The h Bar

MartianCactus

7:00 AM

i will understand what torque really IS real quick

brb

G. Bergeron

ok

user228700

@JohnRennie: Are u interested in (_/\ _) a discussion on fluids and snorkelling and pressure and all?

John Rennie

@Kaumudi.H yes of course. What's the question?

Secret

[Games] Perspective games like Echochrome and its variations often have weird projection rules to ensure each perspective is mapped to a unique block:

i.e. if moved to any of the red arrows, white cube get transported to where the arrows end

user228700

The question...well, erm, is to figure out how far beneath the surface if water I can go snorkelling.

user228700

7:08 AM

Right, so if I go very deep, I can't breathe in the air anymore, because the pressure in my lungs becomes greater than the pressure outside, above the water, yeah?

John Rennie

OK. You can breathe in because your chest muscles can expand your chest and reduce the pressure in it.

Suppose your lungs have some volume $V$ at rest. When your chest muscles expand your lungs to a new volume $V'$ the pressure inside falls to $P'/P = V/V'$ i.e. just Boyle's law.

user228700

Yes, yes, that makes sense.

John Rennie

The reduced pressure causes a pressure gradient between your lungs and the outside air, so air flows from the outside into your lungs.

user228700

Right, yes.

user228700

So if I need to be able to breathe, the pressure in my lungs has gots to be less than 1 atm.

Mew

7:11 AM

yes

John Rennie

The problem is that when you go underwater the water pressure applies a compressive force to your chest, and that reduces the ability of the chest muscles to expand the lungs and reduce the pressure inside them.

Mew

and when u breath out it must be more than 1 atm

user228700

@JohnRennie Right.

user228700

@Mew Huh, right.

John Rennie

Suppose the maximum pressure drop your chest muscles can create is $\Delta P$.

Mew

7:12 AM

@JohnRennie and the other problem is when you breath underwater

you fill your lungs with water rather than air

John Rennie

@Mew snorkling!!

user228700

x'D Dude.

Mew

oh

user228700

@JohnRennie Right.

John Rennie

Then if the external water pressure is much greater than $\Delta P$ it will stop your chest muscles from being able to expand your lungs and you won't be able to breate.

Mew

7:14 AM

@JohnRennie what's the longest straw one can drink from?

user228700

In that case, I can go as far down as $\Delta P/(\rho g)$, yeah?

John Rennie

It'll be a bit deeper than $\Delta P$ because I'd guess some of the water pressure is dissipated in the non-breathing related bits of your chest, but basically yes.

Mew

@JohnRennie is it not possible to breath without displacing water?

that is, could the diaphram move down

and thus still increase lung volume

John Rennie

I think it would be very hard to calculate exactly what the maximum depth is. You'd need to try the experiment.

user228700

@JohnRennie Yes, of course, OK.

user228700

7:16 AM

@JohnRennie Yeah, about the experiment:

John Rennie

@Mew but that pushes your stomach out and still displaces water.

Mew

@JohnRennie does it necessarily though?

user228700

8.01x - Lect 27 - Fluid Mechanics, Hydrostatics, Pascal's Principle, Atmosph. Pressure

►

Mew

I mean I can flex my stomach muslces

and breath without a noticible increase in volume

John Rennie

@Mew and if you do tht your diaphragm can't move.

Basically you're suggesting we could increase our volume without increasing our voulme.

Mew

7:17 AM

so there's no internal cavity for the diaphragm to move into though?

user228700

40:00

Mew

@JohnRennie not at all

@JohnRennie there is empty space in the abdominal cavity already

for the diagraphm to move into

no need for the body itself to expand, the diagraphm can just move into the abdominal cavity

and thus the lung volume increases, while abdominal volume decreases and total volume remains constant

John Rennie

@Mew it's not empty. The only bit of the body that contains a compressible substance (air) is the lungs.

user246160

@Mew ABDOMINAL CAVITY IS NOT EMPTY. IT HAS ORGANS.

