Problem Solving Strategies

satyam kumar jha

03:34

Hello @JohnRennie sir. I had posted this question which I couldn't understand completely even after getting an answer. Can you please take a look at it? (The comments at the end of the recieved answer help to understand my issue better). Here's the link: physics.stackexchange.com/questions/666052/…

Ashish Ahuja

04:48

@JohnRennie hi

John Rennie

@AshishAhuja Hi :-)

Ashish Ahuja

When we talk about a transistor acting as an amplifier in it's active state in the common base configuration, could you explain why the collector current is only dependent on the voltage between the base and the emitter, and not the voltage between the base and the collector? I found this and it makes sense but I cannot understand why the amount of the carriers depend only on $V_{BE}$

John Rennie

The collector current is not dependent on the base emitter voltage, it depends on the base emitter current.

To be honest I don't remember the exact mechanism, but every electron that flows into the base triggers β electrons to flow across the base-collector junction.

Where β is the gain.

Ashish Ahuja

@JohnRennie but the answer I've linked above says it depends on $V_{BE}$..

John Rennie

Well it kind of does because $I_{BE}$ depends on $V_{BE}$, but the actual mechanism is related to the flow of current not the voltage.

Ashish Ahuja

04:54

ok then could you explain why $I_{BE}$ does not depend on $V_{CB}$?

(in the active state)

John Rennie

You can now see why the collector voltage makes no difference. Every electron flowing into the base allows β electrons to cross the collector-base junction. If you double or triple the collector voltage it makes no difference because the current is limited to β electrons for every base electron.

The transistor is acting like a current limiter i.e. it limits the current that flows through it regardless of what voltage you apply.

Ashish Ahuja

@JohnRennie I agree with this that for every electron flowing into the base $\beta$ electrons flow into the collector, i.e $I_C = \beta I_B$, so if you increase $I_B$, $I_C$ should increase. What I don't understand is why increasing $V_{BC}$ will not increase $I_B$

John Rennie

Oh, wait, I misread your question. Sorry :-(

The current doesn't depend on $V_{CE}$.

The collector base junction is like a reverse biased diode. Yes?

Ashish Ahuja

@JohnRennie yes

John Rennie

Ah, I misread your question again :-( I thought we were discussing common emitter.

Ashish Ahuja

05:01

ok :D

John Rennie

I'm not familiar with the common base layout. To be honest I never learned transistors at college. I learned about them because as a boy I used to spend my time playing with electronic circuits. Computers didn't exist in those days so we nerds used to mess around with electronics instead :-)

So my knowledge is based on experience, and common base circuits were rarely used so I never got to learn them.

Ashish Ahuja

oh yes I used to also play around with transistors, nowadays it is much easier I think since you can make circuits on breadboards. Making a one bit adder was probably the best thing I've ever made.

I'm yet to study common emitter, I think I will run into the same issue there. I will come back then.

John Rennie

:-)

Ashish Ahuja

06:11

@JohnRennie hi

same issue in the common emitter configuration, I don't understand why in the active region $V_{CE}$ has negligible effect on $I_C$ (this can be seen in the output characteristic when $I_C$ asymptotes as $V_{CE}$ keeps increasing)

Ashish Ahuja

06:36

@JohnRennie hi, are you there?

Lllt

07:03

The electric flux through a surface is proportional to the number of field lines cutting
the area element of a surface are placed normal to electric field

@JohnRennie Sir is the above definition of flux correct I came up with??

John Rennie

07:23

@AshishAhuja Hi, sorry I had to drop out for a bit. I'm around now for a couple of hours.

@Lllt Hi :-)

Lllt

Hello sir

John Rennie

I'm not sure it's useful to think about field lines in this context. A "field line" is a vaguely defined concept so it's not useful in definitions.

Lllt

@JohnRennie Yes you are right I asked about this on main site and users said the same

@JohnRennie But This term is being used by my teachers and that's causing confusion

Ok let me tell you what is the source of confusion

Can you draw a figure for me??

John Rennie

I have another question to answer at the moment. I'll ping you when I'm free. Explain what figure you want me to draw and I'll do it as soon as I'm free.

Lllt

Ok, the figure consists of a disc , a open hemispherical shell and a rough arbitrary surface having same axis

Please ping me when you are here:-)

John Rennie

08:08

@Lllt Like this?

Lllt

Yes but that arbitrary surface is like deformed hemispherical surface i.e it is 2-D with circular base

John Rennie

Lllt

Yes exactly

The question was to find flux through all these surfaces when there's a uniform electric field left to right

The disc one is easy

It will simply be $\Phi=E\pi r^2$

John Rennie

Let me redraw the deformed surface so it doesn't stick out past the disk ...

Lllt

The hemispherical one will be $$\Phi=E\int ds \cos\theta$$

John Rennie

08:15

The disk is indeed easy, but actually both the others are just as easy.

Lllt

Yes I know, please don't reveal

John Rennie

OK :-)

Your equation applies to all three surfaces, the only difference is the angle θ in the integral.

Lllt

I was told that the same number of electric field lines cross disc , hemisphere , and that deformed one and therefore flux through each is same

Do you think it is a vague explanation?

John Rennie

To see why this is true let's consider the hemisphere.

Lllt

Yes

John Rennie

08:19

Suppose we treat this as a closed surface with a disk on one side and the hemisphere on the other side.

The volume inside the surface does not contain any charges, and field lines can only start or end on a charge. Yes?

Lllt

Hmm we cannot use Gauss' Law :-)

It is one line proof with it.

John Rennie

We are kind of using Gauss's law. The total flux through the surface has to be zero because the charge inside it is zero. Yes?

Lllt

Yes

John Rennie

So the flux that flows in through the disk side has to be equal and opposite to the flux that flows out through the hemisphere side.

Lllt

Yes

John Rennie

08:21

And we have already decided that the flux that flows into the disk side is πr²E

Lllt

Yes

John Rennie

So the flux through the hemisphere must also be πr²E

Lllt

Yes

John Rennie

So there's our answer. And this argument also applies to the deformed surface.

Lllt

@JohnRennie So by easy you meant this method

John Rennie

08:23

Yes, but you got there yourself :-)

Lllt

Yes But I meant something else as following

The flux through hemispherical surface is

The hemispherical one will be $$\Phi=E\int ds \cos\theta$$

$$\int ds \cos\theta$$ means summation of components of infinitesimal areas perpendicular to electric field

Yess??

John Rennie

The areas dS are not perpendicular

dS cosθ is the the component of the area perpendicular to the field.

Lllt

Yes I meant that only

John Rennie

OK then

Lllt

Then $$\int ds \cos\theta$$ represents summation of all component of the area which are perpendicular to the field.

Yes??

John Rennie

08:28

Yes

Lllt

One minute I will draw the diagram

John Rennie

Yes.

Lllt

If we go and stick all the components to circular base , then they completely fill the circle

Yes?

John Rennie

Yes. The summed area normal to the axis just adds up to the area of the disk on the left side.

Lllt

Yes

And again results come out to be E\pi r^2

John Rennie

08:35

Yes.

Lllt

That's why I was saying electric flux represents relayive no. of field lines which pass through the component of area of a elemental surface which is normal to electric field

and not the following one

electric flux represents _relative no. of field lines which pass through the area of a elemental surface _

Am I right??

John Rennie

Every field line that passes through the component normal to the field, dS cosθ, also passes through dS. Yes?

Lllt

Yes?

Oh so both definitions become equivalent.

John Rennie

Yes

That was the point I was making earlier.

Lllt

Yes it's completely clear now

John Rennie

08:39

OK :-)

Lllt

Thankyou @JohnRennie Sir :-)

John Rennie

This stuff can be hard at first, but when you get the hang of it you'll find it's simpler than you think.

