Problem Solving Strategies

AbhigyanC

02:02

@sammygerbil How was this derived?

The solution given is also the same explanation, but how do we do this by just displacing the Bob slightly?

sammy gerbil

02:46

@AbhigyanC The solution was "derived" by the insight of realising that the apparatus is a rigid structure which is constrained in such a way that the bob moves in a plane perpendicular to fixed axis AB and keeps a fixed distance from the axis, like a simple pendulum.

It is not difficult to make a model and play with it. This will help to develop your insight.

invadingdingo

How open is this chat to absolute noob homework help? I'm pretty not good at physics, lol!

sammy gerbil

03:11

@invadingdingo That's what we're here for. :)

The restrictions on the Q&A site do not apply here.

invadingdingo

Ok, so here is the problem I'm given: In the figure two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L. Assume that θ is so small that tan θ can be replaced by its approximate equal, sin θ. Find the magnitude of q in terms of the given variables, using ε0 and g if necessary

And I have absolutely no idea how to even begin.

sammy gerbil

03:50

@invadingdingo Start by drawing a Free Body Diagram for each ball. Identify the forces on it. The resultant force must be zero.

invadingdingo

04:00

So since they're the same charge, they will exert force on eachother, pushing eachother away. Since they have mass, gravity is also acting on them, right? So each ball has 2 forces acting on it?

Wait, 3 since you have tension from the threads.

sammy gerbil

ok do far.

invadingdingo

What should I start doing after that?

I'll call the forces acting on the charge from the other charge F1

Well, F1 and F2

So F1 = -F2 since they're charges of the same value?

sammy gerbil

Resolve forces into horizontal and vertical components. The sum of each must be zero.

harambe

They are of same magnitude

invadingdingo

So for vertical, FsubT1 - mg = 0 and FsubT2 - mg = 0

and for horizontal, Fsub2 - Fsub1 = 0?

sammy gerbil

04:14

You need to make use of the angle theta at which each string is inclined to the vertical.

What is FsubT1? Also note that since the strings have the same length and the balls have the same mass and charge then the arrangement is symmetrical about the vertical, so the electrostatic forces are horizontal.

The electrostatic forces are mutual and therefore always have the same magnitude even when the charges are different.

invadingdingo

FsubT1 is the tension from the string. It

s pointed directly upward

sammy gerbil

04:30

So where are the electrostatic forces F1, F2?

Is Fsub1 the same as FsubT1?

You only need to consider the forces on one ball. The forces on the other will be the same, except the mirror image.

invadingdingo

That's my current diagram

Ok, so if I label the left ball qsub1 and the right qsub2, then qsub1 = -qsub2?

gateprep

04:49

f(x,y)=⎧⎩⎨xy(x2−y2)x2+y20 for (x,y)≠(0,0) for (x,y)=(0,0).
These mixed derivatives \dfrac{\partial^2 f}{\partial x \partial y}
∂x∂y
∂
2
f
and \dfrac{\partial^2 f}{\partial y \partial x}
∂y∂x
∂
2
f
evaluated at the origin (0, 0)(0,0)left parenthesis, 0, comma, 0, right parenthesis turn out to be 111 and -1−1minus, 1 respectively. Computing this is actually pretty tricky, and requires looking directly at the limit-based definition of the derivative. Wikipedia provides a nice explanation, should you find yourself feeling ambitious.

Let me resend

khanacademy.org/math/multivariable-calculus/…

See the link khanadacemy...

See the portion symmetry of second order partial derivatives.

Now under exception example there is a problem which has been solved somewhere

in Math stackexchange by assuming y=mx

I am getting stuck up there

sammy gerbil

@invadingdingo Please review the discussion so far. Your own contributions as well as mine. Aren't qsub1, qsub2 the same as F1, F2? Have you included the tension in the string in your diagram?

gateprep

Let me resend
khanacademy.org/math/…
See the link khanadacemy...
See the portion symmetry of second order partial derivatives.
Now under exception example there is a problem which has been solved somewhere
in Math stackexchange by assuming y=mx
I am getting stuck up there

invadingdingo

How would I find the tension in the string?

