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04:34
Good morning @JohnRennie sir, can you please help in this question?
Hi :-)
Hello sir :)
The way I would look at this is to consider the position of the fringes as a function of angle. That is, suppose we surround the slits by a cylindrical screen rather than a flat screen, then how are the fringes distributed on the cylindrical screen?
The reason for doing it this way is that on the bottom half of the diagram the screen subtends an angle of 90°, so it's easy to count the number of fringes.
Sorry sir, not getting your point
At A the path difference is 5λ, and at O the path difference is zero, and we get a fringe whenever the path difference is an integral number of wavelengths, so including the central fringe, from A to O there will be 6 fringes Yes?
04:52
@PCMSE This diagram shows what I mean.
We get a bright fringe when nλ = d sinθ, and if we look at the range of angles θ can have from A to the maximum angle that hits the screen it is from θ = -90° to θ = +37°.
i.e. sin θ ranges from -1 to +0.6.
OK so far?
(They chose +37° because sin(37) ≈ 0.6)
@JohnRennie yes sir
Now d = 5λ so d sinθ ranges from -5λ to +3λ. Yes?
@JohnRennie sir but then how did the cylindrical surface help us?
@JohnRennie yes
@PCMSE I used it because it made it obvious to me that we need only consider the angle i.e. the distance to the screen doesn't matter. But possibly this is the just the weird way my brain works :-)
All that matters is the angle that the screen subtends.
Yes sir :-)
05:01
Tilting the screen makes it look as if it's going to be a hard calculation, but in fact it isn't as the angle the screen subtends is very simple to find.
Yes sir, so 8 minimas
Yes, 9 bright fringes with 8 dark fringes between them.
Yes sir done, thank you very much :-) having 2 more questions sir, shall I ask?
Yes, go ahead!
The horizontal and vertical thrust
05:10
Do you want to do the bead one first?
Yes sir
It's an interesting question. I don't think I've seen a question like this before. The point is that the elastic cord is stretched by the weight of the bead, so the velocity of the bead relative to the cord is less than v.
Threaded means somewhat like a necklace?
Yes. The bead has a hole through it and the elastic cord passes through the hole. There is friction between the cord and the bead in this hole so the bead slides at constant velocity.
The frictional force is therefore equal to the weight of the bead.
I assume we can neglect the short period the bead takes to accelerate to the speed v, so we just assume the bead starts moving at speed v at the top of the cord.
Note that v is the speed relative to the ceiling not relative to the cord.
Sir also can we neglect the speed of the cord and say it is still v in ground frame as well as cord's?
05:21
No, you will need to take into account the fact the cord stretches as the bead moves down it. This stretching of the cord is what makes the problem hard.
One thing I'm not sure about is exactly what the question means by the stiffness i.e. exactly how this is defined.
Have you been taught that stiffness has a precise definition?
Maybe like a very light spring
I would guess it's like a spring constant i.e. for the whole cord the force when we stretch the cord a distance x is given by F = kx.
@JohnRennie No sir
@JohnRennie yes
This complicates things a bit because when defined in this way the spring constant changes when we change the natural length of the spring.
e.g. if we halve the length of the spring the spring constant doubles.
OK, shall we dive in and attempt the calculations?
Yes sir, let's try :-)
05:31
We are told the bead is a distance x from the ceiling. Since the cord is being stretched by the bead the distance the bead has travelled down the cord is y, where y < x.
So we can consider this as a length y of the cord that has been stretched by the weight of the bead to a length x, i.e. the extension of the cord is x-y.
OK so far?
Just a minute sir
@JohnRennie yes sir
Now, if we take a length y of the cord then its spring constant is k𝓁/y i.e. if we apply a force F to the cord the force is related to the extension by F = k𝓁/y Δy
