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

[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

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 = 1/M http://data.artofproblemsolving.com/images/latex/9/a/a/9aabc0767384d5b00970c757ff71f80cd37d11eb.gif

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?

 

[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.

 

[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 :) !

 

[Jano L.]

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

 

[Nickie]

There appears to be an error in your calculation for the Y-coordinate of the center of mass. The correct expression for Ycm should be 1/M * http://data.artofproblemsolving.com/images/latex/9/a/a/9aabc0767384d5b00970c757ff71f80cd37d11eb.gif , where M is the total mass of the hemisphere. This can be derived by considering the moment of inertia of a solid hemisphere and using the formula for the center of mass in terms of moments of inertia. Your expression for Ycm only considers the mass of the small element formed by the two rings, rather than the entire hemisphere. Therefore, your result of 3R/8 is incorrect. It is important to consider the total mass of the object when calculating the center of mass.