# Homework Help: Center of mass of a solid hemisphere (where's the error)?

1. Aug 21, 2012

### rahul.mishra

On finding the center of mass of a solid hemisphere i came up with some different result.

Here's what i did...

consider a small ring at a distance r from the center of the hemisphere and one more ring at a distance
of r+dr from center of the ring.

let, mass of the small element formed by the two rings be dm
so,

dm = http://data.artofproblemsolving.com/images/latex/1/8/9/189c6482a7f0b088a7d84607b3df5837533486eb.gif [Broken]

Clearly by symmetry w.r.t origin taken Xcm = 0

now, Ycm = 1/M http://data.artofproblemsolving.com/images/latex/f/4/e/f4e81964f7df0126c6c05827f0d5f0b972fc5f45.gif [Broken] = 1/M http://data.artofproblemsolving.com/images/latex/9/a/a/9aabc0767384d5b00970c757ff71f80cd37d11eb.gif [Broken]

but we all know this is maybe absurd... its 3R/8 actually....

don't give me an alternate way..... i know it gives the right result...
but where am i wrong in my soln?

Last edited by a moderator: May 6, 2017
2. Aug 21, 2012

### Jano L.

You made a little error: the thickness of the disk is not

$$r d\theta$$

but

$$r\sin \theta d\theta$$
- the disk has its height in y direction.

3. Aug 21, 2012

### rahul.mishra

ya... got it.. but the thickness of the small element is rcosθdθ and not rsinθdθ maybe. I am quite sure as it gave me the correct result :) !!

4. Aug 22, 2012

### Jano L.

Ah, you're right, it's cos theta.