The h Bar

heather

00:36

=D just made the physics.se close vote review leaderboard (i'm at the very bottom, but it still counts)

DanielSank

00:59

Hi everybody!

@heather Wow!

I'm so, so remiss in doing reviews around here.

I just do not like doing them at all.

heather

there is some crazy stuff that comes up =)

DanielSank

yeah

I just downloaded Rust.

@BernardMeurer and I are going to see if we can write a chat server in rust :D

It would be cool to do crypto too.

So far I've written chat servers in python and Scala.

I feel like that's a good project to get to know a language.

heather

cool! what exactly is a chat server? and what is rust? another programming language?

DanielSank

Well, right now you and I are using a chat server to talk to each other.

It's fun to write a program like this for yourself because it involves a lot of important design principles.

Network communication, asynchronous code, etc.

heather

hmm, interesting

DanielSank

01:13

I should put my python chat server up on github...

That would be a good project for a Saturday.

I'll put that on my list of stuff to do when I'm not busy.

DanielSank

01:40

Yeah @heather if you want a graphical interface it's even more fun.

:D

I'll let you know when I get the code up on github.

heather

@DanielSank, thanks! It'll be fun to take a look.

user228700

Yello everyone :-)

user228700

I've a quick question. The S.I unit of activity of roadiactive decay is Becquerel = 1 decay/second. So that's one nucleus per second, yeah?

heather

@Kaumudi, hello =)

user228700

Hii :-) How's it going?

heather

01:55

Decently well, I suppose

and how are you?

user228700

Same here, thanks :-)

user228700

OK, I realize that mine was a dumb question. Nvm.

user228700

04:16

@DavidZ: Hi :-) I wanted to personally thank you for your awesome answer to this question:

user228700

2

Q: What does the decay constant mean?

In my curriculum, the decay constant is "the probability of decay per unit time" To me, this seems non-sensical, as the decay constant can be greater than one, which would imply that a particle has a probability of decaying in a time span that is greater than 1. Can someone explain this?

DanielSank

04:32

Hi, everybody.

DanielSank

04:43

This nonsense really annoys me.

Superconducting qubits are not like atoms.

That's just how people thought about it back in the dark ages when they weren't used to the fact that quantum mechanics works with non microscopic systems.

When superconducting qubit talks begin with a diagram of a Fabrey-Perot cavity I want to get out of my chair and delete the diagram from the speaker's slides.

vzn

@DanielSank maybe if you eat a lot of turkey & get that tryptophan high going it wont seem so urgent :)

DanielSank

05:02

@vzn -_-

user228700

05:36

I've a quick question again. If given that the activity of a substance is $x$ dps, (w/ nothing else mentioned) does this mean that it is $x$ at $t=0$? (Since activity keeps on changing)

Anonymous

@Kaumudi Use the formula for $N = N_o e^{-\lambda t}$. $x=dN/dt$, and then find $dx/dt$. Is $dx/dt$ constant? You will get your answer.

Anonymous

dx/dt is only for that particular time instant which is given in question.

Anonymous

And even x.

Anonymous

Swapnil Das

Hi, a question for JEE aspirants.

Does Etoos provide test series? Please reply, thanks.

Anonymous

05:45

If nothing else is mentioned you have to assume $x$ at t=0. But mostly they will mention it.

Anonymous

@SwapnilDas Yes, but i never tried them

Swapnil Das

@S007 Are their physics lectures good? Who's the best lecturer?

Anonymous

@SwapnilDas Nitin Vijay..

Swapnil Das

Really? He shouts a lot :P

Anonymous

I know...but I use Physics Galaxy for Physics :)

Swapnil Das

05:47

Same here!

Anonymous

Great :)

Swapnil Das

But Constrained motion.. :(

Anonymous

@SwapnilDas what ?

Swapnil Das

No QueSets for constrained motion.

While it is thoroughly covered in Etoos lectures.

Anonymous

I never used the QuSets :-P

Swapnil Das

05:49

Oh :P

From where do you do problems?

Anonymous

From Cengage or the Etoos Material

Swapnil Das

Etoos Material? Is that good (for mechanics)?

user228700

@S007 No particular time interval is given.

