As part of a problem I need to find the center of mass of a hollow hemisphere.(adsbygoogle = window.adsbygoogle || []).push({});

I did it in the following way.

I considered the lower hemisphere ( around negative z axis )

First by definition, the position vector of the center of mass is obtained as

[itex]\vec{R}[/itex]= [itex]\frac{\int\vec{r}dm}{\int dm}[/itex]

Now I set up the problem in spherical coordinates

http://img855.imageshack.us/img855/4741/spherev.png [Broken]

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Now, [itex]\vec{r}[/itex] = R[itex]\hat{r}[/itex]

where [itex]\hat{r}[/itex] = sin[itex]\theta[/itex]cos[itex]\phi[/itex][itex]\hat{i}[/itex]+sin[itex]\theta[/itex]sin[itex]\phi[/itex][itex]\hat{j}[/itex]+cos[itex]\theta[/itex][itex]\hat{k}[/itex]

And dm = [itex]\sigma[/itex]ds, where ds is the area element, and [itex]\sigma[/itex] is the mass per unit area.

ds = R^{2}sin[itex]\theta[/itex]d[itex]\theta[/itex]d[itex]\phi[/itex]

So then [itex]\vec{R}[/itex]= [itex]\frac{\int\vec{r}ds}{\int ds}[/itex]

i.e (2[itex]\pi[/itex]R^{2}) [itex]\vec{R}[/itex]=[itex]\int[/itex][itex]\int[/itex]R^{3}(sin^{2}[itex]\theta[/itex]cos[itex]\phi[/itex][itex]\hat{i}[/itex]+sin^{2}[itex]\theta[/itex]sin[itex]\phi[/itex][itex]\hat{j}[/itex]+[itex]\frac{sin2\theta}{2}[/itex][itex]\hat{k}[/itex])d[itex]\theta[/itex]d[itex]\phi[/itex]

integrating [itex]\theta[/itex] from [itex]\pi[/itex]/2 to [itex]\pi[/itex] and [itex]\phi[/itex] from 0 to 2[itex]\pi[/itex]

I get [itex]\vec{R}[/itex] = -[itex]\frac{\hat{k}}{2}[/itex]

now I know this is not the right answer, the correct answer is -[itex]\frac{2\pi}{R}[/itex][itex]\hat{k}[/itex]

I know that the problem can be simplified with symmetry and done with much more ease than what I have attempted, but I'd like to know what is wrong with the way I formulated the problem.

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# Homework Help: Center of mass of a hemisphere

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