Mew

@JohnRennie r u sure about that?

MartianCactus

7:19 AM

so i just understood what a center of mass is

John Rennie

Unless you've been eating beans of course, in which case your bowels will be slightly compressible :-)

MartianCactus

now

Mew

@TheStackExchange it's not 100% full of organs

John Rennie

@Mew yes it is.

Mew

@TheStackExchange have you heard of peritoneal dialysis

John Rennie

7:19 AM

There's no empty space in there.

Mew

@JohnRennie have you heard of gas in the stomach?

user246160

@Mew But atleast 95 percent organs. And that extra 5 percent isnt enough for breathing.

MartianCactus

my question is- they say that the center of mass is the point from where all the mass of an object towards all directions is equal

John Rennie

@Mew sounds like a lot of hot air to me ;-)

MartianCactus

ir atleast thats what i understood

but what exactly IS an object?

John Rennie

7:20 AM

The point is there's not much gas in the bowels, at least not compared to the lungs.

MartianCactus

like, there is mass all around me

user246160

@Mew Place a 50 kg load on your chest and try to breathe with only your abdominal cavity ;)

MartianCactus

actually i will let this discussion end first

Mew

@JohnRennie so when I suck in my stomach, is air from my lungs inevitably being expelled?

since there is no room for dcompression in the abdominal cavity?

@JohnRennie so I could try and breath just by sucking my stomach in and out?

or I guess this could just increase the pressure in the lungs

but still

John Rennie

@Mew Yes. If you learning singing at a professional level they teach you to breathe just using your diaphragm.

Mew

7:22 AM

@JohnRennie is it possible for the pressure of the abdominal cavity to icnrease?

to allow me to breath underwater by expanding the diaphramg into the abdominal cavity

and in turn increasing the pressure of the abdominal cavity

John Rennie

@Mew Pressure yes, volume no - or not significantly.

Mew

k

user228700

Alright, so this is what Walter Lewin did:

John Rennie

@Kaumudi.H did we answer your question, or has it been lost in all the discussion of bowel motions?

Slereah

Remember that question I asked about conical singularities a few months ago?

I finally found the answer in some Ellis paper

About bloody time

user228700

7:25 AM

@JohnRennie No, it got lost.

Mew

@JohnRennie have artificial gills been created yet?

John Rennie

@Mew let's sort @Kaumudi.H 's question first ...

Mew

k

user246160

@MartianCactus Object means any object. Could be a block or even your body.

user228700

This is what WL did:

MartianCactus

7:26 AM

whats the definition of an object?

i mean i am sitting on my chair

and my mass is connected to the chair's mass

then what will be the center of mass?

user228700

Mew

@MartianCactus I'd need to know the dimensions of each object and their mass distributions

user228700

John Rennie

@Kaumudi.H I remember doing that in school :-)

MartianCactus

7:28 AM

dimensions?

user246160

@MartianCactus It can be calculated mathematically. It will be a weighted average of the COM of you and the chair.

MartianCactus

so you take out the center of mass of the system?

John Rennie

@Kaumudi.H You suck on the tube and see how big a pressure difference you can create.

MartianCactus

like the system of me and my chair

user246160

@MartianCactus Yes

user228700

7:29 AM

@JohnRennie Yeah.

Mew

@MartianCactus you find out the average position (weighted by mass) of all the particles under consideration

user228700

I don't get it :-|

MartianCactus

and if i get up and seperate myself from the chair then the center of mass wont be there anymore as we are not connected?

Mew

@MartianCactus it is possible to have a centre of mass for any objects under consideration (connected or not)

MartianCactus

all the particles?

John Rennie

7:30 AM

@Kaumudi.H I guess this goes back to the discussion we had a few days ago about pressure differences ...

Mew

@MartianCactus for example, I can calculate the centre of mass for you, and the centre of mass for the chair seperate, and I can also calculate the centre of mass for the person-chair system

MartianCactus

and then if you connect them with a uniform rod, the center of mass will be the same?

