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Centre of mass

  • Thread starter vintwc
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Homework Statement


Let Ω be a tank whose shape is that of the lower hemisphere of radius R. The tank with a muddy suspension whose density ρ is ρ(x,y,z):=e^-h(x,y,z), where h(x,y,z) is the height of (x,y,z) above the lowest point of the tank. Find the center of mass in the tank


Homework Equations





The Attempt at a Solution


First of all, how does one determine the height, h(x,y,z)? I guess it would be R but I am not able to give a reasoning to my guess. I would appreciate if someone could give me a graphical illustration on how to find the limits of integration for this problem as well. Thanks
 
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Answers and Replies

  • #2
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Alright following the my notes so far which had a sort of similar but different question, I guess that the height is z+R? Assuming if this is right, the limits of integration will be [0,R]x[-Pi/2, 0]x[0,2*Pi] (since we are looking at the lower hemisphere).
However, if I try to calculate the first moments M_yz, I got the integrand as 0 at some point. Maybe this indicates that I am moving in the wrong direction?
 
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  • #3
HallsofIvy
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If you set this up so the hemisphere is the lower half of a sphere of radius R with center at (0,0,0), then the lowest point is (0, 0, R) and the "height above the lowest point" of (x,y,z) is R+ z (z is negative, of course).

[tex]e^{-h}= e^{-R-z}= e^{-R}e^{-z}[/tex].

The equation of the hemisphere is [tex]z= \sqrt{R^2- x^2- y^2}[/tex].

Squaring both sides of that gives [itex]x^2- y^2= z^2- R^2[/itex] which is the equation of a cross section of the hemisphere at that height. Since the density is a function of z, that will be helpful in integrating. Since that is a circle for all z, I would recommend doing the integration in cylindrical coordinates.
 
  • #4
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If you set this up so the hemisphere is the lower half of a sphere of radius R with center at (0,0,0), then the lowest point is (0, 0, R) and the "height above the lowest point" of (x,y,z) is R+ z (z is negative, of course).

[tex]e^{-h}= e^{-R-z}= e^{-R}e^{-z}[/tex].

The equation of the hemisphere is [tex]z= \sqrt{R^2- x^2- y^2}[/tex].

Squaring both sides of that gives [itex]x^2- y^2= z^2- R^2[/itex] which is the equation of a cross section of the hemisphere at that height. Since the density is a function of z, that will be helpful in integrating. Since that is a circle for all z, I would recommend doing the integration in cylindrical coordinates.
Ah, too late for correction. But I had the same approach as yours. I was considering cylindrical coordinates but I was having trouble with finding the limits (spherical one as well).
 

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