Find average velocity of a sphere which expands and moves

[Quarlep]

Is this true

 

[haruspex]

Is this true

Almost. The m/2 should be a factor of the whole, and the factor 2 you have in front only applies to the sin term.

 

[Quarlep]

Ok,now

 

[Quarlep]

You told me smthing in post 41 but I don't understand it.Thats my fault.I am in high school and my physics is not enough.Thank you very much for support.

 

[Quarlep]

Here again

 

[haruspex]

Ok,now

Yes, that's good, but you don't need to do an integral at all this way. When you expand the terms, all references to theta should disappear, so all the terms are constants.

 

[Quarlep]

Ok,I did it and I found what I found before.
In post 31 I guess m(v^2+v'^2)

 

[haruspex]

Ok,I did it and I found what I found before.

You mean, you got the same result as integrating around a circle? Good.

 

[Quarlep]

Yeah

 

[Quarlep]

Why don't you guys just not writing the equation

 

[Quarlep]

Please haruspex

 

[haruspex]

Why don't you guys just not writing the equation

I don't understand your question. What is it you want me to do? Seemed to me you had your answer.

 

[Quarlep]

Thats answer describes ring shell I want sphere shell.Whatever you told I can't find the solution cause I am still learning integral and I don't get any classic mechanic lesson (this is not a very hard question ( ring shell))but I don't know any idea about sphere coordinate system and the other things.You tried to tell me answer you want to help me but I am in high school and I am not learning so much detail.I got this project cause I thought I can do that.But I stucked here and I lost a week so I need full equation of sphere kinetic eneregy.I need to move on

 

[haruspex]

Thats answer describes ring shell I want sphere shell.

The equation you got by combining two diametrically opposite points solves both. The dependence on theta canceled out, making the integrand a constant for a given r. Integration is then just a matter of multiplying by the mass - you no longer care whether it's a spherical shell or a ring. All that matters is that it can be expressed as pairs of opposite points, all at the same radius.

 

[Quarlep]

This must be the true answer.It may be complicated but it must be true.

 

[haruspex]

This must be the true answer.It may be complicated but it must be true.

I believe that is the correct double integral for a spherical shell, and as before it will simplify greatly such that the trig terms will obviously vanish when integrated.
(But it isn't any more correct than the much simpler symmetry approach.)

 

[Quarlep]

I ll do the integration and I ll going to tell the answer

 

[Quarlep]

I find a stupid answer m((v')^2+r^2+2( π)^2

 

[haruspex]

I find a stupid answer m((v')^2+r^2+2( π)^2

I had trouble reading your attachment at post #65. In view of what you say now, I studied it again, and it looks like you have r where I would expect to see v.

 

[Quarlep]

Ok yeah yeah you are right.Is that all thing somy equatipn is true If I change r to v ?

 

[haruspex]

Ok yeah yeah you are right.Is that all thing somy equatipn is true If I change r to v ?

There are some more errors. The m at the front should be $\rho r^2$, where $\rho$ is the density. There should be a factor $\sin(\theta)$ inside the integral (or maybe it's $\sin(\theta)$, whichever goes from 0 to $\pi$). This comes from the polar expression for an area element.

 

[Quarlep]

ρr²((v')2+v^2+2π^2)+ integral sinθ from 0 to π

 

[haruspex]

ρr²((v')2+v^2+2π^2)+ integral sinθ from 0 to π

Better, but that 2π^2 term shouldn't be there. Previously, every term had a factor of either v or v', so that should remain true. Instead, I would expect to see a factor π throughout. Please post all your steps.

 

[Quarlep]

Are you there my friend

 

[haruspex]

Are you there my friend

We're in different timezones, I'm sure. But I did reply to your post #72. Can you not see my reply?

 

[Quarlep]

There are some more errors. The m at the front should be $rho r^2$, where $rho$ is the density. There should be a factor $\sin(\theta)$ inside the integral (or maybe it's $\sin(\theta)$, whichever goes from 0 to $\pi$). This comes from the polar expression for an area element.

