Vertical Circular Motion

user228700

Oct 5, 2016 05:36

@JohnRennie Haha, OK :-) So I'm trying to understand the motion of a body going around in a vertical circle.

John Rennie

OK, that should be easy ...

user228700

And from my textbook, I have learned that the minimum velocity of the body at the lower most point must be greater than or equal to $\sqrt{5gl}$ for it to complete the circle/"loop the loop".

John Rennie

If you say so. I'd have to do a few sums to check that's correct.

user228700

@JohnRennie No, that's correct :-) That fact is not even relevant to my question so never mind. Now we're handling the cases in which this velocity is smaller than required, so the body doesn't complete a full circle.

John Rennie

OK

user228700

Oct 5, 2016 05:41

There seem to be two cases for this to happen: either the tension becomes zero first, while the speed remains non-zero or the speed vanishes while the tension is still present. This is correct, yes?

John Rennie

Yes, that's correct, though I'm not sure that's the best way to describe the two cases.

user228700

@JohnRennie OK...how would u describe these two cases then?

John Rennie

Suppose we take the angle of rotation to be zero when the object is at the bottom of the circle.

user228700

@JohnRennie OK.

John Rennie

So the angle is $\pi/2$ when the string (assuming it's a mass on a string) is horizontal

and $\pi$ when the object is at the top of the circle.

user228700

Oct 5, 2016 05:44

OK.

John Rennie

Yes, I've just corrected it:-)

For angles between $-\pi/2$ and $\pi/2$ the component of $mg$ along the string (the radial component) points outwards i.e. stretches the string.

So between these angles (the bottom half of the circle) the string tension can never be zero.

If the velocity is small we just have a pendulum type motion.

user228700

@JohnRennie It points outwards?

John Rennie

Shall I draw a diagram?

user228700

@JohnRennie Hang on, I'll draw it and u tell me where I'm wrong...

John Rennie

user228700

Oct 5, 2016 05:52

John Rennie

You've drawn the diagram for $|\theta| \gt \pi/2$

user228700

Oh, crap, yes. I misunderstood.

DanielSank

@JohnRennie Standard sysadmin operating procedure.

user228700

Okay, I understand your point. This only was my doubt in the first place; where $T$ tends to become zero and where $v$ does and I understand it now.

user228700

Thank you! :-D

John Rennie

Oct 5, 2016 05:55

The point is that the string can't become slack for $|\theta| \le \pi/2$

user228700

@JohnRennie Yessir, understood.

John Rennie

So for the bottom half small $v$ just means we have a pendulum type motion.

user228700

But there is still one thing I don't quite understand...

John Rennie

The string can only go slack for the top half if $v$ is small enough. Specifically if $v^2/r$ is smaller than the radial component of the acceleration due to gravity.

I have to do a quick job for five minutes. Back in a moment ...

DanielSank

Mwaha hahahaha, I have gotten mathjax to work on my public journal!

user228700

Oct 5, 2016 05:58

@DanielSank Congratulations! :-D

DanielSank

@Kaumudi Thank you. It was surprisingly easy.

@0celo7 Got mathjax working on my website. I think you had asked about how to do that...

user228700

@JohnRennie In the diagram that I drew, for the particle's motion in the upper half, there is always a component of weight of the body acting radially inward; shouldn't this act as a centripetal force and propel the body to go around..?

user228700

Oh hang on, I think I have to go back and think about why this tension is acting in the first place to make sense of this.

user228700

@DanielSank Taken back to JR's comment about a "mad scientist" ;-)

John Rennie

There are only two forces acting on the body. The string tension is radial and gravity is vertically down.

user228700

Oct 5, 2016 06:05

@JohnRennie Yes, but this tension is only there because if the pull due to gravity, no?

John Rennie

And the string tension is $mv^2/r$ (plus a gravitational component)

while gravity is of course just $mg$

@Kaumudi no. Suppose you were whirling the body around on the ISS where there is no gravity. There would still be a string tension.

user228700

@JohnRennie Yes, but there is still the force applied by my hand...

user228700

Wait, what? I'm confused. Hang on.

user228700

Ah, OK, confusion cleared. Go on...

John Rennie

The string tension is $mv^2/r + mg\cos\theta$ using my definition of $\theta$

user228700

Oct 5, 2016 06:11

I'm afraid I'm visualizating the problem a little differently; in my mind, I think that since the component of weight acts radially outward in the lower half, there is a chance that the body will stop rotating because the weight component will somehow cancel the tension or something.

