All the parts are correct

I managed to do (c) (D)

How do you find out the radius of curvature.

Im aware of $\mathrm{ a =\frac{V^2}{r}}$

Can't seem to figure out what will be the acceleration in this case

@JohnRennie are you free?

@GaurangTandon mind taking a look?

I'm working at the moment I'm afraid. Sorry :-(

yep seeing

@JohnRennie no worries

hakuna matata

ok i got it

you listening @AvnishKabaj?

Yes

@GaurangTandon

do you know that velocity of center of mass = omega * L / 6 ?

Yup

good

do you know velocity of point A?

4*omegaL/6

keep it simple

V_a=4*V_com

ok?

Ok

and do you know a_A? (acceleration of point A)

V_a^2/.66666L

?

simply L/2 * omega^2

centripetal

getting it?

Ok

But won't we take icr

In this case

you know that velocity of A and its acceleration are both perpndicular

what is icr?

Instantaneous center of rotation

Instead of COM?

you can; but then you'd have to write velocity of A also in the frame of ICR

currently you've taken V_A in ground frame

Icr is 0

what "icr is 0"?

It's speed wrt ground frame

oh you're right :/

well I guess if you write centripetal accn for ICR it should come to be same

indeed, it will be same

Yeah

True

so, query solved?

Radius of curvature

?

R=v^2/a

just plug the values

Ok let me see now

Ping you in a couple of mins

k

@GaurangTandon got it

Thaaaaaanks

sure thing

good

@AvnishKabaj you there?

@GaurangTandon now I am

@AvnishKabaj ok, look at MS Chouhan; Advanced Problems In Organic Chemistry q119 aldehydes and ketones

have you seen that ring thing before/

? the answer is (b) dicyclopropylketone

1 sec

@GaurangTandon saw it seems to be right

clayden link please? :P

Link for what

to read about this in clayden

For the reaction?

Or why is it stable

i originally thought (a) would be the answer

or nvm

i think you're right

(b) is correct

thanks

You've answered the very same question

no this is a carbanion

never mind i understood it

Ok

@GaurangTandon [Question]

@Abcd post and wait 5mins

If $x^2 + x+1= 0 $ then the value of $(x+1/x)^2 + (x^2+1/x^2)^2+....+ (x^{27}+1/x^{27})^2$ is?

(Please keeeeep Mathjax enabled. )

Now, I feel it should just be 0.

$x^2 + 1 = - x$

So every bracket becomes $-1$

@GaurangTandon waiting

oh yes seeing

actually all values values are squared

so they are either positive or zero

@Abcd is the answer 54

but they are clearly not zero

so the sum is non zero

first term is 1

second term is 1

@AvnishKabaj Yes

third term is 8

and so on

Is my method correct @GaurangTandon?

yes, but the sum isn't zero

(-1)^2 = 1

yes

So answer should be 27 na

1, 27 times

third term is 8

*sorry 4

@GaurangTandon Why

take cube of x+1/x=-1

Please type using mathjax if you don't mind.

@GaurangTandon You mean $(\dfrac{x^6+ 1}{x^3})^3$

no I mean $(x+1/x)^3=(-1)^3$

(same thing)

Yes, so its -1?

Its a GP with common ratio $-1$?

but expand it first...

So sum should be $\dfrac{-1(1- (-1)^{27})}{1-(-1)}$

it is $x^3+1/x^3+3(x+1/x)=-1$

@Abcd

rearrange gives $x^3+1/x^3=2$

so its square is 4

@GaurangTandon My point was: $x^6 + 1 = -x^3$

@Abcd it isn't... $x^6+1=2x^3$; did you see my working above?

@GaurangTandon Initial equation is $x^2 +x +1 =0 $, substitute $x= x^3$ in this.

And you will get what I am saying,

$x^6 + x^3 +1 = 0 $

x isn't a variable

this isn't an identity that'll work for any x you feed it

^^^

x is a constant

its value is omega or omega^2

that substitution won't work

getting it?

I seee

@GaurangTandon Yes, now I can do it.

good

great

@GaurangTandon could you explain as to why is the horizontal component of angular momentum Lcos(there)

Or anyone else

If they're free

@AvnishKabaj How to find common normal to two curves?

Write the equation for the normals and compare

@AvnishKabaj See: $y = x^2 +1$, $y^2 = x+1$

What to do after that?

These are two parabolas.

General equation of normal is: $y = -tx + 2at + at^3$

Yes

I hope you are able to read the mathjax @AvnishKabaj

No

@AvnishKabaj But what do I do after that?

Write one of them for

$y = -tx + 2at + at^3

Then?

