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01:33
@JohnRennie can static friction [as the only external force] cause acceleration?
I think it cannot cause acceleration without some other external force or atleast any force in general. .
can you please take a look here I feel like what I said is correct, but dale claims that static friction can cause motion as the only external force. .
01:52
I found a full family of questions where you cant judge where to account for potential due to other sphere or bot
02:34
@PrateekMourya d>>r usually means don't account for potential due to other sphere
I think you are right. they are using d>>r under the assumption that the spheres can be treated like point charges.
 
1 hour later…
03:45
@napstablook if you go strictly by "static friction is the only external force on some closed system", then I don't think it can accelerate the body; because for static friction to exist, an external normal force must exist on the body ($F_s \leq \mu N$), but that then contradicts the first statement (that no other external force exists).
Dale gives an example of a car, so it looks like to me that he has a different interpretation of what you are asking.
 
1 hour later…
04:57
@napstablook in the document see example 3.14 there they took potential due to other sphere too
And other questions are also mixed up like this
Where they either took or not at thier will
Here these are thier solutions and you can see they are all mixed uo
05:27
@napstablook I am inclined to say it can. Static friction is the constraint that forces the object and the surface to move together (i.e have no relative motion). If in a situation the surface is accelerating, then upto a point the static friction will provide the acceleration to the object on the surface , such that there is no relative motion
the classic two body problem comes to mind: Two blaocks on top of each other, and we apply a force F to the lower one.
if the two blocks move together, then the accn is F/m+M
if you draw the FBD of the top block, only horizontal force is static friction, and it is this that accelerates the top block. f=mF/m+M
so in this case , static friction causes the acceleratipn
in both cases , the point is that the surface is accelerating. So, static friction will cause the object to accelerate at the same rate as the surface.
so my final answer will be yes
but you might need to confirm it from @JohnRennie sir
@satan29 but static friction is not the only external force; the normal and the force of gravity do exist, although they balance each other out. napstablook mentions "...static friction [as the only external force]..."
but your statement "as the only external force" is a bit weird.static force requires a surface, so youll inevitably get a normal force as well
(@napstablook)
yes, I agree. Although that's what he mentioned... I guess if you take static friction as the only external unbalanced force, then I agree that it can surely cause acceleration.
@AshishAhuja yeah.
@PrateekMourya yeah this seems vague to me as well. I guess the only way to solve such questions would be by looking at the options and considering whether they have taken the approximation or not.
06:32
youtu.be/1xFRtdN5IJA?t=1396 from 23:16 to 25:03. Lewin says that the needles orient only in two perpendicular directions due to minimum energy considerations; which consideration is he talking about? The energy would be $U = -\mu \cdot B$, so according to energy considerations all the needles should be pointing in one direction. I don't see how energy considerations lead to only two perpendicular directions.
@satan29 this was exactly what I was thinking. The statement I want to make doesn't make much sense. static friction can definitely cause accleration though. The simplest case is rolling ball. '
one case that definitely proves my stat false is the particle spinning along a rough hemispherical bowl.
Looks like I have got to delete my answer.
@PrateekMourya for these qns the more accurate answer is taking the potential due to other spheres into account, so you should always do that. if the answer doesn't match then approximate it out. usually if distance between the spheres is given in options you should take into account the spheres' potentials
06:53
@AshishAhuja i dont know exactly, but we dont have a simple case of a dipole in a B-field, do we
there are multiple dipoles, so I guess well need to account for the dipole interaction with the B field, as well as dipole-dipole interactions
I agree but I would guess that it would give us many different directions, hence I assumed the simple dipole model. Without that assumption, I don't even know where to start.
although it's probably a bad assumption, I agree
@AshishAhuja can you give some more context of the problem? yt is blocked in my PC
23:17
I have another demonstration.
23:19
And another demonstration is to make you sort of see in a non-kosher way magnetic domains.
23:26
But I will tell you why it's non-kosher.
23:29
I have here an array of eight by eight magnetic needles, compass needles.
23:36
And you're going to see them there.
23:40
And I will change the situation so that you have better light.
23:47
And when I have an external magnetic field and I march over here a little and I just let it go,
23:56
and wait, you will see areas whereby these magnetic needles point in the same direction
that's the transcript, sorry for the long message. By array of needles, essentially there are 64 needles arranged in a square.
Lewin holds a bar magnet and goes over the array of needles, making them spin in random directions (and then he moves away)
They settle down again after a while, and the needles point in only two directions, and that too in groups; sort of like magnetic domains in ferromagnetic materials. But I don't see why they point only in two directions.
As in few groups point in one directions, while the others in a perpendicular direction.
Does anyone here know why the compressed object technique used for calculation of moment of inertia does not work for a hemisphere and the sphere while it works perfectly fine for a hollow cone? Thanks! :)
07:19
Note that the technique is that of putting a hollow cone under a hydraulic press and then converting into a disc and since the mass distribution has not changed thus moment of inertia of this disc is same for the cone as well. This is what I wish to question about in relation to hemisphere.
Oh hello there, my namesake
@Rishi you can do the same, except density will be different now unlike a cone
desmos.com/calculator/gtlcfoykn9 I graphed the integral for potential energy between a charge within a charged shell (x is the displacement of charge from centre of shell)
this is given an answer different from gausslaw which says potential within must be constant
why is this happening?
@RishiNandhaVanchi are you talking about the potential of the shell?
PE of the system
ill send again if that wasnt clear
find potential for this system:

