Problem Solving Strategies

RishiNandha Vanchi

02:27

postulates of kinetic gas theory

there were assumptions that defined "ideal" gases

all collision are elastic and all particles of same size right? so even if they collide they only merely exchange their velocity vectors

being identical you cant see a difference

Prateek Mourya

03:48

What if they didn't collide head on

Then they only move perpendicular

napstablook

05:04

@JohnRennie please help me in understanding the meaning of static pressure.

user586228

05:23

To quote from my book there is a quote,"Consider a rectangular parallelopiped with edges dx,dy,dz.The volume in strained condition is $(1+\epsilon_{x})(1+\epsilon_{y})(1+\epsilon_{z})dxdydz$.

John Rennie

@napstablook what's the issue with static pressure?

Prateek Mourya

Hello @JohnRennie sir

RishiNandha Vanchi

@PrateekMourya no

that holds for not head on too

the case you are talking about is the second one in rest

Prateek Mourya

05:38

Sorry my bad

But still in oblique collision they don't do

That's the case for head on collision you are talking

The situation is different in oblique collision

RishiNandha Vanchi

take vectors and solve for it

you are right with oblique collisions, but we arent talking gas particles as spheres

Prateek Mourya

05:54

The velocity is perpendicular to the line of impact don't change

May exchange velocities along the line of impact

RishiNandha Vanchi

yes

Prateek Mourya

But what about the velocities perpendicular to the line of impact

Still their velocities are changed

RishiNandha Vanchi

they dont get exchanged ur right, i got the diagram wrong

Prateek Mourya

The same configuration of velocity is not conserved

If the configuration of velocity was same before and after collision then we can say that the collision doesn't change the impact with wall at

RishiNandha Vanchi

but looking at it from outside momentum along each axis conserved

i shouldve said in net you dont see a difference, i got that part wrong my bad

napstablook

05:57

@JohnRennie I don't get it. I feel like it is the same as hydrostatic pressure.

@RishiNandhaVanchi Ideal gas model does assume the particles to be rigid spheres though.

RishiNandha Vanchi

oh

it says negligible volume right?

napstablook

yes you are correct the negligible volume allows us to assume that all collisions are head on, I think.

RishiNandha Vanchi

nono i got the head on part wrong

but in the end the derivation is based on net momentums, and some ideal considerations, not individual ones

napstablook

are you saying we do consider ideal gases to have oblique collisions? I don't recall doing calculation for them in my classes. . .

I mean the unimolecular and bimolecular collision we see in gaseous state not in physics

RishiNandha Vanchi

no, saying that even if they collide like that we have net momentum vectors undisturbed and so we dont consider those

they dont collide at all?

John Rennie

06:04

@napstablook static and hydrostatic mean the same don't they? Although hydrostatic strictly speaking applies only to water.

napstablook

I think they are treated quite differently. . if you look at the bernoulli equation along a streamline

wait let me take a picture

prnt.sc/wn4pmw

prnt.sc/wn4q46 is the form it is talking about in the last picture

@JohnRennie by hydrostatic I wanted to refer to the pressure pgh which is dealt in statics although I must admit it is misleading

John Rennie

OK, I guess they are distinguishing between an externally applied pressure $P$ and the pressure due to the fluid's own weight $\rho g h$.

napstablook

is that atmospheric pressure?

John Rennie

It could be, but more commonly your system of pipes would be connected to a pump, and that pump would be producing some external pressure

napstablook

prnt.sc/wn4u7x can you take a look at this question? they are treating this value as the pressure if fluid were at rest?

It seems to include pressure by it's own weight too

user586228

06:18

Q2-4 and Q2-5

I need help for these two

To quote from my book there is a quote,"Consider a rectangular parallelopiped with edges dx,dy,dz.The volume in strained condition is $(1+\epsilon_{x})(1+\epsilon_{y})(1+\epsilon_{z})dxdydz$.

@JohnRennie I a=have taken two problems in a single post can you help me out with these two

John Rennie

@napstablook I must admit it's not immediately obvious to me how that calculation works.

user586228

@JohnRennie Did u get my second query?

John Rennie

@user586228 I'm busy at the moment. I'll ping you as soon as I'm free.

user586228

ok no problem :)

John Rennie

06:41

@user586228 For 2-4 I think you just multiply the original tensor by the transformation matrix. The transformation matrix is what has been written in blue pen to the right.

user586228

Yes but I am not convinced about how this matrix was fraed.

framed..