Is this step OK? i.e. you understand that the spring constant depends on the length of the cord?
The symbols in equations are not appearing properly
05:46
It should look like this
Are you using a phone or PC?
Tablet sir
That L isn't appearing
OK I'll use "L" for the length of the cord:
Yes sir, switched to pc now
Does it look OK on the PC?
Yes sir, now it's perfect
05:49
OK :-)
11 mins ago, by John Rennie
Now, if we take a length y of the cord then its spring constant is k𝓁/y i.e. if we apply a force F to the cord the force is related to the extension by F = k𝓁/y Δy
@JohnRennie Yes sir, clear
Now, in this case the force F is the weight of the bead, mg, and the extension is x - y. So we can substitute these to get:
mg = k𝓁/y (x-y)
Yes
Hmm, that's going to give us a quadratic for y
No, ignore me, it's not a quadratic
Yes sir, $$y=\frac{klx}{mg+kl}$$
Sir $l$ is the natural length?
05:54
Yes, and this looks reassuringly similar to the equations in the options so I think we are on the right track.
Yes $l$ is the natural length of the whole cord.
Given our result let's start looking at the options.
Option A is a bit bizarre. If the speed relative to the ceiling is v and and the distance relative to ceiling is x then the time taken is just t = x/v. Yes?
Yes
Since everything is relative to the ceiling there is no complexity here.
Option C, we already obtained
Not quite. The elongation is x - y, and we found y.
But obviously x - y is now easy to find.
@JohnRennie Ah!! Sorry.
06:01
$$ x - y = x - \frac{klx}{mg+kl} = \frac{mgx + klx}{mg+kl} - \frac{klx}{mg+kl} $$
So C is ... ?
Yes, we got C
Back to B?
Yes sir
If we stretch a spring by a distance Δ𝓁 then the PE is ¹⁄₂k(Δ𝓁)²
This equation applies here, but:
1. that constant k has to be replaced by k𝓁/y
2. the extension is the distance given by option (C)
OK so far?
Yes sir
06:12
The algebra is a little messy here but it is straightforward.
We know $y = \frac{kl}{mg + kl} x$
Factor of 2 missing
Yes.
So B is false, but the questioner made it very close to the correct answer just to try and confuse us :-)
Now, D is simpler than it looks though again the algebra is a little messy.
Power is force times velocity, where here the force is the frictional force between the bead and the cord, and we know the frictional force is mg. OK so far?
Yes
The speed is the speed of the bead relative to the cord, so it's d(x-y)/dt. Yes?
Yes
06:19
Hmm, wait, is that correct?
x-y=mgy/kl
Ah, no, I'm wrong.
The distance the bead has travelled down the cord is just y, so the speed relative to the cord is dy/dt.
Yes?
@JohnRennie just a minute sir
Understanding this again
Yes sir understood
We got power dissipated due to friction
Yes :-)
At least I assume by thermal power dissipated the question means heat produced by friction.
So what matters is how far the bead moves over the cord.
because the work done by friction is the frictional force times the distance moved over the cord.
06:34
Sir what for dx/dt?
dx/dt = v
It's very confusing sir but clear now. So $$P=\frac{m^2g^2v}{mg+kl}$$
Sir a problem again
They have given C, D correct but D??
I got $$ P = \frac{mgk\ell}{mg+kl} v $$
@JohnRennie Oh god, yes sir, sorry
@PCMSE I kind of suspected (D) was meant to be true, in which case they are using a different definition of thermal power from us.
But it's not obvious to me what definition they are using ...
I wonder what the sum of the frictional losses and the PE stored in the cord is. Would that be similar to (D)?
Aha! Yes!
That's what D is.
06:46
Sir the power is dissipated in cord?
If $U = \tfrac12 \frac{m^2g^2x}{mg + k\ell}$ then $\frac{dU}{dt} = \tfrac12 \frac{m^2g^2}{mg + k\ell} v$
@JohnRennie Yes sir
@JohnRennie Yes
Add this to the power we calculated and you get D
Sir the power in cord is dissipated or stored?
That's a good question, and I would say it's stored not dissipated.
I think this is an example where the question is misleadingly written.
As the question is written I would say our answer is correct.
06:49
@JohnRennie yes sir, probably because it's getting stretched
