Circular Disk Rotating About Its Diameter

The moment of inertia for the same circular disk rotating about an axis in the plane of the disk, passing through its center, is given by

$\displaystyle I = \frac{M}{\pi R^2}\cdot 4\int_0^{\pi/2} \int_0^R [r\cos(\theta)]^2\, r\,dr\,d\theta
= \frac{1}{4}MR^2
$

Thus, the uniform disk's moment of inertia in its own plane is twice that about its diameter.


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Rectangular Cross-Section
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Circular Disk Rotating in Its Own Plane