Center of mass of a solid hemisphere (where's the error)?

  • #1
rahul.mishra
7
0
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:

Answers and Replies

  • #2
Jano L.
Gold Member
1,333
74
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
rahul.mishra
7
0
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
Jano L.
Gold Member
1,333
74
Ah, you're right, it's cos theta.
 

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