The other one as $x = -ty + 2at * at^3$

The t's are different

I know \

then?

For one of them x=x-1

Yeah yeah I know that

Then?

Basically write the eqns

For shifted origins

Then divide the coeff

For all of them

The ratio of the coeff will be equal

Now you have a bunch of eqns to solve

I see

@AvnishKabaj $a= 1/4$ for both right?

@Abcd yep

@AvnishKabaj I’ve been looking at your problem, and I don’t see why is Lcosθ and not Lsinθ. Are you sure Lcosθ is the right answer?

imgur.com/gallery/0jj3Q

At this perfect inelastic collision the angular momentum is conserved (there is no friction). That means there is no torque.

Yes

The reason why there isn’t is because the reaction force to the collision of the ball points at the same direction that the vector distance?

@JD_PM is that a question or are you trying to teach me something

@AvnishKabaj is a question, since I am not sure if that’s the reason why there is no torque

@JD_PM there's no external torque

Torques is necessary for balancing the momenta

@AvnishKabaj then do you think what I said is correct? About why there’s no torque

@JD_PM I followed you till friction but what do you mean by vector distance

The magnitude of the vector distance is 0.9m as you can see

Is the distance between rod’s CM and collision point

@JD_PM I don't think you can have collisions but no generation of torque

@AvnishKabaj I get a fifth degree equation using your method!

@AvnishKabaj okey, thanks

But there’s a perfect inellastic collision

@Abcd Jesus

@Abcd is 4x + 4y + 1=0 one of the normals?

Question number 39

I simply can't fathom how to go about doing this one

@AvnishKabaj i guess i missed the ping completely :-O

is your query solved?

@JohnRennie Could you help me with a fluids question?

He is offline, @GaurangTandon You can?

I just want to know the error in my working.

@Abcd sure

@GaurangTandon yeah

k great

@GaurangTandon after you're done with Abcd's question

Could you look at that new question I just posted

yep looking

@GaurangTandon Please listen to my method carefully and please ping me whenever you don't get any step.

Please ignore everything in blue, around the diagram.

[Method]:

We have two sites to focus on, site 1 and site 2 from left to right.

First of all, using the equation of continuity:

$A_1v_1 = A_2 v_2$

$\implies v_2 = \dfrac{A_1}{A_2}v_1$

yes go on

$\implies P_1-P_2= \dfrac{1}{2}\rho_w (v_1^2 (\dfrac{A_1^2}{A_2^2}-1))$

@AvnishKabaj Is answer to q37 (C)?

$\rho_w = \text{density of water}$

@Abcd correct

Then:

Let h be the depth at site 1

Let x be the depth at site 2.

Using hydrostatics:

@GaurangTandon yeah (c) ; for 38 (D)

@Abcd where did P_1 and P_2 go in this step?

@GaurangTandon got it?

@Abcd yep go on

$P_1 + \rho_w gh = P_2 + \rho_w gx + \rho_mg(h-x) $

$\rho_m = \text{density of mercury}$

$\implies (h-x)= \dfrac{P_1 - P_2 }{g(\rho_m - \rho_w)}$

@Abcd i don't get this

explain

$\implies (h-x)= \text{required difference in heights}= \dfrac{\frac 1 2 \rho_w (v_1^2(\frac{ A_1^2}{A_2^2}-1))}{\rho_m-\rho_w}$

@GaurangTandon Look

Pressure at depth $h$ should be same on both sides right?

At site 1, depth h, we only have water

yeah but that part above that mercury tube ... it's simply air

there's no water

@GaurangTandon Its water.

water is in the horizontal direction only

@GaurangTandon No, some water enters and rests there.

Moreover, air can't support heavy water

got it???

actually i never recall studying the upside down venturimeter

i did the vertical one instead

lemme check once

Any contribution to make @AvnishKabaj ?

Don't you think water will be there?

@Abcd haven't been following your question

@AvnishKabaj Just read the question and let us know if water will be there in the vertical tube above mercury #please.

@Abcd don't think that's air it's density is less would have bubbled up

@AvnishKabaj Exactly...

well, i haven't ever done an upside down venturi meter, but read this site nptel.ac.in/courses/122103011/22 the result they've derived doesn't seem to match with yours

this is interesting

@GaurangTandon which result?

My equation matches theirs

3rd slide

they haven't taken diff in densities as you did above

@Abcd they didn't get this

@GaurangTandon they haven't even cared to derive this..

They have derived $v_2$

@Abcd i meant you can rearrange that equation to get an expression in $h$

@GaurangTandon ... no ...

