📂Classical Mechanics

Inertia Moments of Disks and Cylinders

Formula

A disk with radius $a$ and mass $m$ has a moment of inertia

• perpendicular to the disk as $I=\dfrac{1}{2}ma^2$.

• parallel to the disk as $I=\dfrac{1}{4}ma^2$.

Derivation

When the axis of rotation passes through the center of the disk and is perpendicular to the disk

Let $\rho$ be the mass per unit area. Then, the mass of the disk is $m=\rho \pi r^2$. Therefore, it follows that

$$dm=\rho \pi 2r dr$$

The formula to calculate the moment of inertia is $\displaystyle I=\int r^2dm$, hence it can be shown that

$$I=\int_{0}^a\rho \pi 2 r^3 dr=\rho \pi 2 \frac{1}{4}a^4=\frac{1}{2}\rho \pi a^4$$

Where $\displaystyle \rho=\frac{m}{\pi a^2}$, thus

$$I=\frac{1}{2}ma^2$$

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When the axis of rotation passes through the center of the disk and is parallel to the disk

According to the Parallel Axis Theorem, $I_{z}=I_{x}+I_{y}$, and whether the rotation axis is along the $x$ axis or the $y$ axis, the shape is the same, therefore $I_{x}=I_{y}$. Thus

$$\begin{align*} && 2I_{x} &= I_{z}=\frac{1}{2}ma^2 \\ \implies && I_{x} &= \frac{1}{4}ma^2 \end{align*}$$

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