Center of mass of hemisphere: substitution of z=rcos(θ)?

  • #1

Homework Statement



I am trying to understand a substitution used to solve for the center of mass of a solid uniform hemisphere as in this post: https://www.physicsforums.com/threa...in-spherical-coordinates.650663/#post-4151797[1]

I completely understand the math and the correct solution. However, what I don't understand is why the substitution z=rcos(theta) was necessary to obtain the result. If you do not include the cosine term, you find that R_cm = 3R/4, but if you do include it, you find that Z_cm = 3R/8. So we have R_cm = 2*Z_cm. It should be the case that R_cm = Z_cm, since both are defined from the same origin (the center of the hemisphere base), right? So they should be the same vector. I suspect my error is in this assumption, but I would like some help here.

The use of a symmetry should help simplify a problem, but it appears that in this case, it changes the answer entirely. If we are agnostic about the axis of symmetry in an integral, why wouldn't we still get the correct answer without using z=rcos(theta)?

Thanks.


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Answers and Replies

  • #2
SteamKing
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Homework Statement



I am trying to understand a substitution used to solve for the center of mass of a solid uniform hemisphere as in this post: https://www.physicsforums.com/threa...in-spherical-coordinates.650663/#post-4151797[1]

I completely understand the math and the correct solution. However, what I don't understand is why the substitution z=rcos(theta) was necessary to obtain the result.
Well, the OP in that other thread was using spherical coordinates in his integration to find Cm, so any z variables must be converted to their equivalents in the spherical coordinate system in order to carry out the integration.

It's like trying to find ∫ y dx. Things go a lot smoother if you know what an equivalent expression for y(x) is in terms of x.
 

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