?

???

@AvnishKabaj Why is N=0 in q38?

how is the smaller cylinder's weight balanced?

@GaurangTandon no gravity

@AvnishKabaj ok ok

i get it

outer cylinder is "being rotated"; so its omega is constant

but to attain equilibiurm, both vel and omega of small cylinder change

and they both change due to torque by friction

Ok

which is $\mu \times N$

Ok

and N is always the centripetal force equal to mv^2/3R

ok till now?

@GaurangTandon ooooooh

Riiiiiight

???

good

What about the steady state

write equation for frictional torque and frictional force, which give you angular accn and tangential decelaration respectively

relate them by time t

at steady state, velocity of cylinder + R\times its omega = omega of bigger cylinder

(i.e. apply equation of no slipping at time t)

Ok gimme a sec I'll ping you

k

where'd @Abcd go?

@GaurangTandon stuff just cancelled out

@AvnishKabaj :(

ok, ping me tomo

i'm doing o-chem right now

have to finish it

Ok

@GaurangTandon R u there?

I think you have gone....

@AvnishKabaj Do you mind pointing out my error

@Abcd studying right now

@Abcd dude i've to finish ochem today

sorry...

Yeah, fine. We'll discuss that later.

@JohnRennie There?

I really feel stuck

@AvnishKabaj did you figure out why it was Lcosθ at your exercise ? Now I’m curious

Hi @sammygerbil :)

Hi

So, this is the problem that I was talking about:

@sammygerbil There's link to mathjax in the room description. I think you used to visit this room long ago when it was called "JEE preparation"

@sammygerbil Is pressure at the top blue point and the bottom blue point same? Is pressure at the top red point and the bottom red point same?

Because I get the right answer only if I consider it to be same.

And not use $P = p_o + \rho_wgh$

@Abcd No those pressures are not the same. I do not think that can be the explanation.

@sammygerbil So the answer of my book is just wrong?

@sammygerbil If you see this: nptel.ac.in/courses/122103011/22 (the third slide). Even this site takes the pressures to be the same.

@Abcd I don't know. I don't see any obvious reason for the difference.

@sammygerbil Thanks for your time on this problem. Btw, how do we make Physicsqandaexchange more known? Only you are the one answering questions there (and thanks a ton for that)...No one else visits the site

@Abcd Kenshin (Einstein on Physics Q & A) is the site moderator. You could try posting a Meta Question on the Q & A site to ask about advertising the site.

Okay

@Abcd In the nptel problem I think the moving fluid is air, so the density is orders of magnitude less than mercury, that is why the pressure is assumed to be same at the 2 blue markers in your diagram.

Oh, I see. (cc: @GaurangTandon)

@Abcd Yes it does look like they have made the same approximation when the moving fluid is water. ie They seem to have used $\rho_w/\rho_m$ instead of $\rho_w/(\rho_m-rho_w)$.

@sammygerbil Yeah... Maybe that's a standard approximation?

@sammygerbil Can I ask you a thing related to pure rolling?

Never mind

@Abcd Sorry I went out to the shops without logging out.

@Abcd Usually the moving fluid is air, and $\rho g h$ for air is very much smaller than $P_2 - P_1$. Water is about 800 times denser than air, and mercury is 13.7 times denser than water, so the approximation is very good for air/water and air/mercury (errors of about 0.1% and 0.01% respectively), but not so good for water/mercury (error of about 7%).

@sammygerbil So I think we should agree upon this being a mistake in my textbook. (though its the one of the most used physics textbooks in India)

@sammygerbil If a body is rotating about an axis and the axis is also rotating with some angular acceleration, is the angular acceleration of the axis counted as a vector part of the angular acceleration of the body?

I feel that it should be counted as that (and I just solved a question and got a right answer using my assumption) but not sure.

@Abcd I agree, giving an answer to 3dp (1.97cm) is not consistent with an error of about 7%. The answer should have been given as 2.0cm.

@Abcd Yes that is correct. Angular momentum adds up like that, and angular acceleration is related to rate of change of angular momentum.

Okay, thanks !!

@Abcd Was that the question about pure rolling?

@sammygerbil No that was a different question. I posted it on Physicsqanda. Should I post it here?

No, I will take a look at it there.

@Blue look at this: chemistry.stackexchange.com/a/92475/51553 ... The user has abused in Hindi. Please flag it immediately for moderator attention.

@Blue ?

@AvnishKabaj @GaurangTandon What has happened to Chemistry.SE? Indians are spamming it now. It aches me to see that happening...

@Blue I hate such dumb head disasters who have no online etiquettes!!