From b01082797c20d562c1037de43a816c850cdb3077 Mon Sep 17 00:00:00 2001
From: eggy <eggyrules@gmail.com>
Date: Wed, 2 Nov 2022 22:56:14 -0400
Subject: [PATCH] ece105: add com

---
 docs/ce1/ece105.md | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)

diff --git a/docs/ce1/ece105.md b/docs/ce1/ece105.md
index 413af55..b10cd83 100644
--- a/docs/ce1/ece105.md
+++ b/docs/ce1/ece105.md
@@ -35,3 +35,30 @@ $$\frac{dp}{dt} = \frac{mdv}{dt} + \frac{vdm}{dt}$$
 Impulse is actually the change of momentum over time.
 
 $$\vec J = \int^{p_f}_{p_i}d\vec p$$
+
+## Centre of mass
+
+The centre of mass $x$ of a system is equal to the average of the centre of masses of its components relative to a defined origin.
+
+$$x_{cm} = \frac{m_1x_1 + m_2x_2 + ... + m_nx_n}{m_1 + m_2 + ... + m_n}$$
+
+To determine the centre of mass of a system with a hole, the hole should be treated as negative mass. If the geometry of the system is **symmetrical**, the centre of mass is also symmetrical in the x and y dimensions.
+
+For each mass, its surface density $\sigma$ is equal to:
+
+$$
+\sigma = \frac{m}{A} \\
+m = \sigma A
+$$
+
+Holes have negative mass, i.e., $m = -\sigma A$.
+
+For a **one-dimensional** hole, the linear mass density uses a similar formula:
+
+$$
+\lambda =\frac{m}{L} \\
+\lambda = \frac{dm}{dx}
+$$
+
+This means that a hole in a rod can use a different formula:
+$$x_{cm} = \frac{1}{M}\int^M_0 xdm$$