Monday, December 6, 2010

Math and Physics ventures...

Well most likely no one will want to see this, but if anyone is interested... Today in Economics I wasn't paying attention to the teacher, but rather I was formulating and deriving the expression for moment of inertia of a hollow sphere. For those of you who don't know, moment of inertia is how hard it is to rotate something about a point or axis.

Well here we go. This probably will be really hard to understand because of the lack of symbols... sorry for the inconvenience.

int = integral; pi = pi; d = density; m = mass; r = radius; I = moment of inertia; variables = phi, theta, p;

I = mr^2

Converting to spherical coordinates and integrating we get:

I = int[int[int[mr^2,V]]] = int[int[int[(p*sin(phi))^2*p^2*sin(phi),p],phi],theta]

p = r in this case.

The tricky part was the innermost integral. I realized after thinking for a while that you can't integrate with respect to p because it is infinitesimally small and I don't need the whole radius. Therefore, I converted it into a trivial sum in order to just use the very edge. I also set up my bounds in order to account for the entire sphere.

int[int[sum[d*p(i)^4*(sin(phi))^3,i,1,1],phi,0,pi],theta,0,2*pi] where p(1) = r

Simplifying to a double integral...

int[int[d*r^4*(sin(phi))^3,phi,0,pi],theta,0,2*pi]

now some trig subs to make this integrable.

d*r^4*int[int[sin(phi)*(1-(cos(phi))^2),phi,0,pi],theta,0,2*pi]

simplifies down to:

d*r^4*int[4/3,theta,0,2*pi] = (8/3)*pi*d*r^4

We can assume a hollow sphere to only have an area: which is 4*pi*r^2
And planar density is mass over area
And now we substitute

(8/3)*pi*(m/(4*pi*r^2))*r^4 = (2/3)*m*r^2

Hooray calculus!