Mew

@MartianCactus if the rod has 0 weight then yes

user246160

@MartianCactus No, then the COM of the rod will also be taken into account (if it has a weight).

MartianCactus

oh so if there isnt any gravity around?

Mew

7:31 AM

0 mass* sorry

MartianCactus

what is COM?

Mew

gravity has nothing to do with it

user246160

@MartianCactus COM doesn't depend on gravity

MartianCactus

oh ok

user228700

@JohnRennie A few days ago? O_o We had a discussion just today!

MartianCactus

7:31 AM

oh

but the rod is uniform

how will that change the COM?

Mew

easy

because

lets say without the rod, the centre of mass is closest to the right of the system

MartianCactus

ok

Mew

and then you add the rod to the middle of the system

MartianCactus

yeah..

Mew

because the rod's mass is being added to the left of the centre of mass, the centre of mass will now shift closer to the left

thus adding the rod has changed the COM even though it is "uniform"

but if the rod is massless, then it would have no effect

John Rennie

7:33 AM

@Kaumudi.H Take this diagram:

user228700

Okay...

John Rennie

This is supposed to show the setup in your picture.

user228700

Hey, @JohnRennie: Will u be here in 1/2 an hour? I'm starving →distracted.

John Rennie

Yes. I'll go make a coffee and do some work. Ping me when you're back.

user228700

OK :-)

MartianCactus

7:35 AM

well then if we make the rod heavier from the right side of the original COM and if we make its mass equal to the mass of the left side of the COM then the COM will stay the same?

Mew

@MartianCactus Let's specifiy the situation more clearly for me to explain:

(1) A person is on the right of a given point

MartianCactus

yes

Mew

(2) A cat is on the left of a given point

MartianCactus

ok, nice example

Mew

we know that the COM of this system is going to be on the right of the given point (because the human is more mass than cat)

MartianCactus

7:37 AM

true

Mew

Now let's add a bar that joins the human and the cat

If the Bar is uniform, the bar's COM will be exactly at the "given point"

MartianCactus

and the distance of the person and cat from the given point is equal but in opposite directions..

Mew

yep

so to work out the new COM

MartianCactus

ok

wait

oh u said the bar's COM

Mew

yeah

MartianCactus

7:39 AM

ok now we will be calculating the system's COM right?

Mew

the Bar will have 50% of it's mass to the right of the point, and 50% to the left of the point

yes

MartianCactus

ok

Mew

to work out the cat-human-bar system COM we can think of it as combinging (1) the cat-human system with (2) the bar system

now the (1) cat-human system has a COM to the right of the point

and the (2) bar system has a COM exactly on the point

MartianCactus

yeah

Mew

so the weighted average of system (1) and (2) will be a point somwhere between the "given point" and the COM of the cat-human system (which is to the right of the "given point")

thus clearly adding the rod has changed the COM

it has shifted the COM closer to the centre

MartianCactus

7:41 AM

so the system's COM will still remain at the right but a little bit towards the left from the cat-human COM

Mew

correct

MartianCactus

ok

now another example

Mew

and how much further will depend on the weight of the bar relative ot the weight of the cat-human system

MartianCactus

yeah ok

now another thing

let's say right now, John Cena is in Canada..

and I am in India

there is nothing connecting us

but we will still have a center of mass right?

Mew

correct

MartianCactus

7:43 AM

and as John Cena is much more heavier than me

the COM will be towards the western side of the globe

Mew

let's say your individual COM if position X and your mass is M1

and let's say John Cena's individual COM is position Y and his mass is M2

MartianCactus

yeah

Mew

then the COM of you and him is (xM1 + yM2)/(M1+M2)

MartianCactus

yeah

Mew

that is, a position weighted by your respective masses

MartianCactus

7:44 AM

but

we didn't calculate the amount of air between us

and the mass of air will contribute to the COM of our system won't it?

