# Contradiction In Deriving Moment Of Inertia Of Disc

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?
• Here is a link I found for correct derivation Nov 23, 2023 at 22:33
• I think it's more suitable for MathsSE than PhysicsSE since the question is on calculus. The only Physics was $dI=x^2dm$ Nov 23, 2023 at 22:36
• Also just to clarify the rods are basically strips with area = $Rdx$ therefore their mass will be $dm=Surface \ density×Rdx$ Nov 23, 2023 at 22:44

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.
• ...the sum of infinitely many such 0's may not stay 0. Think of it with an example : $\lim_{n \rightarrow \infty} \sum_{x = 0}^n \frac{1}{n + x}$. Each term is individually 0 but there are infinitely many terms, and evidently the sum does not approach zero, it rather approaches ln2 (using limit of a sum as integral). Nov 24, 2023 at 17:04