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7:00 AM
i will understand what torque really IS real quick
brb
 
@JohnRennie: Are u interested in (_/\ _) a discussion on fluids and snorkelling and pressure and all?
 
@Kaumudi.H yes of course. What's the question?
 
[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
 
The question...well, erm, is to figure out how far beneath the surface if water I can go snorkelling.
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?
 
7:08 AM
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.
 
Yes, yes, that makes sense.
 
The reduced pressure causes a pressure gradient between your lungs and the outside air, so air flows from the outside into your lungs.
 
Right, yes.
So if I need to be able to breathe, the pressure in my lungs has gots to be less than 1 atm.
 
Mew
yes
 
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
7:12 AM
and when u breath out it must be more than 1 atm
 
@JohnRennie Right.
@Mew Huh, right.
 
Suppose the maximum pressure drop your chest muscles can create is $\Delta P$.
 
Mew
@JohnRennie and the other problem is when you breath underwater
you fill your lungs with water rather than air
 
@Mew snorkling!!
 
x'D Dude.
 
Mew
7:13 AM
oh
 
@JohnRennie Right.
 
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
@JohnRennie what's the longest straw one can drink from?
 
In that case, I can go as far down as $\Delta P/(\rho g)$, yeah?
 
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
7:15 AM
@JohnRennie is it not possible to breath without displacing water?
that is, could the diaphram move down
and thus still increase lung volume
 
I think it would be very hard to calculate exactly what the maximum depth is. You'd need to try the experiment.
 
@JohnRennie Yes, of course, OK.
@JohnRennie Yeah, about the experiment:
 
@Mew but that pushes your stomach out and still displaces water.
 
Mew
@JohnRennie does it necessarily though?
 
Mew
7:16 AM
I mean I can flex my stomach muslces
and breath without a noticible increase in volume
 
@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
so there's no internal cavity for the diaphragm to move into though?
 
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
 
@Mew it's not empty. The only bit of the body that contains a compressible substance (air) is the lungs.
 
user246160
7:19 AM
@Mew ABDOMINAL CAVITY IS NOT EMPTY. IT HAS ORGANS.
 
Mew
@JohnRennie r u sure about that?
 
so i just understood what a center of mass is
 
Unless you've been eating beans of course, in which case your bowels will be slightly compressible :-)
 
Mew
@TheStackExchange it's not 100% full of organs
 
7:19 AM
@Mew yes it is.
 
Mew
@TheStackExchange have you heard of peritoneal dialysis
 
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.
 
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
 
7:20 AM
@Mew sounds like a lot of hot air to me ;-)
 
ir atleast thats what i understood
but what exactly IS an object?
 
The point is there's not much gas in the bowels, at least not compared to the lungs.
 
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 ;)
 
actually i will let this discussion end first
 
Mew
7:21 AM
@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
 
@Mew Yes. If you learning singing at a professional level they teach you to breathe just using your diaphragm.
 
Mew
@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
 
@Mew Pressure yes, volume no - or not significantly.
 
Mew
k
 
Alright, so this is what Walter Lewin did:
 
7:24 AM
@Kaumudi.H did we answer your question, or has it been lost in all the discussion of bowel motions?
 
Remember that question I asked about conical singularities a few months ago?
I finally found the answer in some Ellis paper
About bloody time
 
@JohnRennie No, it got lost.
 
Mew
@JohnRennie have artificial gills been created yet?
 
@Mew let's sort @Kaumudi.H 's question first ...
 
Mew
k
 
user246160
7:26 AM
@MartianCactus Object means any object. Could be a block or even your body.
 
This is what WL did:
 
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?
 
Mew
@MartianCactus I'd need to know the dimensions of each object and their mass distributions
 
7:28 AM
@Kaumudi.H I remember doing that in school :-)
 
dimensions?
 
user246160
@MartianCactus It can be calculated mathematically. It will be a weighted average of the COM of you and the chair.
 
so you take out the center of mass of the system?
 
@Kaumudi.H You suck on the tube and see how big a pressure difference you can create.
 
like the system of me and my chair
 
user246160
7:28 AM
@MartianCactus Yes
 
@JohnRennie Yeah.
 
Mew
@MartianCactus you find out the average position (weighted by mass) of all the particles under consideration
 
I don't get it :-|
 
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)
 
7:29 AM
all the particles?
 
@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
 
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).
 
7:31 AM
oh so if there isnt any gravity around?
 
Mew
0 mass* sorry
 
what is COM?
 
Mew
gravity has nothing to do with it
 
user246160
@MartianCactus COM doesn't depend on gravity
 
oh ok
 
7:31 AM
@JohnRennie A few days ago? O_o We had a discussion just today!
 
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
 
Mew
and then you add the rod to the middle of the system
 
yeah..
 
Mew
7:32 AM
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
 
@Kaumudi.H Take this diagram:
 
Okay...
 
This is supposed to show the setup in your picture.
 
