Contradiction In Deriving Moment Of Inertia Of Disc

Question

Moment of Inertia is given by $dI=x^2dm$ (x being perpendicular distance from axis)

In deriving the MOI of a thin uniform ring about axis passing through its center perpendicular to plane, our teacher asked us to think of ring as made up of $dm$ units each at distance $R$ from the axis therefore the MOI becomes: $I=R^2{\int{dm}}=MR^2$

In finding the MOI of thin uniform disc about an axis through center and normal to plane, I assumed the disc to be made of rods of $dx$ width. Since MOI of rod about an end is $MR^2/3$, the MOI of disc becomes $$I=R^2\int{\frac{dm}{3}}=\frac{MR^2}{3}.$$

However the correct derivations using concentric rings or unequal rods give the MOI as $$I=\frac{MR^2}{2}$$.

1. WHY this contradiction? Why does that logic work for the ring but not disc? I worked on why my derivation is wrong. Thought there was a portion of the rods that was being repeatedly added but their point of intersection is on the axis, so its MOI is zero anyways.
2. What does this show about the nature of calculus or the method of /caution in using it?

Answer 1

I think this question may look like an integration problem but it is actually a physics concept based doubt.

The main problem about using a rod of length dx with one end at the centre is it does not cover the full disc, how? See this image. Notice how even if you take negligible thickness rods, there will be some portion left between them, If you try to cover all the parts near the centre like this section here then moving outward the portion between two consecutive rods is left uncovered and if you try to cover the entire circumference like this section here then they start overlapping moving inward.

Now this is a basic mistake that people do while finding moment of inertia's of certain objects. Here instead of rods which cover the entire centre but don't cover disc area moving outward, we use isoceles traingles (also called wedges) that expand moving ouward so that there is no leftover area between two triangles (visual depiction).

Calculate moment of intertia of an isoceles traingle about an axis passing through the vertex opposite to the base with surface density = $\sigma$, base = b, height = l and angle between equal sides = $\theta$ (which is a challenge in itself because here's a jist of it (here's a photo of the solution because I am too lazy to type it out using mathJAX) : you take a rod at a distance x from the vertex, write it's moment of inertia about an axis passing through it's COM and perpendicular to it's plane and then using parallel axis theorem to calculate it's moment of inertia about the vertex and then integate. For reference : it comes out to be I = $\frac{\sigma bh}{48} \cdot (b^2 + 12h^2)$ and so $b = 2h \tan(\frac{\theta}{2})$, but using small angle approximation ($\theta \to d\theta$) for our next problem (and neglecting $(d\theta)^3$) we get I = $\frac{\sigma h^4 d\theta}{4}$. After you find this, substitute h = R and surface density = $\frac{M}{\pi R^2}$. Now integrate this for $\theta = 0 \to 2\pi$ to cover the whole disc and you will get a the correct value. Or obviously you can just consider rings and solve.