# Converting rotation to linear harmonic displacement -- Does the bar stop moving at the top and bottom of its displacement?

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## Summary:

Does the motion ever "stop"
Basic zero clearance friction-less model of a crank journal turning with an attached bar on that journal.
Bar moves up and down in Sine fashion, components of journal Velocity vector change accordingly (showing clockwise rotation here), and Velocity of the attached bar only has a Y component (up and down).
All simple thus far.

Now, here's the tricky question, does the bar ever stop moving? We know the Velocity of the bar has to pass through zero at Ymax as the bar itself changes direction, but there is no dt where dx=0, or in other terms, for any dt the dx !=0 .

So from a physics perspective, when V=0 does this attribute alone define motion? The acceleration vector is not zero when V=0, so doesn't this tell us the body is still in motion? Surely we can select a finite point in time to see where V=0, but can a point in time define motion, or must dt/dx be the thing that defines motion?

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kuruman
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As far as I'm concerned, an object is at rest when the first derivative of position w.r.t. time (its velocity) is zero. An object is at rest and remains at rest when all the derivatives of its position are zero. When the bar is at Ymax, it is at rest because its velocity is zero. Does this mean that it is not moving? To answer that question, you have to look at some time dt later. If it is still at position Ymax you conclude that it has not moved within time interval dt, else you conclude that it has moved within that interval. So yes, you need a dt to determine motion.

As far as I'm concerned, an object is at rest when the first derivative of position w.r.t. time (its velocity) is zero. An object is at rest and remains at rest when all the derivatives of its position are zero. When the bar is at Ymax, it is at rest because its velocity is zero. Does this mean that it is not moving? To answer that question, you have to look at some time dt later. If it is still at position Ymax you conclude that it has not moved within time interval dt, else you conclude that it has moved within that interval. So yes, you need a dt to determine motion.
Agreed, but if I (you, anyone) choose any dt !=0 , whether it be up to V=0, across V=0, or after V=0, there is no dt that I can find where dx=0.

If it is truly "stopped motion" when V=0, then for how long is it stopped there? If it's not there at V=0 for any period of time (dt) then is it truly a stopped motion? I thought the definition of "stopped motion" was; for some arbitrary dt you have dx=0.

And the 2nd derivative of position is not zero when V(t)=0.

And then I look at this by looking at the journal itself. The journal never stops moving (a constant 1rpm), and if the bar cannot decouple from that journal movement, then how can the bar stop moving?

kuruman
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Agreed, but if I (you, anyone) choose any dt !=0 , whether it be up to V=0, across V=0, or after V=0, there is no dt that I can find where dx=0.

If it is truly "stopped motion" when V=0, then for how long is it stopped there? If it's not there at V=0 for any period of time (dt) then is it truly a stopped motion? I thought the definition of "stopped motion" was; for some arbitrary dt you have dx=0.

And the 2nd derivative of position is not zero when V(t)=0.
Please explain what dt ! means and how it is different from infinitesimal time interval dt. I am not familiar with this notation.

Also, I am not familiar what you mean by "truly stopped" motion. A motion is "truly stopped" for as long as its position remains constant. Is the motion of the bar "truly stopped"? By my definition it is truly stopped for a zero time interval, i.e. instantaneously.

And then I look at this by looking at the journal itself. The journal never stops moving (a constant 1rpm), and if the bar cannot decouple from that journal movement, then how can the bar stop moving?
That's because the velocity of the bar and only the y-component of the velocity of a point on the rim of the journey that is in contact with bar are coupled and the bar is constrained to move only in the y-direction. When the x-component of that point is instantaneously zero, cannot be moving.

As a related example consider the following situation. You throw a ball in projectile motion and you look at its shadow on a vertical wall the plane of which is perpendicular to the projection plane. When the ball reaches maximum height, it is still moving with some horizontal speed, yet its shadow on the wall is instantaneously at rest. If the ball is just dropped from a certain height, Its shadow on the ground is at rest at the same time as the ball is moving.

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I am using dx loosely here just for math purposes. I suppose dy matches the pic better.
In general, dy/dt describes the motion.

I mean dt as in any change in time will describe the start and end positions X0 and X1, whereby dx is the change in position X1-X0

So, the instantaneous views (some arbitrary finite value of t) does not appear to describe if the body is in continuous motion or not, we need to choose some arbitrary change of time t0 to t1 to know if there was any change in position (motion in classical sense) regardless of how small or big the time change is.