gateprep

@invadingdingo This is a classic central force question.Use kq2/r2 action on both the bodies where r(or x in this case) can be found by approximating arc length to that of a straight line provided the given angle theta is small.

invadingdingo

I had F1 and F2 as the forces exerted on the balls by eachother

sammy gerbil

04:57

@invadingdingo It has horizontal and vertical components which are related via the angle theta. These components need to be included when you sum the horizontal and vertical forces on one ball and set the total to zero.

gateprep

@sammygerbil Pls come to my rescue after solving this question...LOL

sammy gerbil

Alternatively if you are familiar with moments you can take moments about the top of the string. The tension in the string acts through this point so it has no moment. Although the other 2 forces are not aligned their moments about the top of the string must be equal because the string is not turning.

gateprep

@invadingdingo From where did u find this problem?

sammy gerbil

@gateprep I shall have to sleep soon. Sorry I don't have the strength to help.

invadingdingo

It's a problem from my physics book the professor assigned, but did it over the weekend with 8 other problems so I'm unable to seek help from the resources the college provides.

We have a whole tutoring staff that can help, but they don't work on weekends :/

Unfortunately I can't seem to get my head around what you guys are suggesting, so I think I'm going to sleep for now and see what I can make of it tomorrow. I've been staring at this stuff fo 7 hours straight now, haha!

gateprep

05:22

@invadingdingo What is the name of the book?

invadingdingo

05:35

chegg.com/textbooks/…

gateprep

05:47

@invadingdingo How much do they charge for getting the solutions of a book

harambe

06:19

@JohnRennie good morning

Are you free for some time

John Rennie

@harambe I'm around Ll morning but I'm working on and off so I won't always be available.

harambe

Okay.... I just had small doubts about angular momentum

John Rennie

Well, you can ask ...

harambe

Why isn't angular momentum mvR here

imgur.com/a/GDNQPNI

I am unable to interpret angular momentum physically

John Rennie

At the instant shown in the diagram the wheel is, in effect, rotating about the point O as well as rotating on its axis. So you need to calculate the two angular momenta separately then add them.

harambe

06:29

So do points that do not lie in axis have angular momentum w. r. t the axis that we have to count

That would mean angular momentum depends on a frame

John Rennie

@harambe correct, angular momentum does depend on the frme.

harambe

So like linear momentum is, is there any frame where angular momentum is zero

Like centre of mass frame has zero momentum

John Rennie

No. Unless you allow rotating frames (which are non-inertial and therefore should be avoided like the plague) there is in general no frame where the angular momentum is zero.

Though in special cases such a frame does exist.

harambe

Oh okay

John Rennie

For example consider a point mass passing you at a distance $d$ and velocity $v$. In your frame the angular momentum is $mvd$.

If the distance $d$ is zero, i.e. if you are in the path of the particle, then the angular momentum is zero.

But obviously that's a special case e.g. it only works for point particles.

harambe

07:27

@JohnRennie what is orbital angular momentum and spin angular momentum

Wikipedia also states it but I didn't get that

Is rotation about axis spin angular momentum like here and the momentum about point O orbital?

Abcd

08:21

@JohnRennie Are you there

John Rennie

@Abcd hi

Abcd

Very quick questions Please see:

@JohnRennie These 4.

John Rennie

For the first question the answer is obviously the density. They start with the same volumes and are both compressed by 50% so they both end at the same volumes.

Abcd

@JohnRennie Why not initial energy???

John Rennie

@Abcd hmm, to be fair that should also be correct since the two samples start out identical.

But then the question doesn't say what it means by identical energy. $U$? H$?

Abcd

08:30

@JohnRennie Or is it that one could be monoatomic and other diamotic? That's why energy is not same?

@JohnRennie obviously internal energy

John Rennie

Ah, yes, that's possible

Abcd

@JohnRennie Okay, next question please.

John Rennie

@Abcd the rod is at equilibrium so the temperature profile will be a straight line within each bit of the rod. The gradient of the staight line will be proportional to the thermal resistivity i.e. inversely proportional to the thermal conductivity.

Abcd

@JohnRennie so?

John Rennie

So the graph will be three straight lines with a higher gradient in the lower conductivity reagion.

Abcd

08:35

@JohnRennie Is D correct?

