# Center of mass of solid hemisphere.

1. Dec 25, 2011

### AlchemistK

1. The problem statement, all variables and given/known data
I've just started with center of mass, and instead of the method in the book, I tried solving the center of mass for a solid hemisphere using angle as a variable, but the answer didn't match.
A small disk of mass "dm" is taken, which subtends an angle "dθ" at the center.The total mass of the hemisphere is "M". Please refer to the attachment for the figure and my full attempt.
The solutions elsewhere use distance from the center of the hemisphere as a variable.
Where have i gone wrong?

2. Relevant equations
y(cm) = ∫y dm / ∫ dm = 1/M (∫y dm)

3. The attempt at a solution

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Last edited: Dec 25, 2011
2. Dec 25, 2011

### Thaakisfox

That is not the right volume element. The radius of the chosen disk is not the "same" at the two sides of the infinitesimal angle as the sphere has a definite curvature. What you calculated would be the volume element of a cylinder.

3. Dec 25, 2011

### vela

Staff Emeritus
That's not the problem. The effect on the volume element due to the difference in radii is a second-order effect, so it is negligible. The problem is the thickness of the slab dy varies with θ, and that isn't accounted for in the OP's attempt.

4. Dec 25, 2011

### Thaakisfox

Yes, Vela that's right :)

5. Dec 25, 2011

### SammyS

Staff Emeritus
In your work, you have that $y=R\sin(\theta)\,.$

So, how is the thickness, dy, related to dθ ?

6. Dec 26, 2011

### AlchemistK

Lets zoom in at the curved part of the disk, since dθ is very small, it can be considered as a straight line, there in the triangle formed, dy = R sinθ dθ. Using this as the thickness and R cosθ as the radius we can find the volume of the disk, but there will be the part that is left in the 3D triangular ring that is formed, how do I calculate that part? (please refer to the new attachment for the diagram)

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