Mew

yes because we choose what objects we want to consider in our calculation

we are calculating the COM of the you-John system, not the you-John-air system

MartianCactus

well in a real life situation

there WILL be air

Mew

yes and there will be the Earth and the Sun and the Moon etc.

MartianCactus

what will we do then?

how do we calculate the COM of the system then?

Mew

we have to define the system we want to consider carefully and define the rest as "exteranl environment"

I can choose whatever objects I want to to include in my COM calc

as long as I'm consistent in my calculation between the system and the environment

MartianCactus

7:46 AM

doesn't the external environment affect the system too?

Mew

it does but you can seperate it tout

that is,

Newton's 1st law is that an object will travel at a constant velocity unless acted upon by an EXTERNAL force

now I can consider any objects I want to constitute the "object" and it's position will be the COM

MartianCactus

in a real life situation, seperating an external environment which affects the system, won't that lead to misfunctioning or something?

Mew

and then this COM will change velocity only if there is a force from the "EXTERNAL" system on that object I hvae chosen

@MartianCactus not at all

MartianCactus

oh

Mew

@MartianCactus Newton's 1st law explains the connection beteween the system you choose and the "exteranl environment" that may act on that system

for example

MartianCactus

7:48 AM

so the system will only change if an external force is applied on the COM?

Mew

correct

or objects that constitute that COM

consider this:

Suppose there is a universe with only 2 balls heading towards each other, and they rebound

the balls are of equal mass

I can consider (A) the COM of both balls combined

or (B) consider COM of one ball and consider the other ball exerting an external force

in the case of (A) the COM is going to be the collision point of the two balls

this won't change, because there is no external force acting on the balls

MartianCactus

yeah

Mew

or (B) there is an external force on the ball and thus the COM of a single ball will change

so we can conclude the COM of the ball-ball system doesn't change, but the COM of each individual ball does

see how you can use whatever objects you want to construct the COM, as long as your consistent with applying Newton's laws and thinking about what is the "system" and what is the "external forces" acting on that system

MartianCactus

this is confusing

how will the COM of the individual balls change?

Mew

because the balls are moving

if the ball is a point particle, the ball position is that ball's COM

so each individual ball's COM is clearly changing as the ball moves

MartianCactus

7:52 AM

oh yeah

Mew

but the COM of the combined ball-ball system isn't moving (Because there is no "external" force on the ball-ball system)

I should say "accelerating" actually

MartianCactus

but the COM of the individual balls is moving all the time

Mew

the COM of the combined ball-ball system isn't accelerating

the COM of the individual ball is acclerating during the collision with the other ball

because the other ball exerts an external force on it

and thus the velocity of one ball can change

but the combined ball-ball system has no external forces and thus has a constant velocity (of 0)\

does this make sense?

MartianCactus

so the COM of individual balls IS moving but it accelerates(or deacclerates i shouild say) only when they collide

Mew

yep

MartianCactus

7:55 AM

oh

Mew

by Newton's 1st las

MartianCactus

now

how will you get the COM of the MartianCactus-JohnCena-air system?

Mew

if you want an exact answer you'd have to know the position of all the air molecules

and their mass

MartianCactus

EVERY air molecule?

Mew

yeah of course

you could estimate by guessing a distribution of air around the world

but this would still be very hard if not impossible

MartianCactus

7:57 AM

oh the area-probablity thing?

Mew

given how variable weather is

MartianCactus

i think it was in my Math book

so we just neglect it normally?

Mew

but it's kind of arbitrary to look at human-human-air system

MartianCactus

or do experiments in vaccum?

Mew

why not look at human-human-air-building-earth-moon-sun?

MartianCactus

7:58 AM

what does arbitrary mean?

Mew

normally you just look at the system whos' change your interested in

MartianCactus

but the sun and moon are not connected by earth in any way

Mew

arbitrary means like why would you consider air rather than water or buildings or birds?

MartianCactus

oh

Mew

@MartianCactus they are connected by gravity

MartianCactus

7:59 AM

and then if we connect a rod between me and Cena

but gravity doesnt have mass

so the COM won't rly do anythign

Mew

you misunderstood

how is air connected to you?

it is connected to you by the EM force

MartianCactus

i mean it IS mass..

