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 tangent line.
Explanation of Method
Rather than integrating directly, the parallel-axis (Steiner) theorem lets the moment of inertia about the tangent be built from two standard, independently-known quantities: the disk's moment of inertia about a diameter through its own center, and the mass times the square of the distance between the two axes.
- Compute the moment of inertia $I_c$ about a central diameter.
- Apply the parallel-axis theorem.
- Verify by direct integration in shifted coordinates.
Step 1 · Central moment of inertia
$$ I_c = \iint_{x^2+y^2\le a^2}y^2\,dA = \int_0^{2\pi}\int_0^ar^3\sin^2\theta\,dr\,d\theta = \pi\cdot\frac{a^4}4 = \frac{\pi a^4}4. $$
Step 2 · Parallel-axis theorem
With mass $m=\pi a^2$ and distance $d=a$ from center to tangent:
$$ I = I_c + md^2 = \frac{\pi a^4}4 + \pi a^2\cdot a^2 = \frac{5\pi a^4}4. $$
Step 3 · Direct verification
With the disk centered at $(0,a)$, tangent $y=0$, and $Y=y-a$: $I=\iint(Y+a)^2\,dX\,dY=\iint Y^2\,dX\,dY+2a\underbrace{\iint Y\,dX\,dY}_{=0}+a^2\iint dX\,dY=\frac{\pi a^4}4+\pi a^4=\frac{5\pi a^4}4$, matching Step 2.
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=1.5$): direct double integration gives $I\approx19.8804$, matching $\dfrac{5\pi a^4}4$ exactly.
Result
The moment of inertia about the tangent line is: