> [!note]+ A Note for SoME4 Reviewers > **For the best experience, please use a computer (e.g. laptop or a desktop)** > > - The top right shows an interactive graph of how concepts connect > - The left panel is for site navigation - you can ignore this > - Underlined text ([[SoME4 Competition Submission - Parallel Axis Theorem|like this]]) links to related concepts > - You don't need to click on them; all definitions are explained > - References to **figures** jump directly to diagrams > **Thank you for reviewing my submission. I hope you enjoy!** >[!info]+ Read Time **⏱ 7 mins** # Opening Motivation Why is it easier to spin a broom around its middle than to spin it on its edge? These simple observations lead to a bigger question. How can mathematics describe the difference between rotating an object at one point versus another? The answer comes from a compact equation known as the parallel axis theorem. # Important Terms To understand why spinning an object at some points is easier than spinning it at other points, the following definitions must be understood. The [[Center of Mass|center of mass]] is a [[Scalar & Vectors|point]] where you can imagine all the mass of the object to be located. For a broom, it lies somewhere near the middle. Mathematically, the center of mass is denoted as $\vec{R_{cm}}$ and has the following equation, $\vec{R}_{cm} = \frac{1}{M} \int r \,dm$. The [[Moment of Inertia|moment of inertia]] is an **object's resistance** to rotating around an axis. The higher the moment of inertia is, the harder the object is to spin. Usually, the moment of inertia is assumed to occur at the center of mass, unless otherwise stated. Mathematically it is stated as $I$ with the following equation, $I = \displaystyle\int r^2 \, dm$. > [!note]+ Center of Mass & Moment of Inertia Diagram ![[pat_5.png]] A broom is calculated for the center of mass. It is located somewhere near the middle of the broom. > Then the moment of inertia describes a spot to rotate the object. Meaning the resistance of rotation is the same anywhere along the dotted line (axis). # Parallel Axis Theorem Suppose you're given an irregular 2D rock. The center of mass and the moment of inertia are known. If the rock is forced to rotate about a different point $P$, its moment of inertia changes. This new moment of inertia at the point $P$ will have an axis parallel to the moment of inertia at the center of mass. Because this new axis is parallel to the one through the center of mass, we can relate the two using the **parallel axis theorem**. Giving the moment of inertia at the point $P$. > [!note]+ Parallel Axis Diagram ![[pat_1.png|450]] > The original moment of inertia is along the center of mass. The parallel axis theorem is a new moment of inertia whose axis is parallel to the original moment of inertia. ## Derivation > [!warning] Assumptions To derive an equation for the parallel axis theorem, a unique scenario needs to be assumed. > Assume you have a rock (which is 2D), where the center of mass, and moment of inertia (at the center of mass) are known. Then we force the object to rotate at another point. Which means there is a new moment of inertia. > Use the diagrams below as a visual aid throughout the derivation. > [!grid] Important Diagrams for Derivations > ![[pat_2.png]] > ![[pat_3.png]] > ![[pat_4.png]] In [[SoME4 Competition Submission - Parallel Axis Theorem#Derivation|Figure 1]], assume that our new moment of inertia occurs at the point $P$. Which is a [[Distance|distance]] $D$ from the center of mass ($CM$), with components $x_{p},y_{p}$. **These distances are fixed**. The new moment of inertia at point $P$ is labelled $I_{P}$ and the moment of inertia at $CM$ is labelled $I_{cm}$ Since our rock is unsymmetrical, we'll need to [[Indefinite Integrals|integrate]] the total mass. So we'll pick out a differential mass, $dm$ a distance $r$ away from the center of mass, with components $x,y$. These distances can change and are variables ([[SoME4 Competition Submission - Parallel Axis Theorem#Derivation|Fig. 2]]). The differential mass, is also a distance $a$ away from point $P$, with components $a_{x},a_{y}$. ([[SoME4 Competition Submission - Parallel Axis Theorem#Derivation|Fig. 3]]). > [!warning] Recall Moment & Further Assumptions Before all the integration starts lets recap the important equations and the goal. > - **The moment of inertia** at the center mass is calculated as $I_{CM}=\int r^2 dm$ > - **The goal is** to find the moment of inertia at point $P$ so $I_{P}=\int a^2 dm$ > - The center of mass in terms of the $x$ and $y$ direction are $\vec{x}_{cm} = \frac{1}{M} \int x dm=0$ & $\vec{y}_{cm} = \frac{1}{M} \int y dm=0$ > If you get confused with the variables refer back to the diagrams above. $ \begin{align*} I_{P} &= \int a^2 \,dm \\ \end{align*} $ Using [[Pythagorean Theorem|pythagorean theorem]], $a^2 = a_{x}^2 + a_{y}^2$ ([[SoME4 Competition Submission - Parallel Axis Theorem#Important Diagrams|Fig. 2]]). As well from [[SoME4 Competition Submission - Parallel Axis Theorem#Important Diagrams|Figure 3]], $a_{x}+x=x_{p}$ and $y+a_{y}=y_{p}$, so then $a_{x}=x_{p}-x$ & $a_{y}=y_{p}-y$. $ \begin{align*} I_{P} &= \int a^2 \, dm \\ &= \int a_{x}^2 + a_{y} ^2 \, dm \\ &= \int (x_{p}-x)^2 + (y_{p}-y)^2 \, dm \quad \text{Expand}\\ &= \int (x_{p}^2 -2x_{p}x + x^2 + y_{p} ^2 - 2y_{p} y + y^2) \, dm\quad \text{ Group}\\ &= \int (x_{p} ^2 + y_{p} ^2) \, dm - \int2x_{p} x \, dm -\int 2y_{p}y\,dm+ \int(x^2+y^2)\,dm \\ \end{align*} $ > [!warning] Recall Remember that $x_{p},y_{p}$ are **fixed variables**, they dont change, and $x,y$ **are variables**. Therefore, fixed variables can be taken out the equation, since they act like constants. $ \begin{align*} I_{P} &= \int (x_{p} ^2 + y_{p} ^2)\,dm - \int2x_{p} x\,dm - \int 2y_{p}y\,dm+ \int(x^2+y^2)\,dm \\ &= (x_{p}^2+y_{p}^2) \int dm - 2x_{p}\int x\,dm - 2y_{p} \int y\,dm + \int(x^2+y^2)\,dm \end{align*} $ > [!warning] Key Reminder Remember that $\vec{x}_{cm} = \frac{1}{M} \int x dm=0$ & $\vec{y}_{cm} = \frac{1}{M} \int y dm=0$ and since $M$ **is the total mass of the object**, it is a **constant**. So the only way for this to be equal to zero is that $\int xdm=0$ and $\int ydm=0$ $ \begin{align*} I_{P}&= (x_{p}^2+y_{p}^2) \int dm - 2x_{p}\int xdm - 2y_{p} \int ydm + \int(x^2+y^2)dm \\ &= (x_{p}^2+y_{p}^2) \int dm - 2\cancel{ x_{p}\int xdm }^{\space0} - \cancel{ 2y_{p} \int ydm }^{\space 0}+ \int(x^2+y^2)dm \end{align*} $ > [!warning] Final Recall Using [[Pythagorean Theorem|Pythagorean theorem]], $x^2+y^2=r^2$ and $x_{p}^2+y_{p}^2=D^2$ ([[SoME4 Competition Submission - Parallel Axis Theorem#Derivation|Fig. 3.]]) Remember that $I_{cm}=\int r^2dm$ where $\int dm=M$ $ \begin{align*} I_{p }&= (x_{p}^2+y_{p}^2) \int dm - 2\cancel{ x_{p}\int xdm }^{\space0} - \cancel{ 2y_{p} \int ydm }^{\space 0}+ \int(x^2+y^2)dm \\ &= (D^2) \int dm +\int r^2 dm\\ &= D^2 M + I_{cm} \\ \end{align*} $ ## Final Result So then our simplified equation for the parallel axis theorem is $ I_{P}= I_{cm} + MD^2 $ Take a moment to appreciate what just happened. The result for $I_{P}$ is unknown. The starting of the equation started with a set of unknowns, $I_{P} = \int a^2 dm$, which expanded into long equations, mixed with knowns and unknowns like this $\int (x_{p} ^2 + y_{p} ^2) dm - \int2x_{p} x dm -\int 2y_{p}ydm+ \int(x^2+y^2)dm$. Then using pure logical assumptions from our ideal diagram and setup **(without using complex algebraic manipulation)** the result was **simplifed to just three terms**. So when you force an object, like our 2D rock, away from the center of mass, it turns out the new moment of inertia is just a simple equation $I_{P}=I_{cm}+MD^2$ # Closing Motivation Let's return to our broom. Spinning it at the center is easy because that's where the center of mass lies. Spinning it on its edge is harder, since the moment of inertia grows by $MD^2$. The parallel axis theorem explains why some axes make rotation effortless, while others resists through a mathematical viewpoint.