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

Question

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.

Answer 1

You have made a simple mistake. Infinitely many rods do not have the same mass distribution as a uniform disc:

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.

So in conclusion, you have failed to consider that disc contain more material around their edge.

Answer 2

I believe the problem is that you cannot approximate the disk as a series of rods joined at one end. This is because there will always be holes, no matter how infinitesimal the rods are (no matter how many rods you use, they will never be touching other than at the center, and the space between them grows as you go out from the center).

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.