# Homework Help: Centre of mass of a solid hemisphere.

1. Jan 3, 2009

### grahammtb

Hi there, I can't get my head round how to do the math for this problem. I'm sure it's not as hard as I think...
1. The problem statement, all variables and given/known data
Show that the CoM of a uniform solid hemisphere of radius r lies at a distance (3/8)r from the centre of the flat face.
You may find it convenient to regard the hemisphere as consisting of a very large number of extremely thin discs of varying radii stacked on top of each other.
2. Relevant equations
I'm thinking I'll need the equation for CoM, involving an integral: R=(1/M)$$\int$$rdm. This I think can be modified to: R=(1/M)$$\int$$r$$\rho$$dV, where rho is the density at radius r.
3. The attempt at a solution
N/A

Thanks very much for any help!
~Graham
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 3, 2009

### Staff: Mentor

So far, so good. To avoid confusion, I recommend you use a different variable of integration than r--let's say y. (Imagine the hemisphere axis to be along the y axis from y = 0 to y = r.) What's the volume of a thin disk located at position y?

3. Jan 3, 2009

### grahammtb

Ok, since the discs are extremely thin, I'd approximate the volume of the disc to be its area. So: A(y) = Pi.(r(y))2.

4. Jan 3, 2009

### tiny-tim

Hi Graham!
You're ignoring the hint …

if you use discs stacked on top of each other (with height z, say), you can go straight to integrating over dz, instead of dV.

5. Jan 3, 2009

### Staff: Mentor

The disks are thin, but not zero thickness! Hint: dV = Area dy.

6. Jan 3, 2009

### grahammtb

Ah I think I see...so if dV = Ady, then I can put that into my original eqn. for CoM? R = (1/M)$$\int$$$$\rho$$ydV. Then R = (Pi.$$\rho$$/M)$$\int$$y3dy...I think

7. Jan 3, 2009

### Staff: Mentor

You're on the right track, but not quite there yet. What's the radius of a disk at position y? (Draw yourself a diagram.)

8. Jan 3, 2009

### grahammtb

yep! It worked! Thanks a lot guys, I'd have spent all afternoon trying to pick my way through :)