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Q. 20.7

Use the perpendicular axes theorem to find the moment of inertia of a thin circular disc of radius r about a diameter.

Verified Solution

By symmetry the moment of inertia about every diameter is the same, say $I_{d}$. This means that, when the disc is in the xy plane with its centre at the origin,

$I_{x} = I_{y} + I_{d}.$

But $I_{z}$ is the moment of inertia about the axis through O perpendicular to the disc and this is $\frac{1}{2}Mr^{2}.$ Hence

$\frac{1}{2}Mr^{2} = I_{x} + I_{y}$

$\begin{matrix} = 2I_{d} & \boxed{I_{x} = I_{y} \text{ and } I_{z} = \frac{1}{2}Mr^{2}} \end{matrix}$

$⇒ I_{d} = \frac{1}{4}Mr^{2}.$