9.6 Center of mass (Page 3/13)

University physics volume 1 Page 3 / 13

which follows because the derivative of a sum is equal to the sum of the derivatives.

Now, ${\vec{p}}_{j}$ is the momentum of the j th particle. Defining the positions of the constituent particles (relative to some coordinate system) as ${\vec{r}}_{j} = (x_{j}, y_{j}, z_{j})$ , we thus have

{\vec{p}}_{j} = m_{j} {\vec{v}}_{j} = m_{j} \frac{d {\vec{r}}_{j}}{d t} .

Substituting back, we obtain

\begin{matrix} M \vec{a} & = \frac{d}{d t} \sum_{j = 1}^{N} m_{j} \frac{d {\vec{r}}_{j}}{d t} \\ = \frac{d^{2}}{d t^{2}} \sum_{j = 1}^{N} m_{j} {\vec{r}}_{j} . \end{matrix}

Dividing both sides by M (the total mass of the extended object) gives us

\vec{a} = \frac{d^{2}}{d t^{2}} (\frac{1}{M} \sum_{j = 1}^{N} m_{j} {\vec{r}}_{j}) .

Thus, the point in the object that traces out the trajectory dictated by the applied force in [link] is inside the parentheses in [link] .

Looking at this calculation, notice that (inside the parentheses) we are calculating the product of each particle’s mass with its position, adding all N of these up, and dividing this sum by the total mass of particles we summed. This is reminiscent of an average; inspired by this, we’ll (loosely) interpret it to be the weighted average position of the mass of the extended object. It’s actually called the center of mass of the object. Notice that the position of the center of mass has units of meters; that suggests a definition:

{\vec{r}}_{CM} \equiv \frac{1}{M} \sum_{j = 1}^{N} m_{j} {\vec{r}}_{j} .

So, the point that obeys [link] (and therefore [link] as well) is the center of mass of the object, which is located at the position vector ${\vec{r}}_{CM}$ .

It may surprise you to learn that there does not have to be any actual mass at the center of mass of an object. For example, a hollow steel sphere with a vacuum inside it is spherically symmetrical (meaning its mass is uniformly distributed about the center of the sphere); all of the sphere’s mass is out on its surface, with no mass inside. But it can be shown that the center of mass of the sphere is at its geometric center, which seems reasonable. Thus, there is no mass at the position of the center of mass of the sphere. (Another example is a doughnut.) The procedure to find the center of mass is illustrated in [link] .

An illustration of finding the center of mass of three particles. Figure a shows the locations of the three particles in he x y plane. m 1 is in the second quadrant. Vector r 1 starts at the origin and extends to the location of m 1. m 2 is in the first quadrant. Vector r 2 starts at the origin and extends to the location of m 2. m 1 is in the fourth quadrant. Vector r 3 starts at the origin and extends to the location of m 3. Vector r 1 is the shortest of the vectors in the diagram, and r 2 is the longest. Figure b shows the vectors m 1 r 1, m 2 r 2 and m 3 r 3. Vector m 1 r 1 points in the same direction as vector r 1 in figure a, but is longer than r 1. Vector m 2 r 2 points in the same direction as vector r 1 in figure a, but is shorter than r 2. . Vector m 3 r 3 points in the same direction as vector r 3 in figure a, but is shorter than r 3. Vector m 1 r 1 is the longest vector in the diagram. Vectors m 2 r 2 and m 3 r 3 appear to be of equal length. Figure c shows the vector sum of m 1 r 1, m2 r 2 and m 3 r 3, which have been drawn in blue and placed head to tail. The red vector m 1 r 1 plus m 2 r 2 plus m 3 r 3 is the vector from the tail of m 1 r 1 to the head of m 3 r 3. Figure d shows the red vector m 1 r 1 plus m 2 r 2 plus m 3 r 3 all divded by the sum m 1 plus m 2 plus m 3. This vector is in the same direction as the vecor m 1 r 1 plus m 2 r 2 plus m 3 r 3 in figure c, but shorter. — Finding the center of mass of a system of three different particles. (a) Position vectors are created for each object. (b) The position vectors are multiplied by the mass of the corresponding object. (c) The scaled vectors from part (b) are added together. (d) The final vector is divided by the total mass. This vector points to the center of mass of the system. Note that no mass is actually present at the center of mass of this system.

Since ${\vec{r}}_{j} = x_{j} \hat{i} + y_{j} \hat{j} + z_{j} \hat{k}$ , it follows that:

r_{CM, x} = \frac{1}{M} \sum_{j = 1}^{N} m_{j} x_{j}

r_{CM, y} = \frac{1}{M} \sum_{j = 1}^{N} m_{j} y_{j}

r_{CM, z} = \frac{1}{M} \sum_{j = 1}^{N} m_{j} z_{j}

and thus

\begin{matrix} {\vec{r}}_{CM} = r_{CM, x} \hat{i} + r_{CM, y} \hat{j} + r_{CM, z} \hat{k} \\ r_{CM} = | {\vec{r}}_{CM} | = {(r_{CM, x}^{2} + r_{CM, y}^{2} + r_{CM, z}^{2})}^{1 / 2} . \end{matrix}

Therefore, you can calculate the components of the center of mass vector individually.

Finally, to complete the kinematics, the instantaneous velocity of the center of mass is calculated exactly as you might suspect:

{\vec{v}}_{CM} = \frac{d}{d t} (\frac{1}{M} \sum_{j = 1}^{N} m_{j} {\vec{r}}_{j}) = \frac{1}{M} \sum_{j = 1}^{N} m_{j} {\vec{v}}_{j}

and this, like the position, has x -, y -, and z -components.

To calculate the center of mass in actual situations, we recommend the following procedure:

Problem-solving strategy: calculating the center of mass

The center of mass of an object is a position vector. Thus, to calculate it, do these steps:

Define your coordinate system. Typically, the origin is placed at the location of one of the particles. This is not required, however.
Determine the x , y , z -coordinates of each particle that makes up the object.
Determine the mass of each particle, and sum them to obtain the total mass of the object. Note that the mass of the object at the origin must be included in the total mass.
Calculate the x -, y -, and z -components of the center of mass vector, using [link] , [link] , and [link] .
If required, use the Pythagorean theorem to determine its magnitude.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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