Hi I see your post but I couldn't answer cause I am in holiday now and here time is 07:20 am. In this post you said $\sin(\theta)$ will be inside the integral but theta goes 0 to pi so here I am confused and I ll going to write pr^2 instead of m ?

 

[haruspex]

Hi I see your post but I couldn't answer cause I am in holiday now and here time is 07:20 am. In this post you said $\sin(\theta)$ will be inside the integral but theta goes 0 to pi so here I am confused and I ll going to write pr^2 instead of m ?

In your integrand you have trig functions of phi and theta. But you must have put the area element as $m d\theta d\phi$. The area element should have been $\rho r^2 \sin(\theta)d\theta\phi$. After simplifying the rest of the integrand, you still have one term left that has trig functions in it. You must multiply the sin from the element with this before integrating it.
This will be a lot easier to explain if you post all your steps as typed equations. Please stop posting images of working, they're too hard to read and too hard to make comments about.

 

[Quarlep]

http://www.HostMath.com/Show.aspx?I...hi)^2+\int_{0}^{\pi}pr^2\sin\thetad\theta\phi look

 

[Quarlep]

I made a mistake in there.Can you write the correct one and send me like this url then I can work on it

 

[haruspex]

http://www.HostMath.com/Show.aspx?IsAsc=True&Code=\int_{0}^{\pi}\int_{0}^{2\pi}(v\cos\theta\sin\phi+v')^2+(v\sin\theta\sin\phi)^2+(v\cos\phi)^2+\int_{0}^{\pi}pr^2\sin\thetad\theta\phi look
$\int_{0}^{\pi}\int_{0}^{2\pi}(v\cos\theta\sin\phi+v')^2+(v\sin\theta\sin\phi)^2+(v\cos\phi)^2+\int_{0}^{\pi}pr^2\sin\theta d\theta\phi$

That equation makes no sense syntactically. It should read
$\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}((v\cos\theta\sin\phi+v')^2+(v\sin\theta\sin\phi)^2+(v\cos\phi)^2)\rho r^2\sin\theta d\theta d\phi$

 

[Quarlep]

Yeah I noticed that pr^2 must be inside the integral I noticed it after I wrote the equation

 

[Quarlep]

Can you check the integral again I think there's a wrong something

 

[haruspex]

Can you check the integral again I think there's a wrong something

Wrong in my version? I don't see anything - please be more specific.

 

[Quarlep]

Theres two integral first one belongs theta second phi but then end you wrote d theta phi

 

[Quarlep]

I find 2pr²((v')²)+2πv²)

 

[haruspex]

Theres two integral first one belongs theta second phi but then end you wrote d theta phi

It's a double integral. Have you not dealt with double integrals before?

 

[Quarlep]

No

 

[Quarlep]

Did you see my answer ?.Or its wrong

 

[haruspex]

Did you see my answer ?.Or its wrong

It's wrong. Please post all your steps.

 

[Quarlep]

Can dtheta and dphi change sides

 

[haruspex]

Can dtheta and dphi change sides

Do you mean, can they be in either order? Yes. It's the order of the integral signs that defines the order of integration (inside first).

 

[Quarlep]

I am in holiday I am not working in computer so I can't write all steps to you right now.We will do first integral (the one which is inside) that's phi but there wrote dtheta so I asked you this

 

[Quarlep]

Can you tell me what was that please the only thing I ll do is put it on program

 

[haruspex]

Can you tell me what was that please the only thing I ll do is put it on program

Sorry, I don't understand what you are asking.

 

[Quarlep]

The answer :):)

 

[haruspex]

The answer :):)

Do you simply want the answer to the original question, by whatever means, or do you specifically want to see how to do it through an elaborate double integral?

 

[Quarlep]

I want to see answer to the original question

 

[Quarlep]

I am working on symbolab but I found nothing

 

[haruspex]

I want to see answer to the original question

Then use my symmetry method. It gets rid of the trig terms and makes the integral trivial.

 

[Quarlep]

I was banned so I couldn't answer over a 10 days I found 4p(pi)r^2((v'^2)+v^2)