Anonymous

Yes. It is great. You should try. BTW all kota institutes have good material. Just do them fully.

Swapnil Das

@S007 Does it have good constrained motion problems? :P

Anonymous

05:51

@Kaumudi Then assume at t=0, nothing else can be done :P. Anyway I would like to see the question :) Is it solvable by assuming at t=0

user228700

If given that "3.7 $\times 10^{10}$ is the activity for 1g of radium", should I assume t=0 is what I asked.

Anonymous

@Kaumudi is the answer coming by assuming t=0 ?

Anonymous

@SwapnilDas yes

user228700

No question, bhai.

Anonymous

@Kaumudi Are you able to solve the problem by using the assumption? Is the answer matching ?

user228700

05:54

There is no problem!

Swapnil Das

Lol.

Anonymous

@Kaumudi Oh. Anyway in most problems the time will be mentioned for x. Don't worry then.. and yes t=0 is a valid assumption....

Swapnil Das

@S007 Does that Complete study material have theory explained in their lectures?

Anonymous

@SwapnilDas yes

Swapnil Das

I'll test a little, please don't mind. Does it have v1+v2/2 = Vp? :P

The etoos pulley shortcut^^

Anonymous

06:00

@SwapnilDas What is v1 and v2 ?

Swapnil Das

Velocity of system 1 and 2.

Anonymous

What type of system ? Can you draw it ?

Anonymous

It looks like a simple constraint relation..

Swapnil Das

Sure.

user228700

@JohnRennie: Morning! :-)

Anonymous

06:03

@JohnRennie Good morn :)

Swapnil Das

@JohnRennie Morning Sir! :)

John Rennie

Morning all :-)

Swapnil Das

@S007 This one.

Anonymous

@SwapnilDas Yes, displacement of the pulley is always average of the displacement of the two blocks..what is the the problem ? :)

Swapnil Das

Is this there in that book?

Just checking its quality. Such stuff is never there in common text books.

Anonymous

06:06

@SwapnilDas Which book ? This is there is almost every jee physics book in the market

Swapnil Das

@S007 That Etoos material.

Anonymous

even in the Physics galaxy lectures it is there

Swapnil Das

Not in DC Pandey I believe :p

Anonymous

oh! The etoos material only has questions

Swapnil Das

Damn.

Anonymous

06:07

The theory is taught in lectures

Swapnil Das

Only questions and 7200 rupees?

That's too much, I believe :P

Anonymous

I use the free study material of etoos physics .... etoosindia.com/studymaterial/jee/studyMaterial.do

Anonymous

They have almost the same questions

Anonymous

In different sequence

Swapnil Das

Free?

Anonymous

06:09

yeah

Swapnil Das

They ask to sign up.

Anonymous

yes you have to sign up

Anonymous

its free

Swapnil Das

Great. Thanks a lot, I'm indebted to you. Have you enrolled in their courses?

Anonymous

I bought the courses

Swapnil Das

06:10

Great. DVDs?

Anonymous

yep

Anonymous

you can watch online also

Anonymous

VOD form

Swapnil Das

Yeah. Thanks a lot.

Anonymous

welcome

Anonymous

06:12

:)

John Rennie

06:23

@Kaumudi You don't have to worry about the time. You're given the activity per gram of radium. At a later time the activity will have reduced, but that's because some radium has decayed so you have less than 1g left. The activity per gram of radium is still the same.

Anonymous

Oh yes, I did'nt notice the "per gram" :P

user228700

@JohnRennie Ah, yes! Me=Dumbass. Thanks! :-)

David Z

07:03

@Kaumudi oh, uh, you're welcome

user228700

07:20

:-P It's very beautiful. I even took the time to try to explain it to my mother!

user228700

Does anybody know how to bookmark conversations on a phone?

user228700

Oh, nvm.

John Rennie

Just out of curiousity: do you prefer using a phone rather than a laptop, or is the availability of laptops a problem?

Anonymous

I am facing a confusion regarding pseudo forces. Suppose I am in a reference frame moving with an acceleration $a_1$ along positive x axis and I am observing an object moving along positive x axis with an acceleration say $a_2$ and mass $m$. And then I wish to draw the Free Body Diagram of the object from my frame. In that case I introduce a pseudo force $ma_1$ pointing to negative $x$ axis.