John Rennie

Yes, D is correct

Abcd

@JohnRennie Answer given is A. D was my choice

John Rennie

Hmm, I would have said B.

Abcd

????

John Rennie

At equilibrium the amount of heat flowing into the hot end has to be the same as the amount of heat flowing out of the cold end. So the total heat flow must be constant thoughout the rod.

If they mean heat flow per unit area then that decreases as we move from left to right as the area increases.

Abcd

08:42

area is not changing

John Rennie

Oops, sorry, I'd moved on to the third question. Let me go back and look at the second question.

I don't see how A can possibly be the correct answer to the second question.

Abcd

@JohnRennie does it have to do something with "unlagged" ?

Nothing is insulated

And $i = \dfrac{-KA dT}{dx}$ is for insulated stuff

John Rennie

Ah ... yes, I completely missed that. Yes, heat will flow sideways out of the rod at a rate roughly proportional to the temperature.

Abcd

@JohnRennie ??

John Rennie

So more heat will be lost at the hot end than at the cool end i.e. the temperature will fall faster with $x$ at the hot end than at the cold end.

Abcd

08:48

not getting

John Rennie

@Abcd if the rod isn't lagged then heat will flow sideways out of the rod into the environment around it.

Abcd

.. Okay leave it then @JohnRennie

@JohnRennie next question please

John Rennie

The answer to the third question is B

AbhigyanC

@sammygerbil Thanks...

It makes a lot of sense now...

Mr. Xcoder

Hi, people! Does anybody know any good astronomy/astrophysics courses for starters?

Courses as in online materials, books etc.

AbhigyanC

08:56

@Mr.Xcoder MIT 8.901

Mr. Xcoder

That's graduate level I think, I'm looking for something a bit more basic

harambe

@Mr.Xcoder maybe try khan Academy?

@JohnRennie are you free now again?

John Rennie

@harambe basically yes. I'm intermittently struggling with a special relativity calculation so I may be distracted occasionally.

BTW has your suspension finished?

harambe

@JohnRennie yea. It ended on Sunday somewhat 2 PM IST

I asked @ACuriousMind for making sure and he gave me permission so it's all right now

Mr. Xcoder

@harambe I'll try that and see if it's what I am looking for. Thanks for the suggestion.

harambe

09:07

Oh no problem

John Rennie

@harambe cool :-)

harambe

@JohnRennie is there a time when you are free from work and also have some spare time to chat?

John Rennie

@harambe I'm free now (mostly)

harambe

Okay.. I just do this want to impose on you that's it.

About the previous question, I did the calculation and got the answer but I didn't get the part where you said we have to include both momentum

I mean is the wheel rotsting on two axis?

John Rennie

Actually that was rubbish, ignore me.

There is only ever one axis of rotation.

An object can't simultaneously rotate about two different axes without tearing itself apart! :-)

Blame it on it being early in the morning

harambe

09:16

Okay.. I was having the same doubt. Two axis rotation is out of topic for us anyways

harambe

09:38

@JohnRennie I tried a similar question

imgur.com/a/FEjFECf

In the frame of axis the mass at the other end will have relative velocity w. r. t the mass in the middle and using m(rxv) gives the answer

John Rennie

Question 3?

harambe

But why are we not including the angular momentum of axis here. We included it in the previous one

Yes

John Rennie

Incidentally, if you could photograph the questions the right way up that would make life a lot easier.

harambe

Oh my mistake

I caught the photo in a hurry

John Rennie

OK, the question is asking about the rotation of the end relative to the particle at the centre. That can get confusing because you're trying to calculate rotation relative to something else that is rotating.

However the angular velocity has to be equal to $\omega$ i.e. the same as the angular velocity of the whole system. So that makes things a lot easier.

harambe

09:43

Yea property of rigid body

John Rennie

So for the particle at the centre, the other particle is moving with angular velocity $\omega$ at a distance of $d/2$.

harambe

Yes

John Rennie

So the moment of inertia of the other particle is $mr^2 = md^2/4$

harambe

Okay

@JohnRennie I get what you are doing. You are transforming the mass kept at the centre to be another axis

And that will give us the angular momentum easily about it?

