doing uniform circular motion. Thanks everyone!

• hutchphd
haruspex
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in light of the transformation of the origin
Sure, I know that works. I can only keep repeating to you: solving the Olympiad question is not the point of the thread. @guv's question is whether the argument provided by the Teacher's Association is valid as given. In particular, it makes no appeal to the fact that the string has zero relaxed length (as you do), and without that condition the claim is false.

Sure, I know that works. I can only keep repeating to you: solving the Olympiad question is not the point of the thread. @guv's question is whether the argument provided by the Teacher's Association is valid as given. In particular, it makes no appeal to the fact that the string has zero relaxed length (as you do), and without that condition the claim is false.
The 2 sentences were in response to the problem, which explicitly specified a zero length spring, so the argument should not be evaluated as if it had been presented without that condition.

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haruspex
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The 2 sentences were in response to the problem, which explicitly specified a zero length spring, so the argument should not be evaluated as if it had been presented without that condition.
I do not accept that at all. It takes a bit of algebra to get from the zero length spring to the deduction that purely vertical velocity corresponds to a purely horizontal acceleration. In the provided solution, no attempt was made to bridge that gap. Rather, the text presents it as a general relationship.
As ever-reliable @TSny observed in post #13, the problem statement may have been intended to specify uniform circular motion, as indeed you read it in post #2. Armed with that, the two sentences are valid.

Other than gravity and the spring force, no forces in the system are specified in the problem; however, the problem asks about the circular orbit, and without any other force subsequent to an initial tangential acceleration sufficient to bring about an orbit, a rotation speed would be sinusoidally periodic ##-## the alternation of direction between toward the ground and away from the ground would not allow a circular rotation to be strictly uniform, but it would be per-rotation uniform ##-## as the magnitude of the velocity changes, then either the radius of the orbit circle or the magnitude of the centripetal force can remain constant, but not both ##-## the solution calls for the point at which the direction of the motion of the mass is exactly vertically up be considered ##-## that's the equilibrium point of the gravitational and spring forces, as stated in the solution.

• hutchphd
TSny
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Other than gravity and the spring force, no forces in the system are specified in the problem; however, the problem asks about the circular orbit, and without any other force subsequent to an initial tangential acceleration sufficient to bring about an orbit, a rotation speed would be sinusoidally periodic ##-## the alternation of direction between toward the ground and away from the ground would not allow a circular rotation to be strictly uniform, but it would be per-rotation uniform ##-## as the magnitude of the velocity changes, then either the radius of the orbit circle or the magnitude of the centripetal force can remain constant, but not both ##-## the solution calls for the point at which the direction of the motion of the mass is exactly vertically up be considered ##-## that's the equilibrium point of the gravitational and spring forces, as stated in the solution.
Following @hutchphd in post #16, let ##\vec R## locate the position of the particle ##P## relative to the fixed point ##A## where the spring is attached to the collar. Define vector ##\vec h = m \vec g /k## which locates a fixed point ##C## relative to A. Define ##\vec r## as the vector locating ##P## relative to ##C##. Note that ##\vec r = - \vec h + \vec R##. The net force on ##P## is ##\vec F_{net} = m \vec g -k \vec R =- k(- \vec h + \vec R) = -k \vec r##. So, the net force always acts toward the fixed point ##C## with a magnitude proportional to ##r##. The net force is a central, attractive force relative to ##C##. So with appropriate initial conditions, the particle can move in uniform circular motion about ##C## with any radius.

• sysprog, Delta2 and SammyS
haruspex
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the point at which the direction of the motion of the mass is exactly vertically up be considered − that's the equilibrium point of the gravitational and spring forces
Why could there not also be vertical acceleration at that point? As you say, that would mean the centripetal acceleration is changing to ensure constant radius, but you have not shown that the centripetal force is constant.

Why could there not also be vertical acceleration at that point? As you say, that would mean the centripetal acceleration is changing to ensure constant radius, but you have not shown that the centripetal force is constant.
Although the problem statement includes nothing regarding an initial impartation of tangential acceleration, the request in the solution that the recipient consider the vertical upward instant, implicitly postulates hypothetically such an initial impetus; along with that, for half of the path of the mass, the force of gravity adds to the centripetal tension, and for the other other half it subtracts therefrom. 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.

Following @hutchphd in post #16, let ##\vec R## locate the position of the particle ##P## relative to the fixed point ##A## where the spring is attached to the collar. Define vector ##\vec h = m \vec g /k## which locates a fixed point ##C## relative to A. Define ##\vec r## as the vector locating ##P## relative to ##C##. Note that ##\vec r = - \vec h + \vec R##. The net force on ##P## is ##\vec F_{net} = m \vec g -k \vec R =- k(- \vec h + \vec R) = -k \vec r##. So, the net force always acts toward the fixed point ##C## with a magnitude proportional to ##r##. The net force is a central, attractive force relative to ##C##. So with appropriate initial conditions, the particle can move in uniform circular motion about ##C## with any radius.
Below is the drawing from the problem statement, with your drawing, and a third drawing that uses some features of your drawing combined with the one from the problem statement: The trajectory of mass ##\text P## about axis ##\text A## is elliptical, and about point ##\text C ## it is circular. The equilibrium point is at ##\text E##. At point a the velocity is entirely inertial. Then from point a to point d, gravitational acceleration translates at least partially into the elongation of the spring to maintain the center of the circle of orbit at point ##\text C##. Then from point d to point a, gravitational deceleration shortens the spring.

The vertical component of the spring force is equal and opposite to the gravitational force only at the equilibrium point. It diminishes as the ##\theta## angle goes from 90° at point ##\text E## to 0° at point a.

Given that ##a_c=−ω^2r## (inverse (toward the center) radial acceleration is proportional to the product of the square of the angular velocity and to the radius of the circular path), ##\cdots## to be continued ##\cdots##

• Delta2