Ashish Ahuja

2 hours ago, by Ashish Ahuja

same issue in the common emitter configuration, I don't understand why in the active region $V_{CE}$ has negligible effect on $I_C$ (this can be seen in the output characteristic when $I_C$ asymptotes as $V_{CE}$ keeps increasing)

Lllt

@JohnRennie True , but you are always there to make them easy for us , Thankyou :-)

John Rennie

@AshishAhuja Hi :-)

This the key point to understand about transistors. They are current limiters.

If the base current is Ib then a transistor limits the collector current to βIb and this does not depend on the collector voltage.

(in the active region)

Ashish Ahuja

08:43

yes agreed but shouldn't changing $V_{CE}$ have an effect on $I_B$?

John Rennie

No. Any current that flows across the (reverse biased) collector-base junction flows straight through the base and out of the emitter.

The base-emitter junction is like a forward biased diode so current flowing both into the base and into the collector flows out of the emitter.

The base-emitter current is determined only by the voltage applied to the base and the resistor in series with the base.

Ashish Ahuja

ok so if we hypothetically change $V_{CE}$ by a large amount, something in the circuit must change right? $V_{BE}$ does not change and neither does $I_B$ and hence neither does $I_C$, so what does change?

John Rennie

The effective resistance of the (reverse biased) collector-base junction changes.

With no base current the collector-base junction is like a reverse bised diode so no current flows through it.

But when we introduce a current into the base this makes the collector-base junction "leaky" i.e. some current can get across it.

Ashish Ahuja

ah yes I see. $R_O = \frac{V_{CE}}{I_C}$ changes, which limits the change in the current $I_C$, right?

John Rennie

In effect a base current reduces the resistance of the junction from ∞ to some finite value.

@AshishAhuja Yes.

Ashish Ahuja

08:51

ok I think I understand now, I was confused because I did not realize that the resistance would change hence it felt weird that $I_C$ was staying constant. Looks like "dynamic" resistance is indeed an apt name.

thanks

John Rennie

OK :-)

sonicsid

Hello.

That is the cross section of a hemisphere, and a charge +q is placed at O.

A second charge Q is placed at one of the positions A, B, C and D. In which of these positions would the flux of the electric field through the hemisphere remain unchanged?

Ah, ignore my question. By framing it in my own words, I found my mistake.

Misinterpreted the question as NET electric flux.

I would be right in saying that the NET electric flux would remain unchanged if Q were placed at any of the four positions, yes?

But flux (not net flux) through the hemisphere would change were Q to be placed at C or D.

Ashish Ahuja

@sonicsid not sure how you're defining net electric flux, but if you're considering the hemisphere to be a closed surface (with a thickness), then the net flux through the hemisphere will always be zero because there's no charge inside the walls of the hemisphere.

sonicsid

The charge q is placed at the center of the hemisphere.

Ashish Ahuja

Oh so you mean like a solid hemisphere?

sonicsid

09:01

So by Gauss's Law, a flux of q/2epsilon would pass through the hemisphere, right?

Uh, I'm not sure. Concepts of Physics part 2, page 140, Q6

Ashish Ahuja

@sonicsid yes I understand what your setup is now, and yes a flux of q/2e would pass through

sonicsid

He only writes "hemisphere"

satan 29

@sonicsid yes, I'd say so

sonicsid

Was my other reasoning correct?

The one about Q's position

and its relation with the net electric flux

satan 29

Any charge not enclosed within the surface will have no effect on the flux

Ashish Ahuja

09:03

well the flux through the hemisphere does not change when you place any charge outside the hemisphere, by gauss's law

I'll leave satan to it :)

sonicsid

@satan29 do you mean to say NET flux? Because surely field lines of the charge's field would still pass through the hemisphere, even if it is placed outside it.

satan 29

the field lines pass through the hemispgere, mind you

field lines enter the surface, but also leave the surface.

sonicsid

I've realised that I am unclear about Gauss's Law. I'll refer to my notes and confirm my understanding here afterwards. Thank you.

caramalizedTomato

09:27

@JohnRennie Hi, sorry wasn't there in the morning

John Rennie

10:00

@caramalizedTomato Hi :-)

caramalizedTomato

So I was asking why do we need the ohm's law here to explain the current?