John Rennie

@harambe $L = I\omega = m \omega d^2/4$

Answer (b)

harambe

09:52

Got it. I just had one question

John Rennie

Yes?

harambe

How is this case different from the previous question I asked

I mean in calculatimg angular momentum

John Rennie

It isn't different. In both cases we need to find the moment of inertia and the angular velocity about our chosen reference point. Then the angular momentum is just $L = I\omega$

For the wheel you used the parallel axis theorem to calculate $I$. For the last question the moment of inertia is simpler because for a point mass at some distance $r$ we get $I = mr^2$.

harambe

@JohnRennie but clockwise and anticlockwise rotation of disc is giving different answers

Why so

John Rennie

10:07

@harambe In the question the wheel is rolling along the floor so the point of the wheel in contact with the floor is stationary relative to the floor.

But if the wheel is rotation in the other direction it is skidding along the floor so the point of the wheel in contact with the floor is moving.

harambe

Okay but what does it tell us

John Rennie

That means you can't simply proceed as if the whole wheel is rotating around the point of contact with the floor.

harambe

About angular momentum

John Rennie

I'd have to thnik about how to do the calculation in that case, but it would certainly produce a different answer.

harambe

Okay so L=I w is valid only for inertial points

That too rotsting about a fixed axis

Got it

harambe

10:30

@JohnRennie I am getting some doubts again

imgur.com/a/Nyq5Qfp

In Q 4,the initial angular momentum is Iw

But for calculating the final angular momentum will the moment of inertia be 2I

John Rennie

Yes. Moments of inertia just add (assuming they are about the same axis). Since the two wheels are identical their combined moment of inertia is twice the moment of inertia of one wheel.

harambe

Oh kay.. I got confused in this adding of moment of inertia

Can't the axis change in these angular momentum conservation questions

Like for Q5 @JohnRennie

I am getting angular velocity to be w/2 considering axis doesn't change

John Rennie

The rotation axis before and after adding the stone is the same axis. Yes?

harambe

Yes

I don't understand why it doesn't change

John Rennie

The moment of inertia of the disk about this axis is ... erm ... I can't remember the equation for the moment of inertia of a disk about its centre.

harambe

10:36

I think I am just getting confused by my own silly doubts

MR2/2

John Rennie

@harambe yes, $I_\text{disk} = \tfrac{1}{2}mr^2 $

And if we can treat the particle as a point mass then its moment of inertia is bust $MR^2$ i.e. $I_\text{stone} = 2m (a/2)^2$

harambe

I proceed we just like that

But these silly doubts just come to me xd.

John Rennie

Yes, the total moment of inertia afterwards is just $\tfrac{1}{2}ma^2 + 2m(a/2)^2 = ma^2$

harambe

And equating it with intial angular momentum gives the required value of angular velocity

@sammygerbil good morning

sammy gerbil

@harambe Good morning. Just taking another look at gateprep's question.

John Rennie

10:53

@sammygerbil would you like to be a room owner here? You seem to be putting a lot of work into answering questions.

sammy gerbil

@JohnRennie What does the job involve?

John Rennie

@sammygerbil nothing really. It just gives you the power to delete posts and kick people out of the room.

I've used the superpowers to delete posts, but I've never had to eject anyone.

sammy gerbil

@JohnRennie ok count me in!

Feeds

John Rennie has added sammy gerbil to the list of this room's owners.

John Rennie

Done. If you ever feel the urge to abuse your new superpowers go right ahead :-)

sammy gerbil

11:15

@gateprep I located the "exception example" on the Khan Academy page, and I found a question about this on Mathematics SE (math.stackexchange.com/questions/2337004/…) but I cannot find the solution using y=mx. I don't understand the thinking behind this substitution.

If the calculation is "pretty tricky" then it is probably too long for this Chatroom. How about posting in Mathematics SE? They don't have the same HW policy as we have in Physics SE.

harambe

11:27

@JohnRennie imgur.com/a/nRXPO7S

Why doesn't applying conservation of energy directly gives the answer

The kinetic energy of the bullet is converted into rotstional energy of rod

John Rennie

The bullet gets embedded in the rod so it is not an elastic collision.

