Problem Solving Strategies

Avnish Kabaj

02:24

@GaurangTandon take a look please

@AnybodyElse any help is greatly appreciated

Gaurang Tandon

02:41

yep in 10mins

Gaurang Tandon

02:55

woah

that's interesting

Gaurang Tandon

03:05

i've no clue how to solve that :(

Avnish Kabaj

@GaurangTandon just help me understand the solution then

I can't even get that gimme a sec

@GaurangTandon this is the solution

Rumplestillskin

03:21

Guys, im looking to derive the Lagrangian for the external post newtonian gravitational field.... The definition the give is as follows: L = 1 - d\tau/dt.

Can anyone explain that to me in laymans terms?

Gaurang Tandon

@AvnishKabaj ok lemme read

@AvnishKabaj what is the "d" in $\tau \propto mgd$ (first equation)? what is "d" supposed to indicate?

Avnish Kabaj

@GaurangTandon I have no idea I though you might know

@Tanuj @Abcd take a look at this question

1 hour ago, by Avnish Kabaj

Tanuj

05:43

@JohnRennie Good morning :)

@AvnishKabaj Is the answer to this 'D' ?

Avnish Kabaj

@Tanuj no it's a

Tanuj

Hmm , I have no idea about it then.

ask JR

Avnish Kabaj

05:58

@JohnRennie if you're free could you help me out with this

4 hours ago, by Avnish Kabaj

Tanuj

@AvnishKabaj actually , my answer is coming out to be 1/4 , no idea how it could be 1/2

Avnish Kabaj

@Tanuj it could be the key is wrong

How did you do it

Tanuj

@AvnishKabaj what book is this ?

Avnish Kabaj

@Tanuj fiitjee GMP

Tanuj

hmm , it can't be wrong then :p

John Rennie

06:06

@AvnishKabaj I reckon the answer is 1/2. Is that correct?

Gaurang Tandon

@JohnRennie yes it's correct; could you please explain How?

Tanuj

Yup , please explain.

John Rennie

I'll draw a quick diagram ...

Avnish Kabaj

@JohnRennie yes

John Rennie

Gaurang Tandon

06:12

oh oh I get where you're going

nice!

John Rennie

The first thing to note is that the square rotating about its centre is equivalent to four small squares rotating about a corner.

Tanuj

I see what you did there

diobuc eulb

ah yes i was looking everywhere for those

the cubes

Tanuj

@diobuceulb those are not cubes , but depends upon your imagination.

John Rennie

So the question is really if take a square rotating about a corner and double its size then how does that change the time it takes to stop. OK so far?

diobuc eulb

06:13

good observation

Gaurang Tandon

@JohnRennie yes

John Rennie

@AvnishKabaj are you OK with the argument so far?

Tanuj

@JohnRennie shouldn't the size be quadrupled and length pf side doubled ?

John Rennie

@Tanuj the word size normally means the linear dimensions e.g. the length or the width. So the size doubles and the area quadruples.

Tanuj

@JohnRennie cool :)

John Rennie

06:19

@AvnishKabaj: hello? You there?

Oh well, I'll give a quick summary anyway

The initial angular momentum is $I\omega$, so the change of momentum when the square comes to a stop is also $I\omega$, and change of momentum is equal to torque times time i.e. the angular impulse.

So we have: $$ I\omega = T t $$ or rearranging: $$ t = \frac{I}{T}\omega $$

So the question is how does $I/T$ change when you double the size.

The moment of inertia scales like $mr^2$ and the mass scaes like $r^2$ so the moment of inertia scales like $r^4$.

Tanuj

@JohnRennie T (Torque) is due to friction , right ?

John Rennie

The torque is force times distance, and the frictional force is $\mu m$ so the torque scales like $mr$ i.e. it scales like $r^3$.

So the ratio $I/T$ scales like $r^4/r^3 = r$

Tanuj

@JohnRennie doubt .

John Rennie

OK?

Tanuj

I solved kinda the same , but didn't take into account , the doubling of size ,

My question is , why will the torque be different for the two cases ?

John Rennie

06:33

The frictional force depend only on the mass because it is $\mu mg$. Yes?

Tanuj

@JohnRennie yes

John Rennie

And the torque is force times distance from the pivot point, so if we increase this distance we increase the torque. Yes?

Tanuj

yes

John Rennie

That's why the torque scales like $r^3$

Tanuj

@JohnRennie Okay , but if I just do something like , $\dfrac{t_{1}}{t_{2}}=\dfrac{T_{2}\cdot I_{1}}{T_{1}\cdot I_{2}}$

And put $I_{1}=\dfrac{ml^2}{6}$ and $I_{2}=\dfrac{2ml^2}{3}$

Can I know how would $T_{2}$ and $T_{1}$ compare ?

John Rennie

06:46

The expression for the torque will be very complicated. You'd have to divide up the square into elements, calculate the torque for each element and integrate them up to get the total torque. Offhand I don't know what the answer is.