John Rennie

Ohm's law always applies whenever we have a current flowing through a resistance.

In this case if we have some current I flowing round the loop, and the loop resistance is R, then there will be a voltage drop V = IR. OK so far?

caramalizedTomato

Yep

John Rennie

And this voltage drop has to be equal to the EMF generated by the magnetic field in the loop, because that's what drives the current round the loop. Yes?

caramalizedTomato

Yes

John Rennie

10:12

And that's what I did yesterday.

We end up with IR = dΦ/dt

caramalizedTomato

Now, I'm having trouble with faraday's law, as it associates the change in flux with some EMF, now that EMF should've been generated due to some electric field which in turn has been generated by the change in magnetic field, so wouldn't it had been more fundamental to use the electric field than the EMF?

What I mean to say, in the formula why is there an EMF term at all?

John Rennie

The problem is that understanding what is really happening needs Maxwell's equations. These are simple than you think but they are well above JEE level.

In particular Maxwell's equations tell us that an electric field is related to a changing magnetic field by $$ \nabla \times \mathbf E = -\frac{d\mathbf B}{dt} $$

Do you know about the curl of a vector field?

caramalizedTomato

Not mathematically, but I do have a basic idea, like it sort of tells us about circulation, or how much curved something is if I am right

John Rennie

Yes, the curl tells us about how the field lines form loops. If the field has a zero curl then the field lines do not form loops but instead they must begin or end on a charge.

But if the curl is non-zero that means the field lines can form loops.

And if a field line forms a loop we can move a charge around the field line and back to its original position. Yes?

caramalizedTomato

Yes

John Rennie

10:27

And the field line tells us the direction the field points, so if we can move a charge around a loop and back to its starting point the charge will be moving in the direction of the field all the way round the loop, and therefore the field will be doing work on the charge all the way round the loop. OK so far?

caramalizedTomato

Yep

John Rennie

And the work done on a charge q is just qV, so if we calculate the work as we move our charge q round the loop we have also calculated the potential difference round the loop. Yes?

caramalizedTomato

What is V here?

John Rennie

V is the potential difference.

If we move a charge q across a potential difference V then the work done is qV. Yes?

caramalizedTomato

Magnetic potential?

John Rennie

10:32

No, electrical potential.

caramalizedTomato

@JohnRennie You were talking about electric potential all along?

John Rennie

Yes

Sorry, I didn't realise there was any confusion about that.

Magnetic potential doesn't exist because magnetic fields are non-conservative.

Well magnetic potential does kind of exist but it's complicated.

caramalizedTomato

but then electric fields have a zero curl so wouldn't the work done be simply zero?

John Rennie

Remember the equation I posted earlier:

13 mins ago, by John Rennie

In particular Maxwell's equations tell us that an electric field is related to a changing magnetic field by $$ \nabla \times \mathbf E = -\frac{d\mathbf B}{dt} $$

The curl of an electric field is -dB/dt

So the curl of an electric field is not zero whenever we have a magnetic field that is changing with time.

caramalizedTomato

Does that mean electric fields can form a loop?

John Rennie

10:36

Yes, and that's exactly what is happening in this case.

That's why current flows round the ring in a loop.

caramalizedTomato

That's something I didn't know, wow

John Rennie

It's following the electric field lines that form loops.

This is the physical basis for Faraday's law.

A changing field creates electric field lines that form loops, so a current can flow in a loop.

caramalizedTomato

@JohnRennie I guess the due to the epsilon of the conductor the field is quiet strong which enables the electrons to flow than in the empty space is that right?

John Rennie

It's simpler than that.

