Find Center of Mass of Hemisphere Homework

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SUMMARY

The center of mass of a hemisphere with radius R is calculated using the formula Cm=(1/m)∫(z dm). The correct approach involves using volume density ρ instead of surface density, leading to the equation ρ = M/(2/3 π R^3). The integral for the z-component of the center of mass is CM(z-comp) = (1/M) ∫(0 to R) π(R^2 - z^2) ρ z dz, which evaluates to 3R/8. This method ensures accurate results for solid hemispherical objects.

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Homework Statement


A hemisphere of radius R, covering +/- R in the x and y directions and 0 to R in the Z direction (only the top half of a sphere centered at the origin)
Find the center of mass.

Homework Equations


Cm=(1/m)[tex]\int(z dm)[/tex]

The Attempt at a Solution


dm is sigma dA, and dA is x dz, where x = [tex]\sqrt{}R^2-z^2[/tex]. The limits are 0 to R, and the answer should be 3/8 R, but when I evaluate the integral I get nothing like that.
 
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since you are trying to find for a hemisphere not for a hemispherical shell, your equation for 'dm' is wrong.
You should consider not the surface density, but instead the volume density ρ since it is a solid object
[itex]\rho = \frac{M}{\frac{2}{3} \pi R^3}[/itex]

You then have

[itex]dm = \rho A dz[/itex] where A is the area of the circular disc centered on the z-axis with radius [itex]r = \sqrt{R^2 - z^2}[/itex].

The final integral then becomes

[itex]CM (z-comp) = \frac{1}{M} \int_0^R \ dz\ \pi (R^2 - z^2) \rho\ z[/itex]

which on solving gives the result [tex]\frac{3 R}{8}[/tex]
 

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