There's a -q shell (insulating, uniform)
and there's a +q point charge within
+q is at x from centre of -q
we'd be getting -kq^2/r by gauss law, field inside =0
but getting something else by integrating kq/r
@AshishAhuja get it?
@RishiNandhaVanchi Okay so since the density is variable I cannot directly write I=mr^2/2 for the hemisphere and would have to integrate for the variable density.?
yeah
which would just become the integral with sphere before compressing it again
07:33
@RishiNandhaVanchi why would the field inside the shell be zero? There is a point charge inside, so surely it can't be zero throughout the shell..?
no i mean, field by shell alone
Like, potential energy of interaction bw shell and point charge = work done in bringing point charge from infinity to x from centre
ahh I got what you're saying now.
How are you getting something else by integrating kq/r?
@RishiNandhaVanchi Okay, Got it Thanks, but I have already typed about three fourths of the post about this question(and it is long) So if you want to answer there as well, feel feee to do so.:=)
The field inside the shell is zero (due to the shell), so at the moment you reach the shell the integral is done isn't it?
(gtg for lunch)
@AshishAhuja yeah
@JohnRennie Are you free sir?
07:44
@RishiNandhaVanchi hi :-) Yes, I'm free.
I'm trying to find potential energy (excluding self energies of the two) of this system sir:
There's a -q shell (insulating, uniform)
and there's a +q point charge within
+q is at x from centre of -q
The field is zero inside the shell so the PE is constant inside the shell.
Where I use gauss law, and bring the +q from infinity to where it is, im getting kq^2/R.
Yes sir
But when I integrate elementary rings im not getting the same
What integration are you doing?
Dividing the sphere into rings?
07:49
Do you want to go through the caculation?
I graphed it sir
I mean how to arrive at that integral. If the integral gives you the wrong answer that suggests the integral is wrong.
@JohnRennie ohhh
Ill check again sir, thanks!
What is t in your equation?
y is position of ring from centre of shell
I was thinking maybe there will be some potential energy because this is how dielectrics polarize and there is normally some energy associated with polarization
Like induced dipole moment in molecules
found my problem
08:03
Thanks @napstablook + @AshishAhuja
08:30
If air is pumped slowly inside a container (fixed volume and poor conductor of heat), will the temperature of the air inside the cylinder increase?
@RishiNandhaVanchi Is your insulator a shell or a solid sphere?
@Wolgwang ever pumped a cycle wheel with air? Try touching the cylinder with the air trapped in it..
Similar principle, the question is about how fast the heat is dissipated..
@SafdarFaisal Temperature increases.
@SafdarFaisal Is there some law that explains this?
Ideal Gas law explains it to an oversimplified model.. P increases T increases.
n also increases.
08:39
that's why it is an oversimplified model. It doesn't account the pressure exerted by the molecules on each other. and total energy in the system increases.
Every time you add air, it's like you are compressing the air that's already inside..
Ohk. Thanks :-)
 
2 hours later…
10:56
A student forgot to add the reaction mixture to the round-bottomed flask at 27 $^{\circ}$C but instead, he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 C. What fraction of air would have been expelled out?
I got Initial volume $v$ and final volume $2.5v$ (assuming pressure remains constant).
477 or 47?
How $V_{\mathrm{Final}}>V_{\mathrm{Initial}}$?
477
@Wolgwang what do you mean?? why as in??
The question says air has expelled out... the final volume of air in the flask should be less than the initial.
V_final is the net volume that the air expanded to..
So 1.5v/2.5v got expelled..
11:07
@SafdarFaisal Ok Got it! Thanks :p
 
1 hour later…
12:12
@AshishAhuja I am not exactly sure how energy consideration lead to these perpendicular directions. If it's any consolation, we don't really go into depth with ferro(you learn that these domains work the way do, and the relation of $\chi$ and T . we only go to some level of depth in para and dia.
and a bit of hysteresis :)
 
1 hour later…
13:19
@SafdarFaisal how did you read JDLEE book if you did? It's just so much i cant have it all in my head.

I also tried to create notes but its like writing whole book again in a note:(
I still need to read it for an exam in 3 days lol.. Funny you asked..
That's why I told you to focus on PSQ from inorganic.. Learn that way..
PSQ?
oh solving questions?
yeah..
Just memorising facts doesn't work..
13:22
Oh yeah i remember you telling...
i will try thanks
and another doubt
Like in coordination compounds how to compare two compounds stability
example?
oh wait oyu are there in chemistry group can we discuss there?
wait i will get the question from my paper
@SafdarFaisal what is PSQ? Practice solving qns?
that can't be it. .
got it
[Co(NO2)]-3 and [Fe(en)3]3+ which is stable
13:26
@ChristinaMelita that looks very borderline tbh
oh how? i have no idea how to check
but I'd go with the order NO2> en in stability
ok but how are saying that?
what factors determines it
its correct btw
there is a spectrochemical series. this is data based so we end up remembering an order
i remember the order.
but isnt it about which ligand is srong and which is weak?
13:28
stronger ligands give a nicer CFSE
@napstablook problem solving questions.
oh ok but does it not depend on metal?
@ChristinaMelita it would depend on Zeff and specifics of pairing energy. Wait a sec
I think I got it. You understand how to write CFSE with pairing energy and t2g-eg spacing in octahedral?
13:32
EAN??
@SafdarFaisal I thought EAN didn't apply to this set?
I have not revised coordinate for a long time tho
@napstablook ya i know
I checked EAN is for only CO and NO rest are inconclusive
EAN of Co is 36, EAN of Fe is 35.. so Co is more stable..
oh but what is EAN and spectochemical series didnt match each other.
like 1 is better in 1 thing and another thing is better in another one
13:36
this came from just pure intuition of what the question setter is asking for.. It's really cheesy...
okk
thank you @SafdarFaisal and @napstablook
@ChristinaMelita ok so Co has t2g(6) eg(o) and Fe has t2g(5) eg(0) if you compare with CFSE for these electronic configs Co gets an extra $0.4\Delta_{0}$
so the answer is justifiable either way. even if NO2 lost in spectrochemical series
13:38
okk
thanks
@napstablook wait you cant actually say it gets an extra 0.4 delta0 right cause delta0 varies? But as NO2 is stronger it doesnt actually matter as it makes it much stable but still just asking.
Is it okay to use a calculator while solving chemistry questions?
Nope..
calc during JEE is a pain..
5
13:56
I had started using the calculator this year only to save time while practicing. Thanks for the heads up :-)
 
7 hours later…
20:43
so hot
start to miss winter
because there is no cooler here.

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