Anonymous

But, do I also have to introduce the pseudo force due to the acceleration $a_2$ of the bod while drawing the FBD ?

Anonymous

07:29

@JohnRennie Could you help with that ^, please :P

Anonymous

@Kaumudi If you are using Chrome Browser, you can convert to PC mode and then bookmark it like you normally do...

John Rennie

I'm not sure I understand what you are asking. However if you draw the FBD in the rest frame then there is just a single force $F = ma_2$. Transforming into your (accelerating) frame just adds a second pseudo force $F' = -m_a_1$. Those are the only forces acting.

user228700

@JohnRennie :-) No, I don't have a laptop sir! :-( I do prefer to use the phone tho, because I have almost everything I need in this little device I can carry around in my pocket.

user228700

@S007 Why dyou think this is needed?

user228700

I've a very quick (I promise! :-P) question about phase diagrams.

Anonymous

07:38

user228700

Anonymous

I mainly faced had the confusion while solving this particular question @JohnRennie

Anonymous

Is the FBD I drew correct w.r.t

Anonymous

07:39

the wire ?

Anonymous

Why or why not ? I introduced a pseudo force ma due to acceleration of the wire towards the left...

John Rennie

@Kaumudi Yes?

user228700

Basically, if we travel upward and hit the graph at a point, does travelling right from there and then hitting the other curve give me the composition of the vapour at that temp. and composition? It doesn't look like that :-/

John Rennie

@Kaumudi Yes.

user228700

Oh :-/

Anonymous

07:42

@Kaumudi Could you please wait a few minutes ? :-P Let John clear my confusions first, please ?

user228700

I was gonna wait but he asked me first :-P

John Rennie

More precisely, if the draw a horizontal line then the points where the horizontal line intersects the lower and upper curves give the compositions of the liquid and vapour at that temperature.

@S007 I think the question is simpler than it seems.

user228700

@JohnRennie Ah, see, this I understand. But I'm asking about hitting the lower curve then travelling straight toward the right and hitting the other curve.

Anonymous

@JohnRennie So, where am I going wrong in the FBD ?

John Rennie

The net acceleration of the bead is just the vector sum of $mg$ downwards and $a$ to the left. Just add the two vectors and you get the net acceleration of the bead.

user228700

07:45

@S007: The wire is moving in a horizontal plane?

John Rennie

I think the question is just asking for the component of this acceleration along the wire.

> hitting the lower curve then travelling straight toward the right and hitting the other curve

Anonymous

@JohnRennie But that won't be in reference frame of the wire! For wire frame we need to add the pseudo force ma towards right, isnt it ?

John Rennie

@Kaumudi Isn't that just drawing a horizontal line and seeing where it hits the curves?

Anonymous

What you are saying is in the ground frame, i think @JohnRennie

user228700

Well, yeah :-P

user228700

07:47

Yeah, I'm inclined to agree with S007...

Anonymous

@Kaumudi It seems like the horizontal plane is the plane of paper where g is acting downwards...if it was on the plane of table g would not appear in the final answer i guess

John Rennie

Let me draw a diagram ...

user228700

OK, I think I've understood the answer to my question.

user228700

@S007 Table? O.o

Anonymous

@Kaumudi I meant any flat horizontal surface where g is into the plane of the paper

John Rennie

07:53

The net acceleration along the wire is $g\cos(90-\theta) - a\cos\theta$. Yes?

Anonymous

@JohnRennie Yes, in ground frame that is okay. But you did not take the pseudo force ma into account which acts towards the right, since we are observing from the wire's frame....

John Rennie

The way I interpret the question is that it's asking for the acceleration along the wire, and that's what I calculate. I note that my solution matches one of the answers ...

Anonymous

@JohnRennie I know, maybe the question is a bit wrong. But I am asking a general question. Suppose we forget the options given for a moment and we have to find the acceleration of the pearl in the wire's frame, do we take the pseudo force ma(towards right) into account or not ? I mean getting the answer to the question isn't my sole objective...I want to be sure that I am getting the correct answer or not....