But ...

The geometry isn't changing. We are only scaling the whole square up by a factr of two.

Tanuj

Yea ,

John Rennie

So we don't need to know the exact equation for the torque. We only need to know how it scales with the size.

Tanuj

@JohnRennie How are we going to calculate torque ?

John Rennie

i.e. we don't need to know $T_1$ or $T_2$. We only need to know the ratio $T_2/T_1$.

Tanuj

I mean $\mu mg$ is the force , but what distance has to be taken ?

John Rennie

06:49

@Tanuj We only need to know the ratio $T_2/T_1$

Tanuj

@JohnRennie I get it , but if in a different question , let's say I'm asked to evaluate the torque for disc 1 and disc 2 . How would that be done ?

John Rennie

Take an element at $(x,y)$ from the corner. The mass is $\rho dx dy$ where $\rho$ is the area density so the friction is $\rho \mu g dx dy$.

The distance is $r = \sqrt{x^2 + y^2}$ so the torque is $$ dT = \rho \mu g \sqrt{x^2 + y^2} dx dy $$

Tanuj

okay

John Rennie

Now just integrate to get the total torque

Tanuj

How do you integrate this ? It has both dx and dy

John Rennie

06:55

It's just a double integral

Avnish Kabaj

@JohnRennie I'm so sorry I got caught up with something

Tanuj

Okay , so it is out of my scope

John Rennie

First take $y$ constant and integrate $dx$. Then do a second integral $dy$.

Tanuj

taking x constant ?

Madhuchhanda Mandal

@JohnRennie Really?

Tanuj

06:56

and then adding both ?

Madhuchhanda Mandal

Is it Physically Rigorous??

John Rennie

@Tanuj once you've done the $dx$ integral that removes all dependence on $x$. The integral is then just a function of $y$.

Tanuj

@JohnRennie I didn't get anything actually , this is my first encounter with a 'double integral'. So , first , taking y as a constant , I integrate with respect to x , what will happen to dy ?

John Rennie

@MadhuchhandaMandal yes it's rigorous. I skipped some steps to keep the explanation simple, but you can easily see it's rigorous.

@MadhuchhandaMandal as I was explaining to Tanuj we can calculate the torque by splitting the square up into infinitesimal elements $dxdy$ and calculating the toque due to each element. If we keep the geometry constant and just double the size then we quadruple the mass of each element and double its distance.

Avnish Kabaj

@JohnRennie thanks a lot

Madhuchhanda Mandal

07:02

@JohnRennie How are being sure Intuitionally(ofcourse without deriving it) that Torque is dependent on r to the First power?

For example, it could depend on Ln(r) as well

Tanuj

@JohnRennie nvm , I don't think I' have to do the double integral for my exam.

John Rennie

@MadhuchhandaMandal torque is force times distance

Madhuchhanda Mandal

@JohnRennie for a point mass

Here it's a distribution of mass

Tanuj

@JohnRennie Thanks a lot ! :)

Madhuchhanda Mandal

With varrying distance

Can we be sure Intuitionally?

John Rennie

07:04

Any distribution of mass $m$ can be thought of as being built up from small masses $dm$

Tanuj

@JohnRennie I have a question to ask.

John Rennie

The torques just add, so $$ T_\text{total} = \sum_i dT_i = \sum_i \mu g \, dm \, r_i $$

And if we double the size, while not changing the overall shape, then each $r_i$ in our sum is doubled.

Madhuchhanda Mandal

Ofcourse, Now r is varrying and so is dm

John Rennie

Yes, each $dm$ is quadrupled.

Tanuj

@JohnRennie One of the following wavelenghts is absent and the rest are present in the X-rays coming from a Coolidge tube. Which one is the absent wavelength ?

John Rennie

07:07

@MadhuchhandaMandal So every term in our sum is increased by a factor of $2^3$

Tanuj

OPTIONS: a) 25 pm b)50 pm c) 75 pm d) 100 pm

Madhuchhanda Mandal

@JohnRennie ofcourse, so we can definitely argue that if for a small square Torque is F(small dimension, Mass ), then for a large square it must be F(Large dimension, Mass)

John Rennie

Yes

Madhuchhanda Mandal

But , is it evident Intuitionally that F(dimension,Mass) is proportional to dimension to the first power?

John Rennie

@MadhuchhandaMandal Force or torque?

Madhuchhanda Mandal

07:10

F is the torque function

John Rennie

$F$ is so widely used for linear force that I recommend using a different symbol for torque. Both $T$ and $\tau$ are widely used to represent torque.

Madhuchhanda Mandal

I used F to represent F(x,y).. :-P

John Rennie

I strongly recommend adhering closely to the usual conventions when doing calculations as it reduces the scope for mistakes.

Madhuchhanda Mandal

Okay .. so T(x,y)

John Rennie

Anyhow, look back at my sum for calculating the torque of a distributed mass. From that it should be immediately obvious that torque scales linearly with dimension.

