Parallel Axis Theorem
Parallel Axis Theorem
The moment of inertia of a rigid body about any axis of rotation is equal to the sum of the moment of inertia about an axis parallel to it and passing through the center of mass and the product of the body’s mass and the square of the distance between the two axes.
$$ \color{red}I=\color{blue}{I_{cm}}+\color{green}{md^{2}} $$
Proof
Arbitrarily set the coordinate axis and let the moment of inertia about the $z$-axis be denoted as $I_{z}$.
$$ \begin{equation} I_{z}=\sum\limits_{i} m_{i} {r_{i}}^{2} = \sum\limits_{i} m_{i} ({x_{i}}^{2}+{y_{i}}^{2}) \label{eq1} \end{equation} $$
Expressing the distance from the origin to any point on the rigid body as the sum of the distance from the origin to the center of mass and the distance from the center of mass to the point, we have:
$$ x_{i}=x_{cm}+\bar x_{i} $$
$$ y_{i}=y_{cm}+ \bar y_{i} $$
Substituting this into $\eqref{eq1}$ yields:
$$ I_{z}=\sum\limits_{i} m_{i} \left[ (x_{cm}+\bar x_{i})^{2}+(y_{cm}+ \bar y_{i})^{2} \right] $$
Expanding and simplifying yields:
$$ I_{z}=\sum\limits_{i} m_{i}({\bar x_{i}}^{2}+{\bar y_{i}}^{2}) + \sum\limits_{i} m_{i}({x_{cm}}^{2}+{y_{cm}}^{2})+2x_{cm}\sum\limits_{i} m_{i}\bar x_{i} + 2y_{cm} \sum\limits_{i} m_{i}\bar y_{i} $$
Let’s calculate each term.
part 1. The first term
$({\bar x_{i}}^{2}+{\bar y_{i}}^{2})$ represents the squared distance from the center of mass to each point. Therefore, the first term is the moment of inertia about the axis passing through the center of mass, that is, $I_{cm}$.
part 2. The second term
$({x_{cm}}^{2}+{y_{cm}}^{2})$ represents the square of the distance between the arbitrary axis of rotation and the axis passing through the center of mass. Thus, the second term is $md^{2}$.
part 3. The third and fourth terms
By definition of the center of mass, the third and fourth terms are 0. Let’s calculate why. The center of mass in direction $x$ is as follows:
$$ x_{cm}=\frac{\sum m_{i}x_{i}}{m} $$
Expanding yields:
$$ \begin{align*} x_{cm} &= \frac{\sum m_{i}x_{i}}{m} \\ &= \frac{\sum m_{i}(x_{cm}+\bar x_{i})}{m} \\ &= \frac{\sum m_{i} x_{cm}}{m}+\frac{ \sum m_{i}\bar x_{i}}{m} \\ &= \frac{ x_{cm} \sum m_{i}}{m}+\frac{ \sum m_{i}\bar x_{i}}{m} \end{align*} $$
Since $\sum m_{i}=m$, the above equation simplifies to:
$$ \begin{array}{llc} && x_{cm}=x_{cm}\dfrac{m}{m}+\dfrac{ \sum m_{i}\bar x_{i}}{m} = x_{cm}+ \dfrac{ \sum m_{i}\bar x_{i}}{m} \\ \implies && 0 = \dfrac{ \sum m_{i}\bar x_{i}}{m} \end{array} $$
Hence, $\sum m_{i} \bar x_{i}=0$, which applies to $\bar y_{i}$ as well.
Combining these results, we have:
$$ I_{z}=I_{cm}+md^{2} $$
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