John Rennie

@S007 that's what I've drawn isn't it? I've divided by $m$ and only drawn the acceleration, but what I've drawn matches what you say.

user228700

Guys, in which direction is the acceleration of the pearl? wrt ground frame?

Anonymous

08:01

@JohnRennie Okay, maybe I am missing something. You did take pseudo force due to wire into account? Then where did you draw the acceleration of the bead towards the left in the diagram ?

Anonymous

Is my free body diagram wrong btw? Why?

Anonymous

I am getting the net acceleration along the length of wire as $g\sin(\theta)-2a\cos(\theta)$....

John Rennie

The only two forces are $mg$ down and $ma$ to the right. I don't see why you've drawn $a$ to the left ...

user228700

^ That's the acceleration of the bead itself, I presume...

John Rennie

08:04

@Kaumudi yes, those are the forces on the bead.

Anonymous

@JohnRennie I drew the a towards the left to take into account that the bead itself is accelerating towards left

Anonymous

And also a pseudo force is acting due to wire's frame

Anonymous

So I am getting 2ma towards right...

user228700

@JohnRennie No, I meant what S007 said just now ^

John Rennie

But in an FBD don't we just draw the forces acting on the object?

And there are only three forces, mg down, ma right (the psudoforce) and N normal to the wire.

I may have misunderstood because I don't think we used FBDs when I was an undergrad. Certainly I had never heard the term until I joined the PSE.

Anonymous

08:07

Anonymous

No, we indicate acceleration separately

John Rennie

I need to go offline for 5 minutes or so. Back soon ...

Anonymous

Most of the times, isnt it? Like for the block m we write mg-T=ma,....in the above diagram

Anonymous

@JohnRennie Sure..we will wait :)

Anonymous

@Kaumudi Do you have any ideas ? :)....I am tearing my hairs over an apparently easy question :-P

Mew

08:09

why is (A) and (C) the same?

user228700

Solving...

Mew

I get (A) and (C)

Anonymous

Okay even I am coming back in 5 mins :-P

Mew

HalO Kaumu

user228700

I'm making a stupid mistake somewhere. I get (A) and (C), only the whole thing is divided by $\cos \theta$ :-/

user228700

08:13

Hey @Mew :-)

Mew

r u doing mechanics now/

user228700

Yeah, I'm trying to solve S007's problem but I keep making a mistake somewhere :-/

Mew

I get the accelerationdown the wire as $g\sin\theta$ and the psuedo force up the wire $a\cos\theta$

so the net acceleration is $g\sin\theta - a\cos\theta$

but this is in the Earth's frame hey

user228700

^ We'll wait for u, Mew :-P

Mew

watday mean

yeah i think my answer is for the acceleratin gframe

John Rennie

08:20

I must be missing something here. Suppose we paint a dot on the wire and put the bead on that spot. The acceleration of the bead wrt the wire is the second differetial of the distance between the bead position and the spot we painted. And that's what the diagram I showed calculates.

Mew

so I go with (A) and (C)

user228700

I don't think I've understood exactly what's happening properly.

John Rennie

I know. I'll make more coffee. That's always a good option when you're not sure what to do next.

user228700

I keep getting $(g \sin \theta - a \cos \theta)/\cos \theta$

Anonymous

@JohnRennie Okay okay! I got it just now. The relative acceleration of the pearl in the wire's frame is zero towards left, so we do not indicate that in the free body diagram. Whenever we draw a FBD, all velocities and accelerations are taken relative to that frame...

Anonymous

08:24

So we show only pseudo force and mg and N in the FBD

Anonymous

If the accelerations were different, then we had to write down something like $a_2-a_1$ towards left

user228700

-__- I thought u understood where u were going wrong, not why we musn't show the acceleration!

Anonymous

@Kaumudi Yes, I was showing the acceleration and that was what I did wrong.

Anonymous

In wire's frame, the pearl doesn't accelerate towards the left...

Mew

THink of it this way dudes

Imagine you are the wire

and you see yourself as stationary, and you see the pearl trying to accelerate towards you at a

it will thus have a component of acostheta up the wire

gravity still acts in the same way, gcostheta down the wire

gsintheta*

thus the net acceleration relative to you, the wire, is gsintheta - acostheta

QED

user228700

08:29

@S007 This is what I had asked before.

