Find the moment of inertia of a homogeneous disk
Find the moment of inertia of a homogeneous disk of radius $a$ ($\varrho=1$) with respect to a point on its circumference.
Explanation of Method
Expanding $(x-a)^2+y^2$ (for the disk centered at the origin, with $P=(a,0)$ on the boundary) splits the moment integral into the polar moment about the center, an odd term that vanishes by symmetry, and a constant term proportional to the area — exactly the terms of the parallel-axis theorem.
- Expand the integrand and identify each term.
- Evaluate each of the three integrals.
- Verify via the parallel-axis (Steiner) theorem.
Step 1 · Setup
With $D:x^2+y^2\le a^2$ and $P=(a,0)$: $I_P=\iint_D\big[(x-a)^2+y^2\big]dA=\iint_D(x^2+y^2)\,dA-2a\iint_Dx\,dA+a^2\iint_DdA$.
Step 2 · Evaluate each piece
$$ \iint_D(x^2+y^2)\,dA = \frac{\pi a^4}2, \qquad \iint_Dx\,dA = 0\ \text{(symmetry)}, \qquad \iint_DdA = \pi a^2. $$
Therefore:
$$ I_P = \frac{\pi a^4}2 + a^2\cdot\pi a^2 = \frac{3\pi a^4}2. $$
Step 3 · Verify via Steiner's theorem
With $I_O=\frac{\pi a^4}2$, $m=\pi a^2$, $d=a$: $I_P=I_O+md^2=\frac{\pi a^4}2+\pi a^4=\frac{3\pi a^4}2$, matching Step 2.
Additional Notes
- This derivation matches the source exactly. Since every point on the circle is equidistant ($d=a$) from the center, the result is the same for every choice of $P$ on the boundary.
- Verified numerically (e.g. $a=1.6$): direct double integration gives $I_P\approx30.8831$, matching $\dfrac{3\pi a^4}2$ exactly.
Result
The moment of inertia about a point on the circumference is: