# Moment of inertia of a disc with rounded sides

I'm trying to formulate a dimensional equation for the MOI of a disc with rounded chamfer, where the chamfer is circular in shape and has radius $$b$$, and the thickness of the disc is the same as the diameter of the disc. The disc spins around the $$z$$-axis that passes through its center of mass. For clarity, the geometry of the object is drawn in the following sketch:

We know the MOI of a simple disc of thickness $$L$$ without the rounded sides is given by this equation:

$$I = \frac{1}{2} MR^2 = \frac{1}{2}\overbrace{\rho \pi R^2 L}^{M}R^2 = \frac{1}{2} \rho \pi LR^4$$

My approach is to slice up my disc into infinitesimally thin discs each of thickness $$dz$$ and integrate them on the $$z$$-axis from $$-b$$ to $$b$$. The radius of each disc can be calculated using Pythagorean theorem: $$\text{radius} = \sqrt{b^2-z^2} + a$$.

$$I=\int dI = \int_{-b}^{b} \frac{1}{2} \rho \pi \left(\sqrt{b^2-z^2}+a \right)^4 dz \;\;\;\;\;\;\;\; \rho, a, b \in \text{constants}$$

It looks like an integral that needs some lengthy trig substitution and I'm not sure if that's even solvable by (my) hand. So I plugged that integral into WolframAlpha which sadly exceeded the standard computation time.

Would someone help verify is that the right approach or is there a better way to solve for the MOI of this shape?

• @Qmechanic, I'm not sure why this is edited and tagged as homework...
– KMC
Feb 13 at 19:19
• Hi KMC. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. Feb 13 at 19:31
• To make the integration easier, you could expand the bracket using the Binomial Theorem. You would then have various polynomial terms which are easy to deal with, plus two integrals involving square roots which will succumb to trig substitution (or WA) Feb 14 at 10:29
• @user3298777 thanks for the tip! After binomial expansion the integration works out in WA. I've also verified it by plugging in some numbers and the inertia turns out to be somewhere between the inertia of a bigger flat-sided disc (R=a+b) and that of a smaller one (R=a) with the result weighted more towards the bigger disc as expected due to r^2 relationship. As simple as your comment may seem, that actually answers my question. No idea why I've never thought of expanding the terms. You can move it to the answer.
– KMC
Feb 15 at 7:37

## 2 Answers

Hint...To make the integration easier, you could expand the bracket using the Binomial Theorem. You would then have various polynomial terms which are easy to deal with, plus two integrals involving square roots which will succumb to trig substitution (or WA).

I leave the details to you!

(Sounds like you've figured it out anyway)

As you don't said anything about density, I assume $$\rho=\rho_0=cte$$, then

$$I = \frac{\pi\, \rho_0}{2} \cdot \int_{-b}^b (\sqrt{b^2 -z^2}-a)^4 \cdot dz$$

Let's simplify a little more, call z=b*t

$$I = \frac{\pi\, \rho\, b^5}{2} \cdot \int_{-1}^1 ( \sqrt{1 -t^2}-\frac{a}{b})^4 \cdot dt$$

Now your integral have only one parameter $$c=a/b$$, and Wolfram will be able to give you an answer easily.

• I edited my question - $\rho, a, b$ are all constants
– KMC
Feb 13 at 19:23