There are electrons in the wire of the loop (i.e. the conduction electrons) but there aren't any electrons floating around in empty space.

caramalizedTomato

yes how can I overlook upon that

John Rennie

10:42

We get a current in the loop because the conduction electrons of the metal that the loop is made from can flow in response to the induced electric field.

caramalizedTomato

@JohnRennie Yes

John Rennie

In empty space we still get field lines that form loops. It's just that there are no charges floating around in empty space to follow them.

caramalizedTomato

Ok I'm clear with this, so you were saying about the work done on the charge will be qV

John Rennie

Yes.

And the work done can also be calculated from the electric field, because the force on the charge is F = qE and the work done if we move it a distance d𝓁 round the loop is dW = qEd𝓁.

caramalizedTomato

yes

John Rennie

10:47

So the total work will be ∫qEd𝓁, and we've already agreed this is also qV. And that tells us that the field and the potential round the loop are related by V = ∫Ed𝓁.

I've got a bit lost now, but I think we started by discussing whether the electric field or the potential was more fundamental.

And my point is that neither is more fundamental because they related by V = ∫Ed𝓁

They are different ways of describing the same thing.

caramalizedTomato

Ok got that

Just one more thing

John Rennie

Yes ... ?

caramalizedTomato

We say EMF, electromotive force but it is a scalar right for it's just another name for potential difference?

John Rennie

Yes.

sonicsid

Yes, it is not an actual force.

John Rennie

10:53

Maybe there is some technical difference between EMF and potential that I've forgotten, but basically they are the same thing.

sonicsid

EMF is the potential difference across the terminals of the battery when it is not connected in any circuit, iirc

caramalizedTomato

yes

John Rennie

I just Googled it, and it looks the same as potential difference to me :-)

caramalizedTomato

more precisely when there's no current flowing

John Rennie

Electromotive force

In electromagnetism and electronics, electromotive force (emf, denoted E {\displaystyle {\mathcal {E}}} and measured in volts) is the electrical action produced by a non-electrical source. Devices (known as transducers) provide an emf by converting other forms of energy into electrical energy, such as batteries (which convert chemical energy) or generators (which convert mechanical energy). Sometimes an analogy to water pressure is used to describe electromotive force. (The word "force" in this case is not...

sonicsid

10:55

@JohnRennie While I know that "electric field" and "electric field intensity" are terms used interchangbly, would it not be a better idea to specify that we are talking about electric field INTENSITY here? Because while an electric field does have physical existence, electric field intensity and potential are both merely quantities associated with the field and neither of those two can be called more fundamental than the other, as you correctly said.

caramalizedTomato

@JohnRennie Thank you so much sir =)

John Rennie

:-)

Don't be too worried if this all seems very confusing. A lot of what we talked about today is more advanced than the JEE.

@sonicsid I would say "electric field", and most physicists I know would say the same. The electric field is a function that gives the value of the electric vector at every point in space.

caramalizedTomato

@JohnRennie I actually did understand, atleast I got the idea if not the math

sonicsid

I see. I should drop this matter since it is more philosophical (for lack of a better word) than practically useful.

John Rennie

@caramalizedTomato OK :-)

sonicsid

11:02

@JohnRennie But the JEE Advanced is rather advanced. If you look at 2020's question paper, there are some questions that were taken from research papers. There was also an interesting question about how Ohm's Law gets modified when the conductor is placed in an external magnetic field.

John Rennie

I doubt you need to understand Maxwell's equations though. Knowing Faraday's law and Ampere's law will answer most questions.

sonicsid

I agree.

John Rennie

In a way it's a shame, because if you go on to do physics at university and learn about Maxwell's equations a lot of stuff suddenly makes more sense.

caramalizedTomato

@sonicsid Actually yes but the models for those question will be heavily contrained, hence can be solved with basic concepts.

@JohnRennie At the moment I can't agree more

It was like epiphany

John Rennie

:-)

sonicsid

11:08

@JohnRennie Well, things making more sense to us also acts as an incentive to pursue higher education from a good institution :D

stuff making more sense by studying at such a university, I mean

Out of A, B, C and D, where can another charge Q be placed such that the flux of the electric field through the hemisphere remains unchanged?