From the math I see there is no derivation that shows dx=0 for any change in time (dt) no matter how small you make the change in time. In others words, for ALL changes in time there will in fact be a change in position, thus proving the body is in continuous motion, even when the Velocity vector becomes zero for zero time period.

For any/all dt where dx=0 = no motion "stopped" for the period dt, which can never be true with sinusoidal driven motion.

No doubt the bar has some incremental points in time where all components of Velocity vector =0, but does that alone define "stopped motion"? Again, I understand "stopped" to mean no change in position for some arbitrary time period. There is no time period where V remains zero, etc. Do I have the definition of "stopped" wrong, or perhaps my math is wrong?

Ok you are talking about a connecting rod?

Does the rod stop moving at a certain point? For that to happen the con rod has to change direction. The crankshaft is always going in the same direction and one end of the con rod is connected to the crank so in that regard, the crank connected part of the con rod never stops moving but the the con rod rod vertical motion component does.

A.T.
Do I have the definition of "stopped" wrong,
You can define "stopped" to mean whatever is useful to you.

Ibix
Please explain what dt ! means and how it is different from infinitesimal time interval dt. I am not familiar with this notation.
I think he's using C/Java/python/etc notation for 'not equal', actually, rather than modifying dt somehow. Read dt!=0 as ##dt\neq 0##.

@RichFoster - learning the basics of LaTeX for mathematical notation is highly recommended. There's a link to a guide immediately below the reply box.

kuruman
Ok you are talking about a connecting rod?
There's no connecting rod in the pic.

I think he's using C/Java/python/etc notation for 'not equal', actually, rather than modifying dt somehow. Read dt!=0 as ##dt\neq 0##.

@RichFoster - learning the basics of LaTeX for mathematical notation is highly recommended. There's a link to a guide immediately below the reply box.
Yeah, sorry, just much faster to != vs having to click the gui button for symbol and then click a symbol. I did not think != (not equal to) would be confusing vs ≠, I guess programming math has given me some habits that do not map 100% to other notation system(s).

kuruman
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Again, I understand "stopped" to mean no change in position for some arbitrary time period.
It's a matter of clarity. As long as you don't use "stopped" indiscriminately when an object has zero velocity, you will be OK. I prefer to use "instantaneously at rest" or ##at rest and remains at rest" to differentiate the two. "Stopped" to me is inherently ambiguous because of its wide everyday use. For example, when you "stop at the store for groceries" are you really at rest with respect to the store or are you in continuous motion up and down the aisles picking up items here and there?

As I sit here composing this post, I am at rest relative to the room for the arbitrary time period it will take me to finish typing. I intend to get up and do something else when I'm finished. I would say that I am at rest, not stopped, for about 10 minutes. As @A.T. posted,
You can define "stopped" to mean whatever is useful to you.
but you also need to be clear to others about what kind of "stopped" you mean if you wish to be fully understood.

It's a matter of clarity. As long as you don't use "stopped" indiscriminately when an object has zero velocity, you will be OK. I prefer to use "instantaneously at rest" or ##at rest and remains at rest" to differentiate the two. "Stopped" to me is inherently ambiguous because of its wide everyday use. For example, when you "stop at the store for groceries" are you really at rest with respect to the store or are you in continuous motion up and down the aisles picking up items here and there?

As I sit here composing this post, I am at rest relative to the room for the arbitrary time period it will take me to finish typing. I intend to get up and do something else when I'm finished. I would say that I am at rest, not stopped, for about 10 minutes. As @A.T. posted,

but you also need to be clear to others about what kind of "stopped" you mean if you wish to be fully understood.
Good stuff.
The context is "motion". Does the motion of the bar ever stop its' motion?
To have "stopped motion" we need to observe (prove) any dt (an arbitrary period of time) where change in position remains zero. Staying in your chair for a period of 10min is "no motion", only when you change position (which takes some time, some time period) do you then have motion.

If I move in this positional fashion, X0-->X1-->X0 in 1sec, change in position appears to be zero in that 1sec (starting and ending position is the same), but there was motion, just make the 1sec (dt) smaller and observe. In that same motion if we look at dt=0.9s we then get position Xt where X0≤Xt≤X1

If something is not at rest (V=0) for a period of time, then is it in motion?
The word dwell has technical definition of "a slight regular pause in the motion of a machine.". A pause in motion supports the math (and "stopped motion"), as it indicates there is no motion for some period of time (the "pause").

How else can you prove there is or is not motion (even if the motion passes over a point where V=0)?