John Rennie

Do you mean you aren't sure how the transformation matrix is derived?

user586228

absolutely

John Rennie

The simple way to derive the transformation matrix is to consider what it does to the unit vectors.

user586228

ok

John Rennie

06:45

We rotate an angle θ anticlockwise about the z axis. Let me draw a quick diagram ...

user586228

ok

John Rennie

When we do the rotation the unit vector (1,0) becomes (cosθ, sinθ). Yes?

user586228

ok ..but explain -sin and cos that part

John Rennie

@user586228 Well the j vector (0,1) becomes (-sinθ, cosθ). Yes?

user586228

Why minus sin $\theta$?

John Rennie

06:53

Look at the diagram

The blue arrow gets rotated so its tip moves left i.e. to -x

user586228

yes ok then..

John Rennie

So we need to find a matrix so that:

$$ \left(\begin{matrix} a && b \\ c && d \end{matrix} \right) \left(\begin{matrix} 1 \\ 0 \end{matrix} \right) = \left(\begin{matrix} \cos\theta \\ \sin\theta \end{matrix} \right) $$

And:

$$ \left(\begin{matrix} a && b \\ c && d \end{matrix} \right) \left(\begin{matrix} 0 \\ 1 \end{matrix} \right) = \left(\begin{matrix} -\sin\theta \\ \cos\theta \end{matrix} \right) $$

Yes?

@user586228 hello?

user586228

ok let me try once

John Rennie

Well look at the top equation

$$ \left(\begin{matrix} a && b \\ c && d \end{matrix} \right) \left(\begin{matrix} 1 \\ 0 \end{matrix} \right) = \left(\begin{matrix} a \\ c \end{matrix} \right) $$

@user586228 Yes?

user586228

ok

John Rennie

07:04

So a must be cosθ and c must be sinθ

Likewise:

$$ \left(\begin{matrix} a && b \\ c && d \end{matrix} \right) \left(\begin{matrix} 0 \\ 1 \end{matrix} \right) = \left(\begin{matrix} b \\ d \end{matrix} \right) $$

So b must be -sinθ and d must be cosθ

user586228

ok

then

John Rennie

So our transformation matrix is:

$$ \left(\begin{matrix} \cos\theta && -\sin\theta \\ \sin\theta && \cos\theta \end{matrix} \right) $$

I've done this in 2D to keep it simple. The k unit vector (0, 0, 1) is unchaged by the rotation, so the full 3D matrix is:

$$ \left(\begin{matrix} \cos\theta && -\sin\theta && 0 \\ \sin\theta && \cos\theta && 0 \\ 0 && 0 && 1 \end{matrix} \right) $$

satan 29

transformation matrix= ([vector where i goes] [vector where j goes] [vector where k goes])

John Rennie

Yes

user586228

ok

:)

Then I flipped the signs so my answer is not right...In the way it was written by the blue pen

John Rennie

07:14

Yes, I've just noticed you swapped the signs on the sinθ terms

I assume q2-5 is the same except the angle of rotation is 30°.

user586228

ok...

What about the next topic

To quote from my book there is a quote,"Consider a rectangular parallelopiped with edges dx,dy,dz.The volume in strained condition is $(1+\epsilon_{x})(1+\epsilon_{y})(1+\epsilon_{z})dxdydz$.

John Rennie

$(1+\epsilon_x)dx$ is just the length $dx$ becomes when it is stretched. Yes?

user586228

Yes..but I do not understand why that is the length

$\epsilon x$ is the true strain along the x direction..

John Rennie

That's how the strain $\epsilon$ is defined ...

$\epsilon_x = \Delta x/x$

user586228

ok

Then?

What next

@JohnRennie You there?

John Rennie

07:32

@user586228 I don't understand what the confusion is. This is just how the strain is defined. Have you not studied strain yet?

user586228

$\Delta x/x$ How do you expand this

John Rennie

If $x'$ is the new length, then $\Delta x = x' - x$ i.e. it is the amount that the original length $x$ has increased.

15 mins ago, by John Rennie

$(1+\epsilon_x)dx$ is just the length $dx$ becomes when it is stretched. Yes?

So if the new length is $x' = (1+\epsilon_x)x$ then $\Delta x = x' - x = \epsilon_x x$

Then $\Delta x/x = \epsilon_x$

Nobody recognizeable

@JohnRennie hi

John Rennie

@Nobodyrecognizeable hi :-)

Nobody recognizeable

I have a question @JohnRennie

John Rennie

07:37

@Nobodyrecognizeable I'm currently answering a question in another room, but you can post your question ad I'll look at it as soon as I'm free.