Anonymous

29 mins ago, by Kaumudi

Guys, in which direction is the acceleration of the pearl? wrt ground frame?

Anonymous

I saw that.

user228700

Yeah, I asked about the ground frame but I was confused about how u guys knew which direction the acceleration is in wrt the wire as well.

user228700

Which is it?

Mew

the pearl accelerates to the right wrt to the wire

Anonymous

08:32

@Kaumudi For that see John's diagram you will get it! The mistake I made was assuming that the pearl accelerates towards left even in the wire's frame.

user228700

Sigh, @JohnRennie: How did u determine the direction of acceleration of the bead wrt the wire?

Mew

this is so easy

the wire is accelerating to the left with respect to the ground (and the bead)

so clearly the bead is accelerating to the right with respect to the wire

Anonymous

@Kaumudi Suppose you are the wire, in which direction will the bead move? Always towards you...but meanwhile even gravity will pull it down...

Mew

@S007 I already said that dude

Anonymous

@Kaumudi See youtube.com/…

Anonymous

08:36

Concept of Pseudo Force (GA_NL18)

►

Anonymous

You will understand the direction of acceleration of bead if you haven't already....

John Rennie

@Kaumudi the bead is constrained to slide along the wire, so the acceleration of the bead in the wire's frame has to be along the wire.

It can't have any component of acceleration normal to the wire otherwise it would move off the wire.

Mew

@JohnRennie she's meaning right or left

As in, the bead is moving to the right relative to the wire

well trying to

and then it is conttrained to move up

so that is the a\cos\theta component

which is subtracted from the g\sin\theta component

user228700

@JohnRennie: Why didn't u show the normal reaction in ur diagram?

John Rennie

@Kaumudi because the normal force is normal to the wire so it can't accelerate the bead along the wire. You're right that I should have included it for completeness, but it doesn't affect the result.

user228700

08:48

How does it not affect the result? It has components $N \sin \theta$ to the right and $N \cos \theta$ upward, doesn't it?

Mew

No

John Rennie

@Kaumudi I was only interested in calculating the acceleration along the wire. If you're calculating the acceleration in a horizontal direction then yes you'd need to include $N$.

Mew

$N$ is only in the direction perpendicular to the wire

Thus doesn't cause the bead to move parallel to the wire

And because the bead can only move parallel to the wire

$N$ is irrelevant

user228700

@JohnRennie Alright, say I include N and every other force too. This is what I get:

John Rennie

@Mew Poor $N$ :-)

user228700

08:52

Mew

@Kaumudi while correct, you should be breaking all forces into components parallel to the wire

Anonymous

@Kaumudi Yes, now find the net force along the length of wire, you will see that components of N cancel out.

user228700

Sigh. OK, thanks guys.

user228700

I have a question. I started solving this only because u guys started talking about it. I don't understand how the eff the wire is supposed to be moving in a horizontal plane.

John Rennie

There is some unspecified external force accelerating the wire horizontally.

Anonymous

09:02

@Kaumudi I think the horizontal plane here is the plane of the paper and gravity is acting downwards. (Not into or out of the plane of paper as you might have thought). And yes, there must be some external force causing the motion.

John Rennie

@Kaumudi That is, the question just assumes something is accelerating the wire horizontally at a constant acceleration.

Mew

it looks like it is moving in the vertical plane to me

user228700

@Mew EXACTLY!!!

Mew

just poorly worded question I think

I mean just look at the options (A) and (C) are the same

user228700

Yeah ^

user228700

09:06

For the first 15 mins I was struggling to understand the friggin' question itself, how it's possibly moving on a horizontal plane, ffs. OK, I'm gonna leave before I get even more frustrated with my brain so u guys don't have to see it.

John Rennie

@Kaumudi ah, OK, I see your confusion. Many years of parsing badly phrased questions helps here.

Anonymous

Well, almost 90 percent questions in JEE are badly phrased, so I am not that surprised :-P @Kaumudi If you saw 2016's MAIN's paper you know what I mean :-P

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