Derive the moment of inertia of a solid, uniform density, disk rotating about a diameter using vertical rods.
Only the radius of the circle and the inertia of a thin rod is given
Inertia of thin vertical rod = m*r^2
σ= area density
M= Total mass
dm= infinitesimal amount of mass
dI= infinitesimal amount of inertia
A= total area
R= radius of the circle
The Attempt at a Solution
This is the method given in my class. Because calculus 1 is the only prerequisite we can not use double integrals.
I of rod = m*x^2
dm=M/(pi*R^2) * 2y*dx
dI= M/(pi*R^2) * 2y*x^2 * dx
From here we have to put y in terms of x which I think must be done using the Pythagorean theorem so:
subbing in we get:
dI= M/(pi*R^2) * 2(R^2-x^2)^(1/2) * x^2 * dx
This is why I get stuck because I am not quite sure how to integrate that. What is should come out to be is I = 1/4 MR^2. If anyone can offer an explanation it would be very appreciated.