John Rennie

The first term is always positive, and the second term is positive for $|\theta| \lt \pi/2$ and negative for $|\theta| \gt \pi/2$

user228700

Damn, do u understand my confusion? I most definitely don't :/

user228700

Okay. Let me ask a simpler question; why would the tension become zero in the first place?

John Rennie

@Kaumudi are you OK with the statement that the tension is: $$T = mv^2/r + mg\cos\theta$$ where a positive value means the net force is radially outwards?

user228700

@JohnRennie Yes, absolutely.

John Rennie

Oct 5, 2016 06:15

So if the tension is zero that means: $$mv^2/r + mg\cos\theta = 0$$

Which happens because $\cos\theta$ is negative for $\theta \gt \pi/2$

user228700

@JohnRennie The tension exists because my hand is pulling on the block using the string, yes?

John Rennie

@Kaumudi that's not a good way to think about it.

The tension exists because the mass wants to go in a straight line.

You have to apply a force to it to bend its trajectory into a circle.

user228700

@JohnRennie This is how I always think of tension; as an opposing force :/

user228700

@JohnRennie OK, so why would it become zero, at any point? (I understand why in the mathematical sense; just not in the physical sense)

user228700

And how does the force applied by my stupid hand factor into all this..?

John Rennie

Oct 5, 2016 06:22

If the string weren't there the body would move in a parabola. Yes?

user228700

@JohnRennie Let's say that $\theta$ (as you have defined it) becomes $\pi/2$. .

John Rennie

OK

user228700

At that point, the tension acts radially inward and the weight acts downward. If this tension were to vanish, the only force acting on the body would be it's weight, downward. So, parabola?

John Rennie

Well, it would move in a vertical straight line, but that is a limit of a parabola.

user228700

I guess I haven't understood why the body flies off tangential to the circle, if the tension vanishes.

John Rennie

Oct 5, 2016 06:27

I wonder if we need to consider the basics of circular motion ...

If the tension vanishes than that's like cutting the string, or as if the string isn't there.

user228700

@JohnRennie No, can u please just explain what causes the body to fly off tangentially?

user228700

Okay, okay. Brain fart. I understand. Sorry.

John Rennie

And if the string isn't there the only force acting on the body is gravity

"Brain fart" - I thought that was a British slang term. I didn't realise it was used in Tamil Nadu as well :-)

Unfortunately I have to go back to work now for about half an hour ...

user228700

@JohnRennie Well, erm, it isn't. I'm more comfortable speaking in English than my native tongue so I have a better vocabulary than my friends who speak Tamil :-)

user228700

@JohnRennie Okay. Do ping me when u're free again.

John Rennie

Oct 5, 2016 07:04

@Kaumudi: I'm back for about half an hour ...

user228700

@JohnRennie OK. Hopefully, we'll be able to resolve this quickly.

John Rennie

Where did we get to?

user228700

My hand. What about the force that I'm applying via the string? Isn't that the the same as tension? If I were always applying that force, tension would never vanish, correct?

John Rennie

It's the motion of the body that produces the force. So it isn't that you control what force your hand pulls with, you just respond to the motion of the mass.

If the body is moving towards your hand, so the string is slack, you can't exert any force on the body whether you want to or not.

user228700

@JohnRennie What? "The motion of the body produces the force"? But the body is moving because I'm applying the force, no?

DanielSank

Oct 5, 2016 07:10

Trying to understand a mass on a string, under influence of gravity?

a.k.a. conical pendulum?

John Rennie

@Kaumudi Newton's third law tells use that the force the mass exerts on your hand is equal to the force your hand exerts on the body.

DanielSank

IMHO I wouldn't say that the motion of the body produces force.

Although, I guess that's certainly one way to describe it...

John Rennie

So it's true to say that your hand exerts a force on the body.

But whether there is a force or not depends on the trajectory of the body.

DanielSank

@JohnRennie (Yeah, constraint forces are really hard to talk about accurately from the point of view of Newton's laws)

John Rennie

If the radial velocity of the body is negative i.e. $dr/dt \lt 0$, then the string will be slack and you can't exert any force on the body.

user228700

Oct 5, 2016 07:13

@JohnRennie Hang on, why are we even considering this case? How can the body move like that? Sure, it can fall down under the effect of gravity but we're not considering a case in which some other force is acting on the body, right? In that case, we could still pull the string, in an attempt to make it taut. Whether it will become taut or not will depend on the magnitude of the other force acting on the body...

DanielSank

(It's coming back to me how confusing I found this when I learned about Lagrange mechanics)

user228700

@DanielSank No, not a conical pendulum. This one's going around in a vertical circle. Those two things aren't the same, right?

John Rennie

You've lost me. The string has a fixed length doesn't it? Aren't we talking about circular motion?

user228700

@JohnRennie Yes.

DanielSank

Conical pendulum

A conical pendulum consists of a weight (or bob) fixed on the end of a string (or rod) suspended from a pivot. Its construction is similar to an ordinary pendulum; however, instead of swinging back and forth, the bob of a conical pendulum moves at a constant speed in a circle with the string (or rod) tracing out a cone. The conical pendulum was first studied by the English scientist Robert Hooke around 1660 as a model for the orbital motion of planets. In 1673 Dutch scientist Christiaan Huygens calculated its period, using his new concept of centrifugal force in his book Horologium Oscillatorium...

^ I think that picture is exactly what you're talking about.

user228700

Oct 5, 2016 07:15

@DanielSank Definitely not what we're discussing.

John Rennie

And the position of your hand is fixed. Your hand isn't moving.

user228700

@JohnRennie OK.

John Rennie

So you can only exert a force on the body when the string is taut.

DanielSank

Oh your thingy is going in a vertical loop?

user228700

@JohnRennie Right, but we can still tug at the string to make it taut, no?

user228700

Oct 5, 2016 07:16

@DanielSank Yep.

John Rennie

You say "tug on the string" ...

Does that mean reduce the string length or move your hand?

Because neither are possible.

DanielSank

I think it would be a lot easier to think about this if you remove the human hand and replace it with a mechanical hinge.

user228700

@JohnRennie I mean reduce the length of string b/w the body and my hand. And that's not possible, why? (I'm really sorry for being so dumb but please bear with me...)

John Rennie

@DanielSank Yes, I'm inclined to agree. Replace your hand with some mechanical pivot point attached to a strain gauge.

@Kaumudi if you do that it wouldn't be circuklar motion because it wouldn't have a fixed radius.

user228700

@JohnRennie Ah, damn. Okay, sorry.

user228700

Oct 5, 2016 07:20

Yes, OK, understood. So my hand or the mechanical pivot or whatever is not doing anything much..? The tension exists because of the motion of the body..? >.<

John Rennie

Yes. As I mentioned above the tension is related to the velocity of the body by: $$T = mv^2/r + mg\cos\theta$$

So the things that determine it are the speed $v$ and the angle $\theta$

And $v$ and $\theta$ are related.

DanielSank

@Kaumudi The tension exists because the mass has a velocity which would take it beyond the range of the string. However, we're operating with the assumption that the string length cannot change. The string can't even extend a little like a spring would. This constraint means that the string exerts whatever force necessary to keep the mass within the length of the string.

user228700

@DanielSank Riight. Okay.

DanielSank

This is a totally weird situation: we have the distance between the pivot point and the mass as a fixed quantity and the forces (tension in the string) are determined in order to fulfill that constraint.

user228700

@DanielSank Yes, I'm finding it difficult to think about tension like this; I've gotten used to thinking about it like a resistive force to some other force, as opposed to these tendencies.

DanielSank

Oct 5, 2016 07:24

In situations like this, the trajectory of the mass is pre-determined, and you can then solve for the forces required to produce that trajectory. This is very confusing because it's the exact opposite of how we're used to thinking about Newton's laws!

user228700

@JohnRennie OK...

DanielSank

@Kaumudi Not surprising at all. This is a very awkward situation and there's an entire theory called "Lagrangian mechanics" designed to make it easier. You'll learn that in university.

user228700

@DanielSank Right. I will keep this in mind from now, thanks :-) Okay, back to the original question. @JohnRennie: So, with the confusion about tension out of the way, can you please tell me just one last time why the component of weight of the body acting radially inward in the upper half is not the centripetal force needed to make it move in a circle?

user116211

Enlightening!!

user116211

@DanielSank o/

user228700

Oct 5, 2016 07:28

@MAFIA36790 What is?

DanielSank

@MAFIA36790 \o

Yes, as I wrote that I realized that this is actually probably one of the hardest things about introductory mechanics.

user116211

I've comprehended the definition of field in terms of ring theory!!

DanielSank

We're introduced to F=ma, and we get this idea that you can take a given force to solve for the position of the particles.

user116211

And yes, that's enlightening for me.

DanielSank

But then, we have to do mental gymnastics to deal with the fact that in so many problems the givens are positional constraints rather than forces.

user228700

Oct 5, 2016 07:30

@DanielSank This. Sigh.

DanielSank

Actually, now that I think about this, the thought processes required to handle this situation are somewhat advanced!

@Kaumudi Yes yes, but you're getting it!

user228700

@JohnRennie: Do u need to get back to work now?

Danu

Hi @Daniel

DanielSank

@Danu o/

user228700

@DanielSank Well, yeah, slowly. I'm glad I'm progressing at least tho.

John Rennie

Oct 5, 2016 07:30

@Kaumudi There are two forces acting on the body. The force due to its circular motion is $mv^2/r$ and the force due to gravity is $mg\cos\theta$

user116211

back to reading...

John Rennie

The tension in the string, i.e. the centripetal force, is the sum of these two forces.

DanielSank

@JohnRennie Urgh, I don't like that wording.

user228700

@JohnRennie What is "the force due to its circular motion"?

DanielSank

^ See? ;)

Danu

Oct 5, 2016 07:32

Circular motion implies acceleration

John Rennie

@Kaumudi Anything moving ina circle continually changes its momentum. Yes?

DanielSank

@Danu We're past that.

Danu

So... there is a force.

What step did I miss? :P

DanielSank

Mass is moving in a circle --> acceleration points toward center (i.e. "centripetal") and has magnitude v^2/r --> total force = m v^2 / r --> weight + tension points inward toward center with magnitude m v^2 / r.

user228700

@JohnRennie No, no, I meant that there is no force associated with circular motion, in and of itself, no? There are other tangible forces that cause the body to move in a circular path, which cause it to accelerate.

DanielSank

Oct 5, 2016 07:34

@JohnRennie It is incorrect to equate the tension and the centripetal force.

At the apex, the weight is centripetal too.

user228700

@DanielSank Yeah, this makes sense.

DanielSank

@Kaumudi So it makes sense now, or...?

user228700

No, see, the original question was, "Where does the speed become zero" and "Where does the tension become zero"?

user228700

Hang on.

user228700

2 hours ago, by Kaumudi

And from my textbook, I have learned that the minimum velocity of the body at the lower most point must be greater than or equal to $\sqrt{5gl}$ for it to complete the circle/"loop the loop".

user116211

Oct 5, 2016 07:38

@Kaumudi so?

user228700

2 hours ago, by Kaumudi

There seem to be two cases for this to happen: either the tension becomes zero first, while the speed remains non-zero or the speed vanishes while the tension is still present. This is correct, yes?

DanielSank

@Kaumudi Is there a link to the complete question?

user116211

I've actually written two posts on it at Phys.SE.

user228700

I didn't post a question. Asked here directly.

user116211

1

A: Vertical Circular motion - Tension and velocity

Condition for traversing the whole circle: Let the particle of mass $m$ is attached to an inextensible string of length $R$ which provides the necessary centripetal force along with gravity. Let the initial velocity be $u$ at the lowest point of the vertical circle. After time $dt$, it moves t...

user228700

Oct 5, 2016 07:42

@DanielSank: As you can see, (after having read those two messages) the original problem was to figure out the condition when the body will not complete a whole circle and will just oscillate. For this to happen, either tension becomes zero first, or the velocity of the body becomes zero. I was trying to figure out in which part of the body's motion these two cases will happen; in the top half/below half. Then, we got to discussing tension and landed here.

user116211

WTH! I didn't use \mathrm; damn T__T

user116211

DanielSank

@Kaumudi ok

user116211

0

A: Conditions for vertical circular motion

What is the configuration needed for a circular-motion? The answer is, there must be an inward(towards the center) force perpendicular to the instantaneous velocity. The minimum velocity required for the bob initially to loop the whole loop is $\sqrt{5gR}$. In this situation, the tension at the ...

DanielSank

Here's a super quick way to figure this problem out:

The mass's weight is in the same direction as the tension when the mass is at the top of its motion.

user228700

Oct 5, 2016 07:45

@DanielSank Yes, and I keep thinking that this component will serve as the centripetal force required to keep the body moving in a circle...

DanielSank

If the tension is zero at the top of the motion, but the mass is still moving in a circle, then it must be the case that $W = m v^2 / r$.

The weight is $W = mg$, so then we get $v = \sqrt{r g}$.

user228700

@DanielSank No, no, I'm not looking to find the values.

DanielSank

@Kaumudi What are you looking for?

user228700

Okay, hang on. Basically, I'm trying to understand this:

user228700

Oct 5, 2016 07:50

user228700

I want to understand why tension becomes zero only in the second half and why speed in the first half. That's all. Sigh :'-(

DanielSank

Ah. This is a different problem from what I thought.

user228700

@DanielSank Yeah, that's 'cause we went into the discussion about tension in b/w...

DanielSank

@Kaumudi Oh give up the teary face :P

Not needed here. You're surrounded by physicists who want to help.

Ok, from what I read, it makes sense.

user228700

@DanielSank :-P I've been at it for close to 2 hours now. Little frustrated with my stupid brain but I couldn't be happier about the "physicists who want to help" bit so :-)

DanielSank

Oct 5, 2016 07:56

At what point do you lose the argument?

@Kaumudi You're not stupid. These problems are hard.

Once you do a few of them, they become considerably less hard.

user228700

@DanielSank So okay, in the above half of the body's motion, the component of its weight acts radially inward, correct?

user228700

@DanielSank Looking forward to that phase :-P

DanielSank

@Kaumudi Yes. The weight can be decomposed into a tangential part, and a radial part. When the mass is above the mid-line, then the radial part is inward.

user228700

@DanielSank Yes, OK. So I keep thinking how this component can act as the centripetal force needed to keep the body moving in its happy little circle.

user228700

I'm still not able to get my head around exactly what the heck happens when "tension becomes zero".

DanielSank

Oct 5, 2016 08:00

@Kaumudi yeah, note that the motion is circular, but not uniform. I hadn't appreciated that until I saw the problem in the book.

@Kaumudi Nothing special. The tension is just zero.

user228700

@DanielSank Yes, the speed keeps on changing.

DanielSank

@Kaumudi Yes.

Now, suppose the speed of the mass is such that the tension goes exactly to zero at the apex.

user228700

@DanielSank So the body stops having the tendency to fly off in the tangential direction?

DanielSank

@Kaumudi That's not how I would think about it.

The mass always accelerates according to the forces on it. No more can really be said.

user116211

Anyways, check my posts @Kaumudi above; after DS explains the matter to you.

user228700

Oct 5, 2016 08:02

@DanielSank No, I said that about tension becoming zero...

user228700

@MAFIA36790 Yes, thank you :-)

DanielSank

@Kaumudi Not sure what you mean.

user228700

@DanielSank OK, see, I meant, when the tension becomes zero, the body stops wanting to fly off tangentially?

user116211

Okay, wait, @DanielSank; @Kaumudi, could you sum up exactly your thoughts briefly?

DanielSank

@MAFIA36790 Thoughts on what?

Right now I'm thinking about my website and @Kaumudi's problem? I'm also thinking about how awesome neotango is.

user116211

Oct 5, 2016 08:07

@DanielSank I mean she seems to be not along the line; that's why asked her to jot down what exactly her thoughts on the problem is.

user116211

@DanielSank checking neotango

DanielSank

@Kaumudi You could say that. However, try to see it like this: what you think of as "wanting to fly off tangentially" actually means "the tension and other forces and keeping the mass from flying off tangentially".

@MAFIA36790 Neotango is a genre. What I linked is just one example.

user116211

@DanielSank okky ;)

DanielSank

@MAFIA36790 Here's my favorite bit of music.

user116211

aha!!

user228700

Oct 5, 2016 08:10

Okay, hang on.

user228700

In the case of a conical pendulum, the pivot applies a force on the body via the string, and we call this force "tension". There are no forces being applied in the horizontal plane if the circle that are tangential to this circle. Hence, the speed of the body doesn't change. The weight of the body is cancelled by a component of tension and all is well.

user228700

Moving on to the case in which the body is being rotated in a vertical circle...

user228700

The pivot applies a force on the body via the string and we call this force "tension"; same as with the conical pendulum. The difference lies in the fact that now, there is another force acting in the tangential direction, in the plane of the circle itself, so the motion is not uniform and the speed of the body keeps on changing, from place to place to place.

