Center of Mass

Every system of objects has a center of mass. It is the weighted average of the location of mass in an object. It is also the point between the parts of an object or between two or more objects where they move with constant velocity if there is no external force acting on the system.

In locating the center of mass of an object or system of objects, the mass of the objects and the length or distance must be clearly identified. Consider the system of objects below.

Given two objects with masses, $$m_1$$ and $$m_2$$, the center of mass is the point in between the objects where they are balanced and move with constant velocity. Mathematically, it can be expressed as

$$x_{cm}=\frac {m_1x_1\; +\; m_2x_2}{m_1\;+\;m_2}$$

In the case of three or more objects lying on the different position, the center of mass along the vertical will also be calculated.

$$y_{cm}=\frac {m_1y_1\; +\; m_2y_2\;+\;m_3y_3}{m_1\;+\;m_2\;+\;m_3}$$

The equation holds for system of four objects and more.

Example 1.

What is the center of mass of the system?

Given:

$$m_1=5\;kg\\ x_1=4 \;m\\ m_2=10\;kg\\ x_2=7\;m$$

Solution:

To solve for the center of mass of the objects, we use the equation $$x_{cm}=\frac {m_1x_1\; +\; m_2x_2}{m_1\;+\;m_2}$$.

$$x_{cm}=\frac {(5\;kg)(4\;m)\; +\; (10\;kg)(7\;m)}{5\;kg\;+\;10\;kg}$$

$$x_{cm}=\frac {90\;kg\cdot m}{15\;kg}=6\;m$$

The center of mass of the system is at $$6\;m$$.

Example 2.

A hammer is rotating at constant velocity. If the hammer's head has  a mass of $$0.7\;kg$$ and length of $$0.03\;m$$ and the handle's mass is $$0.11\;kg$$ with a length of $$0.35\;m$$. Locate the center of mass of the hammer.

Given:

$$m_{head}=0.7\;kg\\ l_{head}=0.03\;m\\ m_{handle}=0.11\;kg\\ l_{handle}=0.35\;N$$

Solution:

$$x_{cm}=\frac {m_{head}x_{head}\; +\; m_{handle}x_{handle}}{m_{head}\;+\;m_{handle}}$$

$$x_{cm}=\frac {(0.7\;kg)(0.03\;m)\; +\; (0.11\;kg)(0.35\;m)}{0.7\;kg\;+\;0.11\;kg}$$

$$x_{cm}=\frac {0.0595\;kg\cdot m}{0.81\;kg}=0.07\;m$$

Example 3.

Solve for the center of mass of the system of masses $$a, b,\;\text{and}\;c$$.

Given:

$$a=20\;kg\\ x_a=1\;m\\ y_a=1\;m\\ b=15\;kg\\ x_b=3\;m\\ y_b=1.5\;m\\ c=8\;kg\\ x_c=4\;m\\ y_c=2.5\;m$$

Solution 1:

Solve for the center of mass along the horizontal using the equation $$x_{cm}=\frac {ax_a\; +\; bx_b\;+\;cx_c}{a\;+\;b\;+\;c}$$.

$$x_{cm}=\frac {(20\;kg)(1\;m)\; +\; (15\;kg)(3\;m)\;+\;(8\;kg)(4\;m)}{20\;kg\;+\;15\;kg\;+\;8\;kg}$$

$$x_{cm}=\frac {20\;kg\cdot m\; +\; 45\;kg\cdot m\;+\;32\;kg\cdot m}{43\;kg}$$

$$x_{cm}=\frac {97\;kg\cdot m}{43\;kg}$$

$$x_{cm}=2.26\;m$$

Solution 2.

Solve for the center of mass along the y-axis by $$y_{cm}=\frac {ay_a\; +\; by_b\;+\;cy_c}{a\;+\;b\;+\;c}$$.

$$y_{cm}=\frac {(20\;kg)(1\;m)\; +\; (15\;kg)(1.5\;m)\;+\;(8\;kg)(2.5\;m)}{20\;kg\;+\;15\;kg\;+\;8\;kg}$$

$$y_{cm}=\frac {20\;kg\cdot m\; +\; 22.5\;kg\cdot m\;+\;20\;kg\cdot m}{43\;kg}$$

$$y_{cm}=\frac {62.5\;kg\cdot m}{43\;kg}$$

$$y_{cm}=1.45\;m$$

The center of mass of the system can be expressed as $$(x_{cm},y_{cm})=(2.26\;\text{m},1.45\;\text{m})$$.