O is the center of the hemisphere here.

John Rennie

If you put a charge at A or B then every field line radiating out from the charge will pass though the hemisphere twice. Yes?

sonicsid

Sorry, I do not follow. Pass throguh it twice?

Right

John Rennie

Like this

The field line intersects the hemisphere in two places.

sonicsid

Ah, and in that case if once the dot product with the area vector element for each field line would be positive and there will be a corresponding negative one too, so flux will remain unchanged?

John Rennie

Correct :-)

sonicsid

11:16

Got it. Thank you!

John Rennie

:-)

Ashish Ahuja

12:03

@JohnRennie hi

John Rennie

@AshishAhuja Hi :-)

Is it quick? I need to go in five minutes.

Ashish Ahuja

I think it is, if it isn't I can come back later. I'll type it out fast

John Rennie

OK :-)

Ashish Ahuja

In a common emitter state of a transistor, it is clear that $I_E = I_C + I_B$. But surely since the dynamic resistances are different, $V_{CE}$ cannot be equal to $V_{CB} + V_{BE}$?

John Rennie

The voltages can be a bit weird in transistors. Vbe is always about 0.6V because it's just a forward biased diode. However Vce can be as low as 0.2V.

satan 29

12:09

Vcb= Vc- Vb, Vbe= Vb-Ve, Vce= Vc-Ve, so i dont see why that equation wont hold?

John Rennie

The seems not to make sense since we end up with Vce less than Vbe, but it happens when a transistor is saturated.

Ashish Ahuja

@satan29 huh, I never thought about it this way.. you're essentially saying that $\int E \cdot dl = 0$ and hence $V_{CE} = V_{CB} + V_{BE}$?

John Rennie

I need to go. I won't be around this evening but I'll be back tomorrow as usual.

Ashish Ahuja

ok

satan 29

ofcourse that has to be the case, no?

Ashish Ahuja

12:13

yes now that you mentioned it it seems to be obvious, I agree. Although I'm not sure exactly what happens inside the transistor on a microscopic scale, it seems unlikely that there's a changing magnetic flux so $\int E \cdot dl$ must be zero.

satan 29

Yes, I honestly dont see how a non conservative E can be stablished.

Ashish Ahuja

not sure what JR was trying to say, but yes I'm convinced now.

Also, btw can you explain how you had calculated (the method) angular momentum of the disk about the center of the semicircle the other day? I've never calculated the angular momentum in such a situation before.

satan 29

$L_{0}= L_{cm} +m r_{0-> cm} \times (V_{cm}-V{0})$

Angular momentum about a point o.

Ashish Ahuja

oh

right yes

I see now, thanks.

satan 29

so taking point o to be the center, we get $L0= Lcm+ m x \times ( \omega \times x)$

which works out to be $I_{com}\omega + mx^2 \omega= I_{o} \omega$, rather unsurprisingly.

Ashish Ahuja

12:21

hmm did you get the correct answer using this about the semicircle afterwards?

satan 29

No :/

Ashish Ahuja

I'll go try it..

satan 29

hang on ...was v supposed to be relative to the surface?

Ashish Ahuja

no

satan 29

ah wait a min.

while calculating the final angular momentum about the center of the semicircle, we need to be careful.

so the point about which we are calculating is moving.

Ashish Ahuja

12:33

yes, so you need to take the relative velocity of the ball.

satan 29

$L0= r \times m (v-v0)$

Ashish Ahuja

It should be v + v0? The disk and the ball will be moving in opposite directions.

satan 29

I was writing a vector equation

Ashish Ahuja

ok sorry nvm

satan 29

final L is mr(v+wR/2) then for the ball

Ashish Ahuja

12:35

yes

satan 29

muR/2= - 3/4 MR^2 w + mR/2(v+wR/2) i believe?

Ashish Ahuja

can you give me a sec..? I need to go for a few mins.

satan 29

sure

Ashish Ahuja

ok I'm back lemme see..