DanielSank

@Kaumudi Yes.

user228700

And then, we're trying to find out where this tension becomes zero but this doesn't mean that we stop applying the force; it just means that the body doesn't want to do the things we were trying to keep it from doing, yeah?

DanielSank

Oct 5, 2016 08:19

> but this doesn't mean that we stop applying the force

user228700

Because we can't apply the force of tension if the bodies actually complies and the string becomes slack, as JR pointed out before.

DanielSank

Not sure what you mean there.

If the tension is zero... the tension is zero.

user228700

@DanielSank Never mind that. Does the next thing I wrote make sense?

DanielSank

Don't think of "applying" force. That is a sure way to get confused. The force is there because the mass would move beyond the length of the string, which is impossible, so the string pulls.

@Kaumudi Then yes.

@Kaumudi Yes.

Tension is a force that happens when a constraint would be violated.

The constraint here is that the mass cannot be farther than $r$ away from the center.

user228700

OK. So now the task is to find out where this tension actually becomes zero.

DanielSank

Oct 5, 2016 08:21

Who starred that? :P

user228700

Me :-P

DanielSank

@Kaumudi Yes, ma'am.

user116211

@Kaumudi Are you working for vertical circular motion?

user228700

@DanielSank :-P OK, and my question is; why does tension become zero only in the upper half aaah!!!

user228700

@MAFIA36790 What dyou mean?

user116211

Oct 5, 2016 08:24

@Kaumudi I noticed some snapshots of vertical circular motion related topic above; so I guessed you are working on it.

DanielSank

@Kaumudi Well, a quick answer is that it's only in the upper half that we stop having a force trying to make the mass go farther than r away from the center!

Below the upper half (i.e. in the lower half) the gravity always is pushing the mass away from the center.

That's just an intuitive way to think about it.

user228700

@DanielSank OHHH.

user228700

I just understood what I was confusing this with!

DanielSank

@Kaumudi Great!

::And the people rejoiced, and they feasted on the mangos and the coconuts, and there was dancing in the streets::

user228700

This is exactly what JR said in the very beginning, but I kept thinking "Hang on, if there is still a component of gravity acting inwards, why won't the body move in a circle?"

user116211

Oct 5, 2016 08:29

Now, that the discussion is finished, let me ask a question:

user116211

Why is $X\times X$ an ordering of $X\,?$

user228700

@MAFIA36790 No, no, not finished yet!

user116211

._.

user228700

@DanielSank Not yet :-P

DanielSank

@MAFIA36790 What does $X \times X$ mean?

user116211

Oct 5, 2016 08:29

Ask! Ask! @Kaumudi!!

user116211

@DanielSank Cartesian Product.

DanielSank

Ah.

user116211

As Kelley asserts, it's an uninteresting ordering.

DanielSank

I dunno... what's an ordering? I guess it's a set of pairs of elements from a set with some transitive property?

user116211

@DanielSank Partial ordering.

user228700

Oct 5, 2016 08:32

@MAFIA36790 Lol xD Okay. @DanielSank:There is some opposing force in the bottom half and so, tension can't possibly be zero. But in the above half, there is no such force, so tension may be zero but what about that radially inward component of the weight? Why is this force not sufficient to keep the body going in a circle..?

user228700

Oh crap.

user228700

I just realized.

user116211

@DanielSank Yes, but reflexive as well as anti-symmetric. Not a strict ordering though.

user116211

@Kaumudi There is no inward component of $mg\,.$

DanielSank

@MAFIA36790 I'm a bit too tired to go looking up the definitions, but I guess in the end it's just an issue of applying the definitions :)

user116211

Oct 5, 2016 08:33

@DanielSank hmm. I am re-reading Kelley.

Slereah

what part of quantum theory forces the dimension of the hilbert space to be infinite dimensional

Is it the momentum as a hermitian operator

user228700

Yeah, I just realized that with the tension gone, there is no need to resolve the weight and all; the weight makes the body fly off, yeah? @DanielSank

user228700

Wait, no.

DanielSank

@Slereah Hilbert spaces aren't always infinite dimensional.

Slereah

But quantum ones are, are they not

I mean you can contrive one to not be I guess

But a real one will be

user228700

Oct 5, 2016 08:36

The body wants to fly off but once it comes down, tension starts acting again, yeah?

user228700

OK, I think I got it.

user228700

@DanielSank@JohnRennie: THANKS SO MUCH!!! :-D

The h Bar

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