Mew

similarly the sun is connected to you by gravity force

MartianCactus

yeah

Mew

so if we are considering air, then why aren't we considering the SUn too?

the sun is just as connected to you as air is

MartianCactus

8:01 AM

but air is not connected to me as tightly as the rod is connected with me and John Cena, so the air doesn't rly make a big defference?

Mew

@MartianCactus as I said, in any physics problem you define what the "system" your interested is, and then define everthing esel as "external"

so we don't include air in the system, we include it in "external"

we are free to chooose it to be system or external whichever we want, Newton's law remains valid

I'm not interested in how the COM of air changes, so why would I include air in the "system"

MartianCactus

and the air doesn't rly exert a BIG external force so the calculation is almost correct?

Mew

what equation?

MartianCactus

i mean the calculation

Mew

my calculation above for the COM of the you-John system ISNT an approximation

it is an EXACT calculation

because the COM is a quantity for your choice of objects

and in this case the choice of objects doesn't include air

the COM in itself isn't a real physical quantity, it's a mathematical tool for simplyfing Newton's equations

and you can use this tool however you want to make whatever particular problem your solving easier

MartianCactus

8:05 AM

and the air exerts no external influence on the me-john cena-rod system?

Mew

and I've never come accross a problem where it was simplest to include "air" in the COM system I define

@MartianCactus of course it exerts force on that system

MartianCactus

but you said an external force can change the COM

Mew

yes

MartianCactus

like in the ball-ball system

Mew

yeah

MartianCactus

8:06 AM

then won't the external froce that air exerts on the system change the COM?

Mew

yes

so the COM wont' be constant over time

MartianCactus

oh

Mew

the acceleration fo the COM of the you-John-rod system will accelerate according to F = ma

whgere F is the force of the air on the John-you-rod system

and a is the acceleration fo the COM of the you-John-rod system

MartianCactus

oh

and let's say now that there is no air on earth

so there is no external force on the dead cena- dead me- rod system

then we add a pivot under the COM

then the rod will remain balanced right?

Mew

the COM has nothign to do with weight

it is to do with mass

and the earth is curved

i'm not sure how such a curved pivot would work

but let's suppose the Earth is flat

and the pivot was placed under the COM then I'd say yes it would be balanced

MartianCactus

8:10 AM

yeah i AM supposing that the surface is flat

G. Bergeron

damn...

MartianCactus

ok

that was a BIG discussion

Mew

yea

MartianCactus

but thanks now i understand :D

Mew

awesome

laterz

MartianCactus

8:10 AM

same o/

user246160

0

Q: Why isn't work energy theorem equation being valid in both the reference frames?

My question will be a follow up to the PSE question Is the energy conserved in a moving frame of reference? In this question I am trying to write the Work-Energy theorem equation from the man's frame of reference. From ground frame if I write the work-energy equation, I know $$\frac{1}{2}mv^2=m...

user246160

Can someone help me with this ^

user246160

Am I calculating the work done by normal force wrong ?

user246160

@JohnRennie You there ? :P

user228700

I'm back, but I need some urgent help.

user246160

8:20 AM

@Kaumudi.H Even I need urgent help!

user228700

Mine is not even related to the subject but it's very very urgent or my brain is going to explode.