Hey, @JohnRennie: Will u be here in 1/2 an hour? I'm starving →distracted.
 
Yes. I'll go make a coffee and do some work. Ping me when you're back.
 
7:34 AM
OK :-)
 
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
 
Mew
(2) A cat is on the left of a given point
 
ok, nice example
 
Mew
7:37 AM
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)
 
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"
 
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
 
ok
wait
oh u said the bar's COM
 
Mew
7:38 AM
yeah
 
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
 
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
 
Mew
7:40 AM
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
 
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
 
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
 
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
7:43 AM
correct
 
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
 
Mew
then the COM of you and him is (xM1 + yM2)/(M1+M2)
 
Mew
7:44 AM
that is, a position weighted by your respective masses
 
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
 
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.
 
what will we do then?
how do we calculate the COM of the system then?
 
Mew
7:45 AM
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
 
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
 
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
 
Mew
7:48 AM
@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
 
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
 
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
 
this is confusing
how will the COM of the individual balls change?
 
Mew
7:52 AM
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
 
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
 
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?
 
so the COM of individual balls IS moving but it accelerates(or deacclerates i shouild say) only when they collide
 
Mew
7:55 AM
yep
 
Mew
by Newton's 1st las
 
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
 
EVERY air molecule?
 
Mew
7:56 AM
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
 
oh the area-probablity thing?
 
Mew
given how variable weather is
 
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
 
or do experiments in vaccum?
 
Mew
7:58 AM
why not look at human-human-air-building-earth-moon-sun?
 
what does arbitrary mean?
 
Mew
normally you just look at the system whos' change your interested in
 
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?
 
Mew
7:59 AM
@MartianCactus they are connected by gravity
 
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
 
i mean it IS mass..
 
Mew
similarly the sun is connected to you by gravity force
 
Mew
8:00 AM
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
 
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"
 
and the air doesn't rly exert a BIG external force so the calculation is almost correct?
 
Mew
what equation?
 
i mean the calculation
 
Mew
8:03 AM
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
 
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
 
but you said an external force can change the COM
 
Mew
yes
 
like in the ball-ball system
 
Mew
8:06 AM
yeah
 
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
 
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
 
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
8:09 AM
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
 
yeah i AM supposing that the surface is flat
 
damn...
 
ok
that was a BIG discussion
 
Mew
yea
 
but thanks now i understand :D
 
Mew
8:10 AM
awesome
laterz
 
same o/
 
user246160
0
Q: Why isn't work energy theorem equation being valid in both the reference frames?

TheStackExchangeMy 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
 
8:19 AM
I'm back, but I need some urgent help.
 
user246160
@Kaumudi.H Even I need urgent help!
 
Mine is not even related to the subject but it's very very urgent or my brain is going to explode.
 
@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
 
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
8:22 AM
@Kaumudi.H :/
 
@Kaumudi.H is there a link I can hear the song?
 
I've heard it before and I thought it goes "You and me one kiss" but that yielded no results :-(
@JohnRennie Uuum. Okay, wait, lemme check...
Uuuugh. Nothing.
 
user246160
@Kaumudi.H -_____________________________________________-
 
:'-( I'm going to spend another 10 minutes looking for it. @TheStackExchange: Go ahead and ask ur question. I'll brb.
@TheStackExchange: WHAT?!?! My head is going to explode if I don't find it now. Godammit.
 
user246160
@Kaumudi.H -_-
 
user246160
8:26 AM
Nothing :P
 
user246160
@Kaumudi.H Let it explode :P .... you will appear in the newspaper tomorrow ! :-D
 
-___- 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
LET IT EXPLODE !
 
user246160
BAM!
 
user246160
8:28 AM
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?

TheStackExchangeMy 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
Someone help me ^ !!!!
 
@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
8:32 AM
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 ?
 
Siiiiiiiigh. Nothing :'-(
 
user246160
@Kaumudi.H Yahooooooooooooooo!
 
8:39 AM
^ Said the sadist.
 
user246160
Yeah, I am a sadist. True :-D
 
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
@Kaumudi.H Shazam is good
 
@JohnRennie sorry, nobody actually understands QFT
 
user246160
8:41 AM
There are several other apps too
 
It's hopeless
 
@Slereah yes, I was being excessively optimistic :-)
 
Stick to GR
GR is great
 
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.
 
or toothpaste
I found a GR article on singularity classification
 
8:43 AM
Anyway. @JohnRennie: For how much longer will u be around?
 
It's pretty cool
 
@Slereah the disribution of liquid crystal phases present in toothpaste still isn't fully understood. Or it wasn't when I last looked.
2
 
Do you happen to know the stress energy tensor of toothpaste
 
user246160
@Kaumudi.H You need a doctor....really :)
 
@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.
 
8:44 AM
@TheStackExchange Everybody is crazy in their own ways.
@JohnRennie Oh, that's perfect. I'll be back after some Chemistry!
 
@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 ...
 
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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