@JohnRennie got a question for ya ;)

one sec

How do I solve this question using Kepler's law? Would someone please help?

@JohnRennie For this question right here

For #5, is Isphere = 7/5MR^2 ? and Irod = 1/12ML^2 + M(L/2 + 2R)^2 ?

@Hema Are more than one correct in the question?

@amanuel2 what is L?

@AjayMishra ah i posted it earlier for a previous question

thats not what im solving, but values are underneath

If I have to solve it, I would use, For rod $$ I_{edge} = \frac{ml^2}{3} + 4mr^2 $$

@AjayMishra How'd you obtain an answer for Hema's question, out of curiousity?

and you are right about the sphere part.

@kylecampbell , Since The satellite is stopped, there won't any force tangential to the orbit, Am I right?

Yup

@AjayMishra Can we apply this ?? Isn't the parallel Axis theorem applied only thru. COM?

I think, this would be the simple situation like that of falling bodies, and the only difference would be variable $g$?

Sure, so how did you find an answer in terms of $T$?

@AdvilSell Indeed, you are right!.

@amanuel2 What is the problem when you plug values in your equation and enter the answer, is it showing wrong?

@kylecampbell What is the problem in using $ \frac{dv}{dt} = \frac{GM}{r^2} $ ?

Nothing

Then I will using that equation.

Alright, but where is the period of the original orbit factoring in?

Are you going to integrate and then sub $v=r{\omega}=r\frac{2\pi}{T}$?

I'm just curious what your method is.

No, I would solve the differential.

Hm, solve for $v$?

Ya.

And then?

Solve for t

And then?

I don't think you would immediately get an answer in terms of T.

What is the problem?

Both answers b and c are in terms of $T.$ Are you saying neither of them are true?

No.

Okay, so that's all I'm really asking. Are you using an approximation to say that either both or neither of them are $\approx t$? Or did you express your $t$ in terms of $T$ somehow?

Wait, I am just solving this here.

@kylecampbell Are you there?

Sure, but what about in terms of the period. I don't see anything wrong with that! I'm just asking because the question has the time of the fall $t$ in terms of the period of the orbit $T.$

I have to approximate that.

Are you just approximating?

Right, that's what I figured.

As there are no term of $\pi$ in the answer.

Have you solved this?

@JohnRennie Been waiting for you prof!

got a quick question if u available

@amanuel2 hi :-)

Yes I'm free.

Do you remember my previous question or have some idea of it?

The sphere and rod?

Yeah! Well,

This is one of the questions it wont check so I wanted to make sure i got the right answer

For #5, is Isphere = 7/5MR^2 ? and Irod = 1/12ML^2 + M(L/2 + 2R)^2 ?

@amanuel2 Yes, that looks correct to me.

Do you want me to do the calculation and check your answer?

Nah, I just wanted to see if i had the right idea of both of the spheres moment of inertia. No problem on adding them up to calculator

Ok if that seems good i got one more question

The only part i dont understand now is when to use 1/3ML^2 and 1/12ML^2

For the Center of mass moment of intertia of the rod

For example here for some reason you use 1/3? i.imgur.com/YX3TKtP.png

Is it because it specified "left end" ?

If you are using the parallel axis theorem then you have to start with Icm because the theorem only works when you start with Icm.

But the moment of inertia of a rod about the end is one of those standard formulae that you should know i.e. it is Iend = 1/3 mL^2.

I didn't use the parallel axis theorem to derive this because I already knew it.

@Hema there's a sneaky trick you can use. The period of an orbit depends only on the semi major axis and not on the eccentricity.

The fall from a distance $R$ directly onto the planet is the extreme limit of an elliptical orbit with a semi major axis or $a = R/2$. So it has the same period as a circular orbit with a radius of $R/2$.

@JohnRennie Yeah it shows us here

We used 1/3 for the other one because the axis was at the very left end of the rope right?

@amanuel2 correct

In this specific situation the axis cant be in the right end. Correct?

@JohnRennie

@amanuel2 which diagram am I looking at now?

@JohnRennie Nah im talking about right end in general. I can see left end by having the axis be at the very left of rod, but then at the right its connected to the sphere. So i was assuming it cant have a right end?

The object is made up from the rod plus the sphere. You treat the two parts separately when calculting the total moment of inertia. So if the axis is at the point where the rod joins the sphere then it is at the right end of the rod (and the left side of the sphere).

@JohnRennie would you please give a diagram? I'm having a little trouble visualising why R/2 is the semi major axis and not R/3. Isn't the planet located at one of the ellipse's foci?

@Hema sure. Give me a moment and I'll draw a diagram now.

Ohhh ok

@JohnRennie Oh alright thanks :)

@Hema here

This is what you get if you start at the same point and progressively reduce the initial velocity.

@JohnRennie got it thank you :)

hello @JohnRennie

@user8718165 hi

@JohnRennie Morning!

@Dante morning :-)

I didn't even understand what the question is trying to ask.

brb

@Dante it's certainly a weird question. As far as I can tell the situation is this:

So the plank is accelerating in the -ve y direction, and because it is frictionless the disk stays where it is. In the rest frame of the plank the disk would be accelerating in the +ve y direction at $a$.

So presumably at time $t$ the distance to the disk is $s = \tfrac{1}{2}at^2$ and the velocity of the disk is $v_y = at$. Then you need to calculate the instantaneous centre of revolution.

back

@JohnRennie I'll have to revise that concept, will get back to this question later

@JohnRennie Take a look at this.

@Dante yes?

Maximum elongation is when $V_a=V_b$ right?

Okay, it was a calculation mistake, I got the first part, on the second part now (not in the image), will ping you if any difficulty.