John Rennie

@Kaumudi.H sounds like you need a doctor not a physicist :-)

user246160

@Kaumudi.H Okay stopping the explosion is more important than my question, apparently :P

user228700

Please, does anybody here watch Comedy Central? They have a line-up for Winter and they showcased it with a particular song. I HAVE TO KNOW WHAT THAT SONG IS.

user246160

@Kaumudi.H :/

John Rennie

8:22 AM

@Kaumudi.H is there a link I can hear the song?

user228700

I've heard it before and I thought it goes "You and me one kiss" but that yielded no results :-(

user228700

@JohnRennie Uuum. Okay, wait, lemme check...

user228700

Uuuugh. Nothing.

user246160

@Kaumudi.H -_____________________________________________-

user228700

:'-( I'm going to spend another 10 minutes looking for it. @TheStackExchange: Go ahead and ask ur question. I'll brb.

user228700

8:25 AM

@TheStackExchange: WHAT?!?! My head is going to explode if I don't find it now. Godammit.

user246160

@Kaumudi.H -_-

user246160

Nothing :P

user246160

@Kaumudi.H Let it explode :P .... you will appear in the newspaper tomorrow ! :-D

user228700

-___- Quit making fun of me. This is a proper crisis for a person like me.

user246160

@Kaumudi.H No, I am not making fun. I am SERIOUS!

user246160

8:28 AM

LET IT EXPLODE !

user246160

BAM!

user246160

Meanwhile let me ask my questions

user246160

:P

user246160

0

Q: Why isn't work energy theorem equation being valid in both the reference frames?

My question will be a follow up to the PSE question Is the energy conserved in a moving frame of reference? In this question I am trying to write the Work-Energy theorem equation from the man's frame of reference. From ground frame if I write the work-energy equation, I know $$\frac{1}{2}mv^2=m...

homework-and-exercises newtonian-mechanics energy reference-frames work

user246160

Someone help me ^ !!!!

John Rennie

8:32 AM

@TheStackExchange I've edited your question to include the image from the original question and the equation numbers. You can use \tag{n} to typeset an equation number (n)

user246160

@JohnRennie Oh...thanks for the edit :)

user246160

Okay I'll use \tag{n}

user246160

next time

user246160

But any idea about the question ?

user246160

Is my work calculation wrong for normal force ?

user228700

8:38 AM

Siiiiiiiigh. Nothing :'-(

user246160

@Kaumudi.H Yahooooooooooooooo!

user228700

^ Said the sadist.

user246160

Yeah, I am a sadist. True :-D

user228700

I don't give up that easy tho. I'm going to download "Shazam", the app, go back to the channel, wait for them to showcase that lineup again and find it. I'll do it in the evening.

user246160

I am not suffering alone in the world and I am feeling so happy about that now :P.....that feeling when you are staring at a problem for 2 hours and not able to solve it

user246160

8:41 AM

@Kaumudi.H Shazam is good

Slereah

@JohnRennie sorry, nobody actually understands QFT

user246160

There are several other apps too

Slereah

It's hopeless

John Rennie

@Slereah yes, I was being excessively optimistic :-)

Slereah

Stick to GR

GR is great

user228700

8:42 AM

This is too important. It's not even that that song is absolutely amazing or anything. Like I said, I think I've heard it before, even! But still, I have gots to know.

Slereah

or toothpaste

I found a GR article on singularity classification

user228700

Anyway. @JohnRennie: For how much longer will u be around?

Slereah

It's pretty cool

John Rennie

@Slereah the disribution of liquid crystal phases present in toothpaste still isn't fully understood. Or it wasn't when I last looked.

Slereah

Do you happen to know the stress energy tensor of toothpaste

user246160

8:43 AM

@Kaumudi.H You need a doctor....really :)

John Rennie

@Kaumudi.H I'm here till 12 ish UTC (another 4 hours or so). But I'll be busy for the next half hour or so.

user228700

@TheStackExchange Everybody is crazy in their own ways.

user228700

@JohnRennie Oh, that's perfect. I'll be back after some Chemistry!

John Rennie

@TheStackExchange I have to dash off and do some work, but note that the horizontal distance the block moves is different in the two frames. This matters because work is $int Fdx$.

Calculating how far the block moves in the rest frame of the moving observer isn't trivial ...

Slereah

what would be the metric generated by a tube of toothpaste

Assuming an axisymmetric tube

What the hell

Ellis just made a parody of "Dialogue Concerning the Two Chief World Systems" in his paper

Sagredo and Salviati are discussing quasi-regular singularities

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