________________________________________________

Let me give example of when there would be paused motion. Just make the bar opening clearance bigger than the journal diameter, now when the bar reaches Ymax the bar will remain at V=0 for some period of time (aka "no motion") until that slack is taken back up and bar pulled in the opposite direction.

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sophiecentaur
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This idea occurs all over the place and people seem to be after finding a paradox somewhere. There's only a problem if we insist that a velocity of Zero is something special.
I can imagine two learned ancient Greeks having a long discussion along these lines about a point on a cart wheel that's rolling along. That point stops once every revolution but the cart is going at uniform velocity.
Mercene studied the Cycloid which describes the above motion and so did Galilieo.

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jbriggs444
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How else can you prove there is or is not motion (even if the motion passes over a point where V=0)?
It depends on what you mean by "motion".

If you want mathematical proof, you need precise mathematical definitions. Once you have those precise mathematical definitions in hand, you will find that there is no measurable physical distinction to be made.

We do not have the ability to make infinitely precise measurements to disambiguate between infinitely precise distinctions in definitions.

One might argue that the bar never had an precisely defined position in the first place. Nor did it ever have a precisely defined velocity. Questions about whether its velocity was ever zero do not arise -- it didn't have a velocity.

What the bar does have is an approximate position and an approximate velocity. That's good enough for practical purposes.

kuruman
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I can imagine two learned ancient Greeks having a long discussion along these lines about a point on a cart wheel that's rolling along. That point stops once every revolution but the cart is going at uniform velocity.
Even more mind-boggling is that, when a train is moving in a given direction in a straight line, there are points on this train moving in the opposite direction. If the train is moving into the screen, point A (see figure below) on the rim of the wheel is moving out of the screen. At that instant, there are other points on the wheel that are moving in the same direction as the train and point(s) that are at rest relative to the track.

hutchphd
Even more mind-boggling is that, when a train is moving in a given direction in a straight line, there are points on this train moving in the opposite direction. If the train is moving into the screen, point A (see figure below) on the rim of the wheel is moving out of the screen. At that instant, there are other points on the wheel that are moving in the same direction as the train and point(s) that are at rest relative to the track.

View attachment 267771
Why is that mind-boggling? Sure, for some time t0 that specific point has V vector where only one of it's two components =0, but for some arbitrary small t, t1 = t0+t, at t1 the V vector of that moving point on wheel once again has two components. The V vector was and never is zero, even if the observation point is continuous, like making point-A 0,0 of a XY coord system, if we look at 0,0 at all/any time t, the only thing that seems odd is the direction of the V vector, but it's never zero. But we can also make that same observation for any fixed point that the wheel itself passes through, and for any fixed point (eg; 0,1 or 0,1.1, or 0,1.5) at any time t the V vector will remain constant (as long as wheel rpm remains constant). Heck, the wheel is solid at the center, so what's the momentum at the very center of that wheel, must be zero at radius =0, no mass when radius =0, yes? But yet there is mass at the center?

I don't see that train wheel troublesome at all.

I am asking about when some mass has a property of V=0 at some arbitrary time t, is that property alone good enough to know the mass has no motion. I am arguing it is not.

Can a velocity vector have direction with magnitude =0 ?

kuruman
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I am asking about when some mass has a property of V=0 at some arbitrary time t, is that property alone good enough to know the mass has no motion. I am arguing it is not.
For an extended rigid body, V = 0 is not a property of the body's mass but of a specific point on the body. What do you mean when you say "the mass has no motion"? Some points on the train mass move forward, some backward and some not at all. According to you, does the "train mass" have motion or not?
Can a velocity vector have direction with magnitude =0 ?
If a vector has zero magnitude, in what direction does it point?

jbriggs444
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I am arguing it is not
Arguing definitions is a fool's game.

hutchphd and sophiecentaur
Arguing definitions is a fool's game.
By way of the math.
Does the math I have shown not prove there is continuous motion even when there is an observation at some finite time where V=0?

For an extended rigid body, V = 0 is not a property of the body's mass but of a specific point on the body. What do you mean when you say "the mass has no motion"? Some points on the train mass move forward, some backward and some not at all. According to you, does the "train mass" have motion or not?

If a vector has zero magnitude, in what direction does it point?
Mass has no motion - The position remains constant for any arbitrary time period. If it never changes position then it has no motion.

jbriggs444
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By way of the math.
Does the math I have shown not prove there is continuous motion even when there is an observation at some finite time where V=0?
There is no physical question here. You would have to define what you mean by "continuous motion".

