Finding the coordinates of the center of mass

[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.

 

[Limecat]

The denominator is just the mass of the hemisphere.

 

[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?

 

[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

 

[aks_sky]

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

 

[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.

 

[aks_sky]

yes please need a bit of clarification.

 

[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?

 

[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)

 

[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.

 

[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... ;)

 

[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

 

[aks_sky]

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

 

[Limecat]

No, if you do it by x y z, you got to 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.