@satan29 yes that is exactly what I got as well but there's a problem I think

satan 29

yes?

Ashish Ahuja

12:42

The energy conservation equation remains the same irrespective of whether you calculate L about the com or the semicircle center

satan 29

yes that correct, and thats what I was thinking too... the L consv equation should finally match with that about the center..

Ashish Ahuja

so to get the same solution the L conservation equation should end up being the same as well, but it does not in this case.. plus we have three variables (u, v, omega) and 2 equations

@satan29 yes

I think our equation for L is correct, but we just have too many variables so this method does not work

satan 29

no why? I think we can still solve it

we get v-u= R/2 w (3x-1) where x= M/m

Ashish Ahuja

..how? you have 3 variables with 2 equations

satan 29

from energy cons. we can get v^2-u^2= f(R,x,w^2)

dividing, we get v+u = f(R,x,w)

v+u/v-u = f(x)

am i missing something?

@AshishAhuja v and w are the only variables

Ashish Ahuja

12:47

sorry it's hard to follow through on your exact calculations.. what is the final result you're getting?

@satan29 R

satan 29

do you agree with my equation for v-u ?

Ashish Ahuja

give me a second I need to actually go through the calculation.

@satan29 I agree till here at least

satan 29

from energy cons, v^2- u^2 = 3/4 x R^2 w^2

dividing, we get v+u= 3/2 wR (x/3x-1)

Ashish Ahuja

@satan29 agreed till here now

satan 29

ohhhhh wait i messed up my energy conservation calculation

mv^2- mu^2= MR^2/2 w^2

so v^2-u^2=x/2 R^2 w^2

now dividing, v+u= Rw (x/3x-1)

so v+u/ v-u = 2x/ (3x-1)^2

sure, we can solve the equations, but its going to give the wrong answer :/

Ashish Ahuja

12:54

@satan29 isn't u the initial velocity, so it should be mu^2 - mv^2

satan 29

oop

s

okay, add a - sign

v+u/v-u = -2x/(3x-1)^2 then, we can get v in terms of u easily..but its wrong.

actualy if you think about it, how the hell will you get a square root?

oh actually wait nvm,

Ashish Ahuja

I got a square root when doing the calculation about COM, so the answer can't be wrong.

satan 29

yeah yeah the angular momentum consv equation in the COM frame will not have v-u term

Ashish Ahuja

@satan29 there should be an x in the numerator here?

satan 29

x/ 3x-1

Ashish Ahuja

13:01

ok yes my bad

satan 29

I cant find a flaw in this approach though.... L should be conserved about the center of the SC too

Ashish Ahuja

the calculation is surely correct, so either L is not conserved or we are calculating it wrong about the center of the semicircle..

satan 29

Although the center of the SC is an accerating frame, the pseudo force at evvery instant passes through the point, so it wont produce torque at any instant

hang on though

W is changing through the course of the mmotion, so the table has alpha as well

so the center of the SC is accelerating tangentially as well.

the corresponding pseudo force will produce torque, that will change the angular momentum by $\int \tau dt= k \int \alpha dt= k \Delta \omega= k \omega_{f}$

Ashish Ahuja

ok wtf is going on... this goes straight over my head but does it work?

satan 29

'm a bit busy right now so I have to go, but this might be it.

Ashish Ahuja

13:11

ok. Even I'm going for dinner now, please do let me know if it works.

satan 29

@AshishAhuja when we use accelerating frames for analysis (like the center of the semicircle), we must account for Pseudo forces and its effects.

and when we say L "about" a point, we actually mean L in frame of that point

so when analysing L from an accelerating frame, we must be wary of torques that can be produced by pseudo forces.

Ashish Ahuja

13:40

@satan29 yes I understand this. $k$ is the moment of inertia about that point?

sonicsid

16:05

Does anyone here know of some reliable source for HC Verma solutions?

of the subjective problems at least

Buraian

16:44

there is a soltn manual

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