If it was an elastic collision the bullet would bounce backwards off the rod.

harambe

Okay

So how should I proceed for this

I have calculated velocity of centre of mass of the system after impact

Will angular moment be conssrved

John Rennie

Yes, just use conservation of angular momentum. The moment of inertia of the bullet is $mL^2$ and its angular velocity is $\omega = v/L$.

harambe

@JohnRennie one little doubt

The velocity of centre of mass is in ground frame

John Rennie

You don't need the velocity of the centre of mass.

This isn't an inertial frame because the pivot is going to exert a force, so I wouldn't try and use the COM frame for this one.

harambe

11:38

But doesn't hinge make the pivot inertial

John Rennie

The hinge is going to exert a force, so the frame isn't intertial. Luckily the pivot doesn't move so even though it exerts a force it doesn't do any work.

Just calculate the angular momentum about the hinge just before and just after the collision and that will give you $\omega$ after the collision.

harambe

By frame do you mean centre of mass frame

@JohnRennie got the answer

But I still don't understand which frame were you talking about

Maybe centre of mass frame?

These frames are very confusing to me

John Rennie

When you're dealing with rotational motion do all your calculations in the rest frame of the axis of rotation.

John Rennie

12:00

I'm off now. I'll be back briefly later today then tomorrow morning.

harambe

16:19

@JohnRennie A quick question

It isn't nevessarily for a continious function to be differentiable at every point, right

Or does continious function guarantee derivability

John Rennie

@harambe correct. For example a triangular wave is continuous but not differentiable everywhere.

harambe

Yea I just saw this in some book but I think they explained it wrong. Thanks for clarifying

Abcd

19:19

@sammygerbil Are you there?

sammy gerbil

@Abcd yes. Good evening.

Abcd

@sammygerbil Hi, do you know about liquid solutions, the correpsonding laws, entropy, Van't hoff factor etc?

sammy gerbil

@Abcd Almost nothing. I took A level chemistry but that was a long time ago.

Abcd

@sammygerbil Ohkay fine... No expert there to help us in chemistry room :(

sammy gerbil

@Abcd There is a room for JEE chemistry? Is that The Periodic Table?

Abcd

19:25

JEE Chemistry Club

For Mathjax see: chemistry.meta.stackexchange.com/questions/89/…...

@sammygerbil Yes, this one^ (not periodic table)

Chakrapani N Rao

Hey guys do you have any answer to this

-1

Q: Can this thought experiment successfully prove that HUP is epistemological and not ontological?

Here's a thought experiment. Let's assume that we have the most cutting edge technology at our disposal. We need an electron detection sphere, as small as possible, it must detect a single electron when it hits the surface. Now assume that we introduce an electron in the sphere, as slowly as po...

sammy gerbil

@ChakrapaniNRao I think that would get more discussion in the hbar. We're mostly concerned with examination questions here.

@ChakrapaniNRao I don't think your idea of "stunning" an electron would work. Electric fields are conservative, even strong ones. Just because you confine an electron does not guarantee that you remove its kinetic energy. When you "confine" it you are just forcing it to move about in a smaller space. It keeps its kinetic energy and goes back to its previous motion when you release it.

Chakrapani N Rao

19:44

@sammygerbil thanks for responding, I have tried hbar in vain

The idea was never to stun an electron, it was to create a very slow moving electron but was said in an ideal sense. Like if you try to stun an electron you must practically have a very slow moving electron, that was the point

sammy gerbil

@ChakrapaniNRao And it doesn't make sense to use the tag for Classical Mechanics at the same time as Quantum Mechanics unless you are asking about the transition between quantum and classical domains.

Chakrapani N Rao

@sammygerbil very good observation, that's done purposefully, but if I reveal the intent of doing it, people will shut me down

but what do you think of the main question

?

sammy gerbil

20:45

@ChakrapaniNRao Sorry I do not understand your argument, nor the leap into philosophy at the end. It needs much more explanation, eg what you mean by epistemological and ontological and how you reached this conclusion. Raising philosophical issues in Physics SE and assuming users know what these terms mean rarely gets a good response. If you can use plain English you have a much better chance.

JEE Chemistry Club

Transcript for

Problem Solving Strategies