@JohnRennie Sir are you there?

@Knight I'm busy on Monday mornings I'm afraid. I'll probably be finished in a couple of hours.

@JohnRennie Okay sir :)

@JohnRennie Hello sir :-)

Are you available now

@Jasmine @Knight hi :-) I've finished work now.

Hey @JohnRennie I need help!

You logged on just in time :)

Its related to elliptical orbits, and the semi major axis..

The semi-major distance is half the long axis of the ellipse, but in an orbit the Sun is not at the centre of the ellipse. It's at one of the focus points.

OOOOOH

The long axis of the ellipse is $R+r$, so the semi-major distance is $(R+r)/2$.

dang

thanks lol

:-)

@JohnRennie Hello :-)

@JohnRennie I wanted to discuss with you the process of firing electrons on metal plates and the consequent ejection of X-rays

@Jasmine OK. As in an X-ray tube presumably?

@JohnRennie Yes

I dont understand what exactly is happening

I have a question as well

Should I share it first

Yes, let's have a look at the question.

An X-Ray tube is operated at 150Kv at 10mA . if only 1% of electric power supplied is converted into X-Ray ,the rate at which the target is hitted is (in Cal/s

I dont understand exactly is done in the experiment

OK. What happens is that high energy electrons are fired at the metal target and they collide with electrons in the metal atoms and eject them. So you are left with metal ions with a hole where their electron used to be. OK so far?

@JohnRennie okay

Any electron can be ejected, but what is of interest here is when an electron is ejected from the 1s orbital. For historical reasons the 1s is known as a the K shell, so you'll often see these described a K electrons.

When this happens an electron drops down from the 2s orbital into the 1s orbital, and the energy difference is emitted as a photon.

So the end result is that we get a photon emitted with an energy equal to the 2s -> 1s transition.

Okay

So in your question 1% of the electrons eject a K electron and cause an X-ray photon to be emitted.

What exactly is happening here

@JohnRennie or can be an L electron

@Jasmine yes, although the L peak is usually lower amplitude and often absent. But I don't think it matters for this question.

@JohnRennie yes

We are told the tube is operated at 10mA, and I assume this means the beam current is 10mA. That means 0.01 coulombs per second hit the metal. Divide 0.01 coulombs by the electron charge and you have the number of electrons per second hitting the target.

I make this $6.24 \times 10^{16}$ electrons per second.

So is it like only 1% electrons are ejecting X-rays and what about 99% are they completing the circuit

So is it like the current due to 99% electrons = 10mA

The other 99% either go straight through the metal target without hitting anything, or they lose their energy in mutliple low energy collisions that don't emit an X-ray.

@JohnRennie Ok

Anyhow, if only 1% of electrons produce an X-ray that's $6.24 \times 10^{14}$ photons per second.

Then 10mA is due to flow of which elctrons

A current is the flow of electrons. Normally we have electrons flowing in a wire, but if you have electrons travelling through a vacuum then that is a current as well.

In an X-ray tube electrons are emitted at one end then accelerated towards the metal target, where their collisions with the metal emit X-rays. So there is a current flowing along the tube due to the electrons travelling from the end of the tube along the tube to the metal.

@JohnRennie Can you draw a diagram please

Let me see if I can Google one ...

From this page

The electrons are emitted at the cathode and they travel through the vacuum inside the X-ray tube and hit the metal target (the tungsten target in this case).

@JohnRennie where is the external battery and current

@JohnRennie okay so the current is due to the fired electrons going to collide

And the battery is to accelerate the electrons

The external circuit looks like this. The battery is very high voltage. 150kV in this case.

The electrons flow in the direction shown by the blue arrows.

The current shown by the blue arrows is the 10mA current.

@JohnRennie but isnt 1% wasted as X-ray

@Jasmine no, the electrons don't disappear if they collide and cause a photon to be emitted. They just lose energy in the collision and then carry on round the circuit.

@JohnRennie ohhhhh ! Ok

But I read on collision, the photons transfer all their energy to electron in photoelectric effect

And electrons are like photons

Yes, so the electron loses its kinetic energy and stops. But the 150kV battery is driving the electrons round the circuit, so that electron carries on round the circuit.

@JohnRennie Okay

Photons can just disappear, but electrons can't.

@JohnRennie Ok!

Technically we say that photon number is not conserved. The reasons why are a bit onvolved.

@JohnRennie Ok

Electrons carry an electric charge (of $-e$) so if they just disappeared charge wouldn't be conserved. Photons have no charge so it doesn't cause any problems with charge conservation if they appear and disappear.