And then you have the problem that you cannot measure V accurately enough to know whether it is ever zero. The mean value and intermediate value theorems cannot save you -- they require a differentiable function or a continuously differentiable function respectively. There is no guarantee that physical position is differentiable (nor even a guarantee that it is continuous or well-defined).

If you want to argue that there are strictly monotone increasing functions f(x) where f'(x) can be zero, that's certainly true. And trivial. ##f(x) = x^3## is one such. ##f(x)=\tan x - x## is another.

Probably you want to say something like:
Under the laws of Newtonian mechanics, a particle has a well defined position at all times. That position is a continuously differentiable function of time, p(t).​
A particle is said to be in "continuous motion" if, for every pair of times ##t_1## and ##t_2## with ##t_1 < t_2##, there is a time ##t_3## between ##t_1## and ##t_2## such that ##p(t_3)## is different from both ##p(t_1)## and ##p(t_2)##​
In more English-like phrasing, "if you look at a continuously moving particle twice, it will have moved in between".

You could then proceed to prove that if ##p(x) = \sin x##, the particle qualifies as "continuously moving" despite the fact that ##p'(\frac{\pi}{2}) = 0##.

The point, however, is that such a "proof" would not prove anything physical. It would just prove that this particular mathematical function satisfied that particular mathematical definition. The conclusion you obtain is contingent on the definition you choose, not on the physical facts of the situation. Someone else who defines "continuous motion" to mean that ##p'(x)## is different from zero for all x would draw a different conclusion.

Arguing definitions is a fool's game.

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hutchphd and Nugatory
(1) And then you have the problem that you cannot measure V accurately enough to know whether it is ever zero

(2) In more English-like phrasing, "if you look at a continuously moving particle twice, it will have moved in between".
(1) I am taking no measurements, purely the simple math on paper.

(2) I don't yet know that the particle is in continuous motion, I am trying to prove that is it. Saying I am observing a continuously moving particle while at the same time trying to prove that it is continuously moving is a paradox, no? Yes, sin x qualifies as continuous, but I am looking to prove that, etc.

The look-twice observation has some pitfalls in it. What if I look at it before V=0 and after V=0 where V=0 is in-between the two observations. There are several things that could happen in-between, one of them being V=0 for no period of time (like sin x), and another where V=0 for a period of time (I dunno, maybe an odd sin x function where every π the curve has zero slope for a small time period (or π/500)).

My proposal for proof of continuous motion is to prove there is no time period of the function in which position has not changed. This proof has no regard for Velocity. Like looking at every infinitely small yet finite time period, and for each see if the position has (or has not) changed.

And just to note, I am purposely not using any specific function to define the motion because the proof should be for all functions that can define position w/ respect to time.

But just to add some twist, you said sin x is continuous (I agree), so then at some time t where V=0 does that change the fact that sin x is continuous? Or may I ask, where in the function is it not continuous.

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A.T.
I don't yet know that the particle is in continuous motion, I am trying to prove that is it.
You just defined "continuous motion" such that the particle satisfies the defintion:
...proof of continuous motion is to prove...
There is not much to prove here.

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jbriggs444
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My proposal for proof of continuous motion is to prove there is no time period of the function in which position has not changed.
The "look twice" definition is identical to this. You say that there is no period where position has not changed. I say that in all periods, position has changed (and give a specific meaning for that). Six of one, half dozen of the other. But they are both definitions. They are not proofs of anything.

jbriggs444
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But just to add some twist, you said sin x is continuous (I agree), so then at some time t where V=0 does that change the fact that sin x is continuous?
Or may I ask, where in the function is it not continuous.
I had not noticed this remark until now.

When we say that "sin x is continuous", there is a agreed-upon and quite specific mathematical definition for what that means. https://en.wikipedia.org/wiki/Continuous_function#Definition_in_terms_of_limits_of_functions

A function f is "continuous at a point c" if and only if:

1. c is within the domain of f
2. f(x) has a limit as x approaches c
3. That limit is equal to f(c).
[Note that there is a corner case for "isolated points" that we need not worry about].

A function f is "continuous" if it is continuous at every point in its domain.

The question of whether f'(x) = 0 does not arise. However, one can prove that if f'(x) exists then f is continuous at x.
Or may I ask, where in the function is it not continuous.
As I interpret this remark, you think that I am saying that f(x) = sin x might not be "continuously in motion". Yes, that I do think that. As long as "continuously in motion" is not defined.

Having a definition of "continuous" is not the same as having a definition for "continuously in motion".

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vanhees71