Find average velocity of a sphere which expands and moves

[haruspex]

I didnt understand.

Take an element on the shell in spherical polar co-ordinates. Write out its KE. If theta is the angle to the line of movement of the shell's mass centre, integrate in a band of width $rd\theta$ for $\phi$ from 0 to $2\pi$.

 

[Quarlep]

Its too complicated isn't it ?

 

[Quarlep]

I have no idea how did I get this equation but here it is

 

[haruspex]

I have no idea how did I get this equation but here it is

You can make it simpler by recognising that the KE only depends on theta, not phi.

 

[Quarlep]

You can make it simpler by recognising that the KE only depends on theta, not phi.

Ok, I ll make it simpler but my equation is full correct isn't it ? No mistake Even multiply integral with 2 ...

 

[haruspex]

Ok, I ll make it simpler but my equation is full correct isn't it ? No mistake Even multiply integral with 2 ...

Not really. Not sure what your $v_{\theta}$ and $v_{\phi}$ terms are. If they're vectors, they should all be inside the first squared term with v' and v_r. I.e. the overall velocity is the sum of four vectors, v' and three velocities relative to v'. But then $v_{\theta}$ and $v_{\phi}$ would both be zero. If they're not vectors, maybe you intend them as the scalar magnitudes of those vectors, in which case the same comment applies.
So in your equation, throw those two away and expand the remaining squared term. Two of the resulting terms will be independent of theta and phi. The third one, the dot product of v' and v_r, will be a function of theta. The integral is then easy.

 

[mfb]

Spherical coordinates make this way more complicated.

There is a nice theorem about the kinetic energy of a system if you know the kinetic energy in its center of mass system and the velocity of this center of mass. Both are easy to find here.
If you don't want to use this theorem, you can split the sphere into two parts and derive a special case of the theorem for this sphere.

 

[Quarlep]

How I need help.My math is not good enough

 

[haruspex]

There is a nice theorem about the kinetic energy of a system if you know the kinetic energy in its center of mass system and the velocity of this center of mass. Both are easy to find here.
If you don't want to use this theorem, you can split the sphere into two parts and derive a special case of the theorem for this sphere.

Sure, but the challenge is to resolve the apparent contradiction that Quarlep came up with by using integration methods.
I gave Quarlep an easy way using symmetry, but he/she seems to want to do it using a more general approach. Certainly it looks like there is merit in Quarlep having a work-out in integration.

 

[Quarlep]

I tried to to use symmetry but again I don't know how to do it.I am worling on

 

[haruspex]

I tried to to use symmetry but again I don't know how to do it.I am worling on

As I said, just add the KEs of two diametrically opposite points in the shell.

 

[Quarlep]

Yeah you know that I used it find KE of ring shell

 

[haruspex]

Yeah you know that I used it find KE of ring shell

Do you mean in post #31? I didn't realize that's what you had attempted to do there. If it is, you didn't do it right. The two cos terms should have opposite signs and cancel.

 

[Quarlep]

Theres one cos isn't it.Or I couldn't see

 

[haruspex]

Theres one cos isn't it.Or I couldn't see

The sign of the cos term for one point will be opposite to that on the diametrically opposite point.
You should have $(v'+v_r\cos(\theta))^2+(v'-v_r\cos(\theta))^2$, plus the y direction terms.

 

[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 ?