# Center of mass of a hemisphere

As part of a problem I need to find the center of mass of a hollow hemisphere.

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

$\vec{R}$= $\frac{\int\vec{r}dm}{\int dm}$

Now I set up the problem in spherical coordinates

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

Now, $\vec{r}$ = R$\hat{r}$

where $\hat{r}$ = sin$\theta$cos$\phi$$\hat{i}$+sin$\theta$sin$\phi$$\hat{j}$+cos$\theta$$\hat{k}$

And dm = $\sigma$ds, where ds is the area element, and $\sigma$ is the mass per unit area.

ds = R2sin$\theta$d$\theta$d$\phi$

So then $\vec{R}$= $\frac{\int\vec{r}ds}{\int ds}$

i.e (2$\pi$R2) $\vec{R}$=$\int$$\int$R3 (sin2$\theta$cos$\phi$$\hat{i}$+sin2$\theta$sin$\phi$$\hat{j}$+$\frac{sin2\theta}{2}$$\hat{k}$)d$\theta$d$\phi$

integrating $\theta$ from $\pi$/2 to $\pi$ and $\phi$ from 0 to 2$\pi$

I get $\vec{R}$ = -$\frac{\hat{k}}{2}$

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

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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vela
Staff Emeritus
Homework Helper
I get $\vec{R}$ = -$\frac{\hat{k}}{2}$
now I know this is not the right answer, the correct answer is -$\frac{2\pi}{R}$$\hat{k}$
This can't possible be the correct answer. The units aren't correct.

This can't possible be the correct answer. The units aren't correct.

sorry I made a mistake, the correct answer is- $\frac{2R}{\pi}$$\hat{K}$

You can arrive at that answer with this argument, the sphere can be thought of to be made of a number of semicircular line segments of radius R ( or taking one and rotating it about the negative Z axis )

So the C.M of the sphere will be the same as one such segment, since the rotation occurs about an axis through the center of mass.

http://img97.imageshack.us/img97/3853/semit.png [Broken]

now, if I consider a horizontal strip, the CM of the strip is on the Z axis at a distance Rcos$\theta$ and the mass dm = 2$\lambda$dl , where $\lambda$ is the linear mass density, $\lambda$ gets canceled out and the cm is at

$\vec{R}$ =$\frac{1}{\pi R}$$\int$2R2cos$\theta$d$\theta$

dl = Rd$\theta$

$\theta$ varies from 0 to $\frac{\pi}{2}$

then $\vec{R}$ = $\frac{2R}{\pi}$

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vela
Staff Emeritus
Homework Helper
The radius of the circular strip isn't R. It's only equal to R when theta is pi/2.

The radius of the circular strip isn't R. It's only equal to R when theta is pi/2.
you misunderstand what i mean, I am taking a semicircle of radius R and rotating it , think of it like an electric drill with a leaf as a semicircle,

when it spins it will form a hemisphere right?

P.S

This is not what is happening in the integral, i just used the notion of rotation to infer that the CM of both the semicircle and hemisphere have to be the same

vela
Staff Emeritus