Madhuchhanda Mandal

07:16

Well, multiplying each term of the summation by a Scalar

Right?

John Rennie

Yes

Madhuchhanda Mandal

And using the bijection argument

John Rennie

bijection argument ?

Madhuchhanda Mandal

That is , Mapping each point of the original figure to a point in the New scaled figure

And as every point of the new figure can be traced back to a point of the original figure

That is as Inverse of the mapping exists

So the mapping must be a bijection

And thus Cardinality of the domain and Codomain of the mapping must be same

Which is indeed true

Because as we are considering Point mass

The Cardinality is infinite in both the case

(Original body and scaled body)

John Rennie

The sum is a finite sum of discrete elements, so we don't need to worry about cardinalities or rigorous handling of infinitesimals.

And I'm being a typical physicist in assuming that we can move to the infinitesimal limit and change the sum to an integral :-)

Shrug :-)

Madhuchhanda Mandal

07:28

:-P

John Rennie

@Tanuj I don't understand the question. An X-ray tube normally produces $K_\alpha$, $K_\beta$, etc, but there is no simple relation between the wavelengths.

Tanuj

hmm yea

@JohnRennie if I was somehow able to calculate the minimum wavelength , then I could get the answer , but then the applied voltage is not given

John Rennie

@Tanuj I guess they must mean the shortest wavelength is absent because I can't see any argument for the presence or absence of other wavelengths.

Tanuj

why would that be ? ( shortest wavelength absent)

John Rennie

Because the shortest wavelength has the highest energy, and would be missing if the accelerating voltage was too low.

Tanuj

07:33

@JohnRennie okay.

Madhuchhanda Mandal

@JohnRennie Hey !! I found a way to bypass the double integral !!!

John Rennie

@MadhuchhandaMandal Yes?

Tanuj

@MadhuchhandaMandal It'd be interesting to know.

Madhuchhanda Mandal

Tanuj

@MadhuchhandaMandal So finally you get torque proportional to distance of center of mass from the axis of rotation ?

Madhuchhanda Mandal

07:39

I think it can be used to calculate individual Torques... (But it will be very inefficient to calculate the ratio as it is unnecessary . Cause the method @JohnRennie pointed out is much easier)

Tanuj

@MadhuchhandaMandal wouldn't the torque be 0 in the first case according to your expression ?

Madhuchhanda Mandal

@Tanuj in that case the Splitting of OP is not valid

I mean OC is null vector

I'm assuming they are non coincidental

Tanuj

I don't know how that can be true , but okay

Madhuchhanda Mandal

Anyway, i may have messed up something too.. i need to double check it

Tanuj

How did you write the torque due to friction ?

Madhuchhanda Mandal

07:44

Which one?

Tanuj

Both

Madhuchhanda Mandal

Proportional to Mass × daxiscom

So Ma/√2 in second case

In first case , split the square into four pieces

Tanuj

hmm , I really don't get how you got that - T=Mass × daxiscom

Madhuchhanda Mandal

I posted the derivation 🤔🤔

Tanuj

@MadhuchhandaMandal I know but , I didn't get it.

Madhuchhanda Mandal

07:48

Oh.. Which area?

Tanuj

most part , one question though.

Is it going to be always true ?

T=Mass × daxiscom

Madhuchhanda Mandal

I need to recheck it

I can't tell anything for sure..until I double check the derivation

It might be erroneous

Well I think it will be

I complicated the derivation for no reason

Wait lemme put it simply

Tanuj

cool

Madhuchhanda Mandal

It won't be .. sorry

I checked the derivation

Sum(|Vector|) not equal to |Sum(Vector)| I made that stupid mistake

Stupid me

😢

Tanuj

Ah

Madhuchhanda Mandal

08:11

But wait.. something cool is waiting

Lemme cook

Tanuj

Alright . :)

Manolis Lyviakis

08:31

orning

morning

suppose i have a metal shell with negative charge

what happens if i put a non conducting sphere inside it with oposite charge?

suppose the charge in the non conducting sphere is uniform

Avnish Kabaj

08:46

It stays there?

Gaurang Tandon

09:00

@ManolisLyviakis if you put it at the centre, nothing should happen; everyone stays where they were

Manolis Lyviakis

yeap

just figured it out

also what do we mean the Electric field inside a conductor is zero

do we mean

inside inside

or on the surface of the conductor

suppose i ahve a metal shell

that has 2 radius

one is the radius of the emptyness inside

what i mean is

even if i put a charge on the inside of the metal shell

its electric field will be zero?

Manolis Lyviakis

09:55

im consufed

suppose i have a ballon with uniform charge in the surface

and that baloon is getting bigger

the change inside stas zero

the charge outside far away

is $kQ/r^2$

but what is the change on the surface

i dont get the question

isnt infinte on the surface

?

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