@JohnRennie Okay

Are you happy you understand the question now?

@JohnRennie yes :-)

OK, I need to work for a few minutes but I'll be back soon if you want to ask anything further.

@JohnRennie yes I had more to say

Please ping me when you are free

The answer given is 355cal/s

I got the answer no worries :-)

I had another question as well

@Jasmine hi, I'm back.

@JohnRennie hello, just wanted to know if I can solve the above question using gauss law ^

I can't see a way to do that using Gauss's law. As a general rule Gauss's law is most useful when we have some symmetry we can exploit to get the answer, but I can't see that here.

I think you have to do the double integral.

It's a semi-infinite wire. Will it help if we assume another semi-infinite wire giving a symmetric condition sir?

@JohnRennie can we use Gauss law for the infinite line charge

Since its semi infinite

For an infinite line charge the field lines are radial so the flux through the surface would be zero.

That just tells you that if you split the line charge into two semi-infinite parts then they contribute equal and opposite fluxes through the surface.

@JohnRennie Okay

Then we can simply skip this one

It's an annoying calculation, but not an especially difficult one.

@JohnRennie I would like to try that later

I would consider an annulus of radius $r$ and width $dr$ and integrate along the wire to calculate the flux through the annulus.

I had one more question, I just wanted to seek your idea on that

OK ... ?

I just need a hint on this

I tried it and was getting random wrong answers

Shouldn't the rod remain horizontal during free fall?

The string is attached to the midpoint of the rod, so the rod will be pivoting about it's midpoint.

Sorry, I thought we're cutting the string at t=0.

You just need to calculate the torque and the moment of inertia, the the angular acceleration is $\alpha = \tau/I$.

@JohnRennie about which point though.. alpha for all points should be constant

My interpretation of the question is that the rod is being held at the end to keep it horizontal. The word release just means letting go of the end so the rod starts rotating, not cutting the string.

@Jasmine the string is attached to the midpoint of the rod, so initially the mid point remains stationary i.e. the rod rotates about the mid point.

So the torque is $\tau = mg\ell/2$

@Jasmine: Sorry to interrupt, is the answer A; If so I could try to help after John Rennie sir.

@JohnRennie Oh!

@GuruVishnu yes

A C

And the moment of inertia is about the mid point so $I = \tfrac{1}{12}M\ell^2 + m(\ell/2)^2$

@JohnRennie can we calculate torque and moi about any point as alpha for the rod is going to be constant

@Jasmine My advice; Calculate torque about the suspension point to not include the tension's torque; it'll cause a lot of tension :)

And also to eliminate torque on the metal rod due to gravity.

@Jasmine why would you do the calculation for any point other than where the string is attached? It would just make the calculation harder.

@JohnRennie Okay,, btw I was messing up with centre of mass

@GuruVishnu yes

You do need to use the centre of mass for the second part (C and D).

Yes I see , I got it

Once you know $\alpha$ you calculate the linear acceleration of the COM due to the rotation.

@JohnRennie yes :-)

Then the linear acceleration of the COM is $g - T/(M+m)$

Yea I got it :-) Thank you @JohnRennie

:-)

@JohnRennie: Hi sir. Are you free now?

@GuruVishnu hi :-) Yes I'm free.

@JohnRennie oh no ! i got a wrong answer to the X-Ray question.. The answer is 355 cal/s I am getting 3.55 cal/s

How can it be !!

@Jasmine what are Cal/sec? Calories per second?

@JohnRennie yes

Hmm, my answer is also 3.55 like yours.

I think the book is wrong.

Probably we are misinterpreting the question

@JohnRennie Probably

I think in the question , target is the metal surface

Means 100% electrons hit the metal surface so the answer comes 355calorie/s

The question says 1% of the power is emitted as X-rays.

@JohnRennie but as you said all the electrons are going in the circuit , 1% just knock out the electron

The power supplied is $P=IV$ where $I=0.01A$ and $V = 150000V$ so the power supplied is $1500W$. 1% of that is $15W$, which is 3.55 cal/sec.

Means the metal surface is being hit by all electrons but only a percent is knocking out electrons

Can you post a picture of the question so I can see exactly what it says?

@JohnRennie This was given by our teacher in class not sure where he took this from..

But is my interpretation wrong because if I do it like this I get the correct answer

@JohnRennie I found ..

@JohnRennie google.com/amp/s/amp.doubtnut.com/question-answer-physics/…

Well the electrical power being supplied is 1500W and that is 355 cal/sec.

I remember my sir had also told the same solution

Ahh, it says the rate at which the target is heated.