@JohnRennie Means only C?
I would say only C is correct.
Considering stored, the change is just mg-2kl
Well we worked out earlier, when we were checking B, that: $$ U = \tfrac12 \frac{m^2g^2x}{mg + k\ell} $$ Yes?
Yes
And differentiating this gives the power going into stored elastic energy.
06:54
Yes
$$ P_{stored} = \tfrac12 \frac{m^2g^2}{mg + k\ell} v $$
yes sir, probably a question error
Yes, I think so.
It's a fun question though. Surprisingly tricky.
Sir can we take a break of 30 mins. I'll quickly have my lunch, really exhausted after this question?
@JohnRennie Yes sir :-)
@PCMSE OK :-) Ping me when you want to pick this up.
06:58
@JohnRennie Okay sir, thank you very much :-)
07:31
Sir I am done, sorry to keep you waiting
@JohnRennie
Hi :-)
This question?
Yes sir
They are asking for horizontal and vertical thrust
Again it's not a question I've seen before, but my immediate reaction is that if we consider a height x above the base we can find the angle of the curve at that height from the derivative of the curve.
Pressure always exerts a force normal to the surface, so we can find the angle of the force and therefore it's horizontal and vertical components.
Though now I think about it there is an easier way to do this ...
In fact a very easy way
I'll draw a diagram
On the right of the dashed line we have the parabolic wall, and on the left we have the rest of the water above the flat base.
OK so far?
07:47
Yes sir
Consider the horizontal forces. The water to the left of the dashed line exerts a horizontal force on it that we can easily find by integrating the pressure ρgh. This is the force I've labelled F. OK so far?
Yes clear
Are my messages getting delayed @JohnRennie?
@PCMSE Yes
I am messaging instantly but it shows 3 mins gap
Just a minute sir
Your last two messages came through immediately, but the one before that took several minutes.
07:53
Rejoined now sir
OK :)
Since the water isn't moving there must be an equal and opposite force to the left from the water in the curved part of the dam. Yes?
Yes
I think rhogh/2*A
And that means there must be the same horizontal force on the curved part of the wall. This way the net horizontal force is zero, which it has to be since the water isn't accelerating.
Yes
So the horizontal force on the curved part is easy and doesn't require any difficult calculation.
The vertical part of the force is also easy because it has to be equal to the weight of water in the curved part of the dam. Yes?
07:59
Yes, I think $F_{horizontal } =\rho gh A/2$
@JohnRennie yes
So can you take it from here?
The horizontal force is correct sir?
@JohnRennie To find weight of water in the curved part do we need to first complete rectangle and then find area through integration and then use unitary method?
@Lllt Just integrate to find the area under the curve and subtract it from the rectangle.
@JohnRennie Yes
08:03
Or I guess you could rotate the curve 90° and integrate y = √18x instead
You'd get the same answer either way.
Ok
@PCMSE No,
@PCMSE This diagram:
Suppose we consider a width W of the wall, then the area of the wall between the horizontal dashed lines is W dh
So the force is dF = PWdh = ρghW dh
Integrate this and we get F = ¹⁄₂ρgWh²
Yes?
Yes got it sir
Done for today sir, thank you very much for your time :-)
Bye sir @JohnRennie
 
4 hours later…
11:45
Suppose a motorcycle starts at $v_i$ and then skids over a distance of$ d$, ending the skid at $v_f$. Find ${v_f}^2 - {v_i}^2 $ in terms of the distance $d$ and the deceleration a.
Answer assumed that acceleration is constant. I don't see that acceleration is constant is mentioned in the question. Does to skid mean that acceleration is constant?
 
1 hour later…
13:03
"I don't see that acceleration is constant is mentioned in the question."

Why? As the motorcycle is deaccelerating, a force is acting (that is kinetic friction.) and kinetic friction is constant as the object's mass is not changing.
 
4 hours later…
16:45
@BannedUser the only force is kinetic friction f, which is constant, so a=f/m is also constant

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