# Finding the coordinates of the center of mass

1. Apr 2, 2009

### aks_sky

A solid hemisphere of radius r given by:

x2+y2+z2= r2, z ≥ 0

Suppose the density of the hemisphere is the constant δ. The coordinates of the center of mass (a,b,c) are given by:

a=(∭ xδdV) / (∭ δdV)

b=(∭ yδdV) / (∭ δdV)

c=(∭ zδdV) / (∭ δdV)

Find a, b ,c

** So far what i have tried is that i have tried integrating the function and also i have used spherical co-ordinates but i am still not sure how to actually divide by the constant when trying to find a. Just a bit confused there.

2. Apr 2, 2009

### Limecat

The denominator is just the mass of the hemisphere.

3. Apr 2, 2009

### aks_sky

yup that is true. But would i need to integrate that in respect to anything. since its a constant i can take the limits of integration as spherical co-ordinates?

4. Apr 2, 2009

### Limecat

No you don't need to do any integration for the denominator. The density is constant.
The mass of the hemisphere is just the volume of a sphere divided by two, and then multiplied by the density

Last edited: Apr 2, 2009
5. Apr 2, 2009

### aks_sky

ohh sweet.. so basically i would integrate the numerator as a volume integral and the denominator stays constant.

6. Apr 2, 2009

### Limecat

My physics is a bit rusty... but I'm not sure how I would do that in 3 separate components like you've written there... but then I can't quite get the math to work out right doing a single triple integral in spherical coordinates. I know that the centre of mass should be above the middle of the flat face of the hemisphere though (if you think of it as laying flat)

Edit: nevermind. I figured it out. Ask if you need any clarification.

Last edited: Apr 2, 2009
7. Apr 2, 2009

### aks_sky

yes please need a bit of clarification.

8. Apr 2, 2009

### Limecat

I've only done it in spherical coordinates... probably not the best way if the density isn't a constant.
But since the density IS constant, you can assume that the centre of mass lies somewhere along the vertical axis.
To find the centre of mass of something, you need to do (m1x1 + m2x2 + ....) / (m1 + m2 + ...)
Well you know the denominator now and it's a constant so you can just have it off to the side. The numerator is an integral in spherical coordinates (that's how I did it)

So instead of x, I used r. so you integrate mr. that's delta times dV
dV in spherical coordinates is (r^2) sin(phi) dr dphi dtheta (sorry I don't know LaTeX)

So delta*r*dV is delta*(r^3)*sin(phi) dr dphi dtheta.

You know the limits of integration, yeah?

9. Apr 2, 2009

### Limecat

denominator:
$$\frac{\delta 4\pi r^3}{3}$$$$\frac{1}{2}$$

numerator:

$$m_{1}r_{1} + m_{2}r_{2} + ...$$

$$m = \delta V$$; $$dm = \delta dV$$ (since $$\delta$$ is constant)

so now numerator becomes

$$\int\int\int rdm = \int\int\int r\delta dV$$

$$dV$$ for a sphere is $$r^2 sin\phi drd\phi d\vartheta$$ and $$\phi$$ is the polar angle (some people use $$\vartheta$$ as the polar angle instead)

Last edited: Apr 2, 2009
10. Apr 2, 2009

### aks_sky

yup so the limits of integration are from 0 to r which is the radius for dr, and i am guessing for the middle integral dphi i would get 0 to x since we are integrating that part and dtheta would be 0 to 2pi.

11. Apr 2, 2009

### Limecat

for $$\phi$$ not quite. $$\phi$$ goes from zero (which is at the top of a sphere, if we're thinking of a full sphere for example's sake) to pi (which is the bottom of the sphere)

In this case, we have half a sphere so... ;)

12. Apr 2, 2009

### Limecat

oops. I left out a $$\delta$$ in the numerator. Make sure you have a $$\delta$$ there when you do the actual calculation. I added it to the reply

13. Apr 2, 2009

### aks_sky

ohhh ok.. thanx a lot for all da help!! and also i guess for y and z, i use the other parts of the spherical co ordinates..

14. Apr 2, 2009

### Limecat

No, if you do it by x y z, you gotta do it a bit differently.. I explained in my other reply that you don't have to do it in x y z (unless you want to be extra rigorous with your math).

It's because the mass is uniformly distributed, which means that the centre of mass is directly above the middle of the flat face of the hemisphere (if the hemisphere is laying flat on a level surface)
so the resulting r you get is just the distance above the bottom. no x or y component. just z. r starts at the 'centre' of the sphere, but in this case you only have the top half so that's why you just integrate $$\phi$$ from 0 to pi/2 instead of from 0 to pi.