Quarlep
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Please haruspex
The discussion focuses on calculating the average velocity of a spherical shell that expands and moves, specifically addressing the velocities involved: expansion velocity (v) and movement velocity (v'). Participants clarify that while expansion velocities can cancel out in certain contexts, the average velocity remains dependent on the movement velocity v'. The kinetic energy of the shell is also discussed, emphasizing that energy is a scalar quantity and does not simply cancel out like vector quantities. The conversation highlights the need for proper mathematical treatment, including the use of double integrals in spherical coordinates for accurate calculations.
PREREQUISITESStudents studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for examples of complex motion and energy calculations involving spherical objects.
I don't understand your question. What is it you want me to do? Seemed to me you had your answer.Quarlep said:Why don't you guys just not writing the equation
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 said:Thats answer describes ring shell I want sphere shell.
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.Quarlep said:This must be the true answer.It may be complicated but it must be true.
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 said:I find a stupid answer m((v')^2+r^2+2( π)^2
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 said:Ok yeah yeah you are right.Is that all thing somy equatipn is true If I change r to v ?
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 said:ρr2((v')2+v^2+2π^2)+ integral sinθ from 0 to π
We're in different timezones, I'm sure. But I did reply to your post #72. Can you not see my reply?Quarlep said:Are you there my friend
haruspex said: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.
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.Quarlep said: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 ?
That equation makes no sense syntactically. It should readQuarlep said: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##
Wrong in my version? I don't see anything - please be more specific.Quarlep said:Can you check the integral again I think there's a wrong something
It's a double integral. Have you not dealt with double integrals before?Quarlep said:there's two integral first one belongs theta second phi but then end you wrote d theta phi
It's wrong. Please post all your steps.Quarlep said:Did you see my answer ?.Or its wrong