# Trying to get moment of inertia of a disc using moment of inertia of a rod

I know how moment of inertia of a disc is calculated using the usual way, but just for fun, I tried this way which is rather giving incorrect answer. I don't know what's the flaw in this and thus would like to have you help me figure out the flaw.

So I imagined that a disc can be taken as a collection of many thin rods with one end joined. The arrangement can be imagined as spokes of a wheel as shown

Now, moment of inertia of a rod about an axis through one end perpendicular to it is $$ml^2/3$$ where m is mass of rod and l is it's length. Using this moment of inertia of the arrangement which tends to be a disc as number of spokes increases will be simply the sum of moment of inertia of each spoke giving $$Ml^2/3$$ if the total mass of arrangement is $$M$$.

Now l is nothing but radius of our disc and thus it gives moment of inertia of a disc as $$MR^2/2$$.

I don't know what's the problem in my line of reasoning, but the answer is clearly incorrect.

Consider a thin ring of thickness $$\mathrm{d}r$$ at the distance of $$r$$ form the center of the disc. Its mass is $$m=h2\pi\rho r \mathrm{d}r$$ (simply the volume of the ring, multiplied by its uniform density, $$h$$ is the uniform thickness of the disc) Mass of such rings increases linearly with their radius.
On the contrary, a "ring" cut out from your rod-wheel has a constant mass, independent of its radius: $$m=\rho NA\mathrm{d}r$$, Where N is the number of rods, $$\rho$$ the density, and A the cross section of a rod.
The better way to do what you propose is to break the disk up into little wedges. Each wedge will have a moment of inertia of $$\frac12R^2dm$$. Integrating over this will give you the correct moment of inertia.