Nobody recognizeable

2

@JohnRennie this one ^^

John Rennie

07:51

@Nobodyrecognizeable it's effectively just two spherical capacitors in parallel, well two halves of spherical capacitors in parallel.

Manoj Ghosh

@JohnRennie I am also interested in above problem

@JohnRennie please explain how to calculate capacitor for halves shperical cell ?

John Rennie

@ManojGhosh to a good approximation it is just half the capacitance of a complete spherical shell.

There will be fringe effects where the two halves join that mean it isn't exactly half, but it will be very close to half.

Manoj Ghosh

@JohnRennie that means just half capacitance of the sphere for each dielectric and they are in parallel, right ?

John Rennie

08:11

@ManojGhosh Yes

Manoj Ghosh

@JohnRennie look at that

John Rennie

@ManojGhosh That looks right

Manoj Ghosh

@JohnRennie thank you

user586228

13:10

@JohnRennie Please solve Q2-5 I am not getting te correct answer ..I do not know why

I tried a number of times

John Rennie

15:51

@user586228 I Googled for this and found these equations:

And I think these are the same equations written in a slightly different way:

user586228

ok ..that is correct it is for Mohrs' circle

Thanks a lot for the concern

I rused rotation of matrices there..

Why should that not work here?

John Rennie

Don't know, sorry, it's not a subject I have ever studied.

user586228

ok...

Now I got the correct answer

Thankyou@JohnRennie

Prateek Mourya

16:27

Hello @JohnRennie sir

John Rennie

@PrateekMourya hi :-)

Prateek Mourya

Sir is mixture of ideal gas also ideal gas if yes then how can we prove it

John Rennie

Give me a few minutes, I'm in the middle of eating lunch.

Prateek Mourya

Ok

Bohemian relativist

@JohnRennie it has been 5: 35 pm but you are still eating lunch?

John Rennie

16:36

16:35 in the UK

Jasmine

@JohnRennie Evening Sir :-)

John Rennie

@Jasmine hi :-)

Bohemian relativist

@JohnRennie it should be time to eat dinner.

Jasmine

Are you there for some time

John Rennie

@Jasmine about another hour.

Bohemian relativist

16:40

I think I still like Asian cuisine more.

Europeans generally probably don't cook as their food is mostly cold.

John Rennie

@PrateekMourya hi, are you there?

user102532

16:56

0

Q: Confusion in resultant velocity and relative velocity?

A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is. I drew the diagram like this (blue one)and got answer as north east or east north direction. My questions are It is normal right if relative velocity and...

vectors

Pls check this everyone

John Rennie

@user102532 haven't we discussed problems like this? You write the velocities as vectors, then subtract the train's velocity so find the speed of the car relative to the train.

user102532

I know

but I have a doubt as well. Pls check the red diagram

John Rennie

So the speed of the train is (v, 0) and the speed of the car is (0, v). Yes?

user102532

Yes

John Rennie

So if we subtract the velocity of the train then the car's velocity becomes (-v, v)

So in the train's rest frame the car is travelling north-west.

user102532

17:00

One sec

What about the way I drew ? I got your point

And the red diagram doubt.

i understood by maths from your point @JohnRennie

John Rennie

It isn't obvious to me what you have drawn ...

user102532

17:23

V of car - velocity of train = relative velocity of car w.r.t train

That is what I have drawn @JohnRennie

But in the red diagram which is from online , they changed the. Direction V of train without even changing its magnitude.

@JohnRennie

John Rennie

@user102532 So the vector you have drawn to the north-west is the velocity of the car relative to the train?

user102532

I have drawn is vector to north east

yes.But north east

Red diagram from online have drawn north west

John Rennie

Is the is "red diagram from online"?

user102532

Yes

Blue is mine

I redrew the diagram from online ofc

John Rennie

user102532

17:28

both of them are mine . 2 blue diagrams

John Rennie

Your diagram is the correct approach. You have drawn the car's velocity (0,v), and you are adding the negative of the train's velocity (-v, 0) i.e. subtracting the train's velocity.

user102532

Yeah

but answer is wrong

John Rennie

You just haven't added the two vectors correctly.

user102532

Also , I had one more question.Why is V TC I wrote not the resultant vector

How should have I joined then

@JohnRennie I got it.I drew the tails and head wrong.

I got north west

Thank you for your trick you solved question with and help

John Rennie

@user102532 :-)

user102532

17:35

Thank you so much

so kind of you . Take care

Transcript for

Problem Solving Strategies