That's the power being supplied minus the power lost as X-rays.

@JohnRennie I am sorry Sir

:-)

Lol I had been wasting so much time on this

So it's $(1500 - 15)/4.184$

@JohnRennie yes I get it now

I cant believe this lol this is so silly

Hello, I was wondering if I can get help on an exercise that I am developing?

@Eliot Go ahead.

Its the same exercise that I posted about on physics stack exchange here: physics.stackexchange.com/questions/531521/…

I'm just not sure how to go about setting up the equations for force since it is a free body diagram, and I'm not sure how to solve for beta in terms of theta

@Eliot The first question about dimensions of $c$. No you are adding $r^2$ to $r$ which is not possible unless $r$ is dimensionless.

so would c be dimensionless? How would you get units of kg/m^3 then?

@Eliot The formula is already dimensionally inconsistent regarding $r^2+r$. You cannont make it dimensionally correct by suitable choice of dimensions for $c$.

But that is the density function of the cylinder. Is it not possible to have that as the density function?

PS see the following to use Chatjax in Chatrooms : chemistry.meta.stackexchange.com/questions/89/…

I'm creating this problem for students. It is not a homework problem

@Eliot You could include a factor $a$ in front of $r$ so that $ar$ has the same dimensions as $r^2$.

Then you can add them.

and a would be an arbitrary constant correct?

Yes.

Ok. Should I redo the problem to find the moment of inertia and the center of mass before I come to discuss with you again the rest of the problem?

*with the factor $a$ included

The work done by force $F$ is simply the constant force $F$ times the distance moved by the COM of the small cylinder in the direction of the force.

But you would have to do an integral in order to find that work, since the cylinder travels on a curved surface

I don't understand the purpose of your problem. If you want to find the work done by $F$ between 2 positions of the cylinder you don't need the moment of inertia.

No. If the force $F$ is constant then the work done by that force is simple to calculate.

I'm trying to find an expression for $F$ in terms of other constants, which is why I have the force equations and the torque equation

Why do you need to apply a force? The cylinder will roll down because of gravity.

Without a force, the cylinder would roll down because of the static friction force. Gravity would not apply a torque on the cylinder initially.

And your right, if $F$ is constant then you don't need an integral.

However, I'm trying to show students when the integral should be applied, so $F$ should be a varying force on the curved surface of the hemisphere

Are you teaching physics or mathematics? You are making the problem unnecessarily complicated.

If you want to use an integral why not have the cylinder rolling on a flat surface?

You have a point. I probably shouldn't give this to students, but I'm curious as how to solve this on my own

Before solving it you need to define it precisely. If you are trying to find the work done by $F$ between 2 positions of the cylinder then you need to define $F$ as a function of $t$ or arclength $s$ or angle $\theta$. Whereas if you want to find how $F$ should vary with $t$ or $s$ or $\theta$ then you need to know how much work it has done.

Ok I see. I thought I could just solve for $F$, just like the traditional horizontal force applied on a person on a swing problem given to university students as an exercise.

@Eliot There are already many similar problems online with worked solutions. They have been tested. It is better to use one of them rather than invent your own problem.

Ok but if I continue with my own problem, does it still make sense to say that the horizontal force can only be applied until $$\arccos(\frac{R1−R2}{R1+R2})$$? A horizontal force wouldn't be able to be applied anymore since the hemisphere would get in the way, I think.

@Eliot Forces can be applied even when objects are in the way. For example a push can become a pull, or the force could be gravitational or electrical so that it works through objects.

But if the problem specifies that a human was pushing on it, that would be the limit at which a human can apply a horizontal force, correct?

@sammygerbil

@Eliot Not necessarily. In physics classes you can state that a force is applied at a particular point in a particular direction. You don't need to state how the force is applied, or explain how it gets round objects.

@sammygerbil But if the force were to be initially applied such that it points towards to geometric center, the wouldn't that mean that after a time $t$ it no longer points towards the geometric center while it is rolling?

@Eliot It is your problem. You can define the force in any way that suits you. You can define the direction in which the force to point in any direction you choose as well as the point at which it acts. You can make the conditions change every second if that's what you want.

So I am allowed to define the force as always acting on the geometric center?

@sammygerbil

Yes. And always acting horizontally.

@sammygerbil Ok then that simplifies the problem greatly

Sorry I was just trying to make the problem as realistic as possible

But now I see that I don't have to

No you don't have to.

You can define the conditions to avoid complications.

Ok thank you so much for helping me develop this problem.

It was very helpful

ok