OK :-)

@JohnRennie How did they find energy stored in spring in terms of K?

I got till the part that at max elongation, $V_a=V_b=\dfrac{4V_o}{3}$

I get energy stored in spring as $\dfrac{4m(V_o)^2}{3}$

using energy conservation.

@Dante Momentum and energy must be conserved. From momentum conservation you can work out what the velocity is at maximum elongation, i.e. when $v_a = v_b$, then calculate the KE. The PE is then the total energy minus the KE.

Yeah..I did that. But I have no idea how to express the potential energy in terms on 'K' like they've given in the rightmost column.

@JohnRennie Hello?

@Dante initial velocity of B is $2v_0$ so the momentum is $4mv_0$

Yes

When $v_a = v_b = v$ the momentum is $3mv$ so $v = \tfrac{4}{3}v_0$

ye

Initial energy = $\tfrac{1}{2}mv^2 = \tfrac{1}{2} (2m) (2v_0)^2 = 4mv_0^2$

When $v = \tfrac{4}{3}v_0$ KE = $\tfrac{1}{2} (3m) (\tfrac{4}{3}v_0)^2$

KE = $\tfrac{8}{3}mv_0^2$

yes

So the PE is $\tfrac{4}{3}mv_0^2$

yeah

ISWYM there's no obvious way to relate this to the spring constant

I see.

Presumably we have to calculate the extension of the spring ...

Oh, ok

I get you

Let me check

Well, yes it looks like extension, but $\dfrac{k}{m}$ should have been $\dfrac{m}{k}$ there

@JohnRennie good morning

For how much time are you still free

@Scáthach hi

@Scáthach about another hour.

Okay.

@JohnRennie

@Dante hi

When does A acquire max velocity?

Second part of the question.

@Dante if we work in the centre of mass frame then in this frame the initial velocity of A is $-\tfrac{4}{3}v_0$ and the initial velocity of B is $\tfrac{2}{3}v_0$.

The COM frame is moving at $\tfrac{4}{3}v_0$ relative to the lab frame.

Half an oscillation later the velocities are reversed so $v_a = \tfrac{4}{3}v_0$. This is the maximum velocity of A in the COM frame because it's the point where the spring PE is zero and A is moving forward. If we add back the frame velocity we get $v_a = \tfrac{8}{3}v_0$.

Oh, cool. I had never used this kind of approach before!

Thanks!

@JohnRennie Hi

@Scáthach Want to ask?

@Dante hi

Guess no,

@JohnRennie How is reading of voltmeter $-2E$ after switch is opened?

I have no idea. Can the circuit be redrawn to simplify it?

Does this give any idea?

Are the wires joined in the middle or do they just cross there?

Joined.

OK so I can redraw it as:

Is that right?

Right

So with the switch closed the voltages are:

yes

The bottom left capacitor is fully charged and the top right capacitor has no charge.

right

Both inductors have the same current flowing through them so they have the same energy stored in them.

It is far from obvious to me what happens when the switch is opened ...

Hmm, even I'm not able to think what actually happens...

@AjayMishra Hi there. Any idea about this?

Oh....wait

@JohnRennie

Every inductor will generate potential difference of E around it right?

Tell me one thing, will potential at the 2 marked fragments be same after key is pulled out?

Ah, nvm, it's really confusing.

@Dante , Did you too find reasonable that one capacitor would be fully charged, and other wont?

Yes....anything wrong with that?

no.

Just after disconnection, the capacitor would work as battery, right, in the lower circuit?

Yes

And inductor would be try to sustain the current?

Yeah

wont it too work as a battery?

yes

Wont, there be two battery then, in series?

no.

Very confused about that..

If we treat those things as batteries, a same fragment is getting 2 values of potential when I'm trying to solve...

What you think, inductor care about current or potentials?

current?

Sorry for overanthropromization, but how lower inductor will realize that there is not battery anymore, as there is a capacitor, which will sustain the current[Just after the disconnection].

Yes

what yes? I asked a question.

Oops, I meant, yes, I think the lower inductor won't generate any emf, since current will be continued by the charged capacitor.

and what about upper inductor?

what you think?

It will starting generating emf since current through it will start decaying

Emf of?

3iR?

iR Ig

?

iR, I guess.

Are you treating upper capacitor as open wire or simple wire?

Short circuit, since charge on it is zero.

I'm not sure if it should be iR, I just guessed it. Not able to figure out the reason.

Option 3 is right.

Ah,

Wasn't able to figure out if capacitor's plates will be at same potential

Hence I asked here

Thanks a lot dude! Forgot to treat it as short circuit

?

I said I got it. I had forgotten the fact that uncharged capacitor can be treated as a closed wire just when charge begins to build up.

So, you had treated the upper inductor and the lower capacitor as a battery?

Yes.

Now, I see, it is a pretty simple problem.

If the junction (middle part) was at potential x, the top point will be at potential x-E because of the top inductor and the bottom point will be at x+E because of bottom capacitor.

Is that the right way to look at it?

ya,

current through if the current through top inductor would be $i$ toward the junction, then the same current $i$ would emanate from the bottom capacitor toward the junction.

Umm, why that? shouldn't current through inductor be 1/3 rd of it?

Current was initially $i = \frac{E}{R} $ right?

through the inductor

right

Through the both inductors, I guess, then As the lower inductor and capacitor is in the same circuit, the current through lower inductor would be and through the lower capacitor would be same, right?

And for the upper inductor, initially current was $i$, which is what would be after disconnection.

I messed it up, I guess.

I didn't get it....since same potential drop 'E' should be there in both loops, and the resistance value in the upper loop is thrice, I thought current through it must be 1/3 rd.