# 11. Mass and Center of Mass

## b. Center of Mass

## 2. Bar with Non-Uniform Density

On this page, we compute the center of mass of a bar whose mass is not uniformly distributed over the bar, i.e. when the linear density \(\delta(x)\) is a function of the position \(x\) along the bar.

Find the center of mass of a bar between \(x=a\) and \(x=b\) whose linear density is \(\delta(x)\).

As when we computed the total mass, we first cut up the bar into \(n\) small pieces each of width \(\Delta x=\dfrac{b-a}{n}\).

Since the density is not constant, we approximate the density of the
\(i^\text{th}\) piece as \(\delta(x_i^*)\) where \(x_i^*\)
is a point in the \(i^\text{th}\) region. Then the mass of the
\(i^\text{th}\) region is \(m_i=\delta(x_i^*)\Delta x\).
If we think of the bar as a balance beam as on the
previous page, with masses
\(m_1,m_2,\ldots m_n\) located at \(x_1^*,x_2^*,\ldots x_n^*\)
then the center of mass is approximately at:
\[
\bar{x}\approx \dfrac{M_1}{M}
=\dfrac{\sum\limits_{i=1}^n m_i x_i^*}{\sum\limits_{i=1}^n m_i}
=\dfrac{\sum\limits_{i=1}^n x_i^*\delta(x_i^*)\Delta x}
{\sum\limits_{i=1}^n \delta(x_i^*)\Delta x}
\]
As the length of each piece gets smaller, the approximation gets better.
So the center of mass of the bar is obtained by taking the limit as the
number of pieces becomes large (\(n\to\infty\)) and
hence the length of each piece gets small (\(\Delta x\to 0\)):

(Note: The limit of a quotient is the quotient of the limits.)
\[
\bar{x}
=\dfrac{\displaystyle\lim_{n\to\infty}\sum\limits_{i=1}^n
x_i^*\delta(x_i^*)\Delta x}
{\displaystyle\lim_{n\to\infty}\sum\limits_{i=1}^n
\delta(x_i^*)\Delta x}
\]
The limit of the sum in the denominator is the integral
\(\displaystyle \int_a^b \delta(x)\,dx\) which we recognize as the
total mass:
\[
M=\int_a^b \delta(x)\,dx
\]
The limit of the sum in the numerator is the integral
\(\displaystyle \int_a^b x \delta(x)\,dx\) which is called the
first moment of the mass:
\[
M_1=\int_a^b x\delta(x)\,dx
\]
In terms of these, the center of mass is:
\[
\bar{x}=\dfrac{M_1}{M}
\]

When you hear the word **first** moment, you naturally ask:
“What are the second and higher moments?”

The second moment of the mass is:
\[
M_2=\int_a^b x^2\delta(x)\,dx
\]
and similarly, the
\(k^\text{th}\) moment of the mass is:
\[
M_k=\int_a^b x^k\delta(x)\,dx
\]
These are used in various physical applications, for example, the
second moment occurs in the formulas for
moment of inertia
on the next page.

The center of mass of a bar with linear density \(\delta(x)\) between \(x=a\) and \(x=b\) is: \[ \bar{x}=\dfrac{M_1}{M} \] where the total mass is: \[ M=\int_a^b \delta(x)\,dx \] and the first moment of the mass is: \[ M_1=\int_a^b x\delta(x)\,dx \]

Find the center of mass of an \(8\,\text{m}\) bar whose linear density is \(\delta(x)=(1+x)\dfrac{\text{kg}}{\text{m}}\) where \(x\) is measured from one end as discussed in a previous example.

The total mass was found to be \(M=40\,\text{kg}\). The first moment is: \[\begin{aligned} M_1&=\int_0^8 x\delta(x)\,dx =\int_0^8 x(1+x)\,dx \\ &=\left[\dfrac{x^2}{2}+\dfrac{x^3}{3}\right]_{x=0}^8 =\dfrac{64}{2}+\dfrac{512}{3}=\dfrac{608}{3} \text{kg-m} \end{aligned}\] So the center of mass is: \[ \bar{x}=\dfrac{M_1}{M}=\dfrac{608\,\text{kg-m}}{3}\dfrac{1}{40\,\text{kg}} =\dfrac{76}{15}\text{m}\approx 5.07\text{m} \]

You should always check that your center of mass is within the bar. In this case, \(\bar{x}\approx 5.07\,\text{m}\) is within the \(8\,\text{m}\) bar.

Now you do it:

Find the center of mass of a \(10\,\text{ft}\) bar whose linear weight density is \(\delta(x)=\dfrac{1}{1+x}\dfrac{\text{lb}}{\text{ft}}\) where \(x\) is measured from one end as discussed in a previous exercise.

Note: If we are given a weight density instead of a mass density, then we use the weight \(W\) and its first moment \(W_1\) instead of the mass \(M\) and its first moment \(M_1\).

\(\bar{x}=\dfrac{10-\ln(11)}{\ln(11))}\text{ft} \approx 3.170\,\text{ft}\)

The total weight was found in a previous exercise to be \(W=\ln(11)\,\text{lb}\). The first moment of the weight is: \[\begin{aligned} W_1&=\int_0^{10} \dfrac{x}{1+x}\,dx =\int_0^{10} \dfrac{1+x-1}{1+x}\,dx \\ &=\int_0^{10} 1-\dfrac{1}{1+x}\,dx =[x-\ln(1+x)]_0^{10} \\ &=[10-\ln(11)]-[0-\ln(1)] =10-\ln(11)\,\text{ft-lb} \end{aligned}\] So the center of mass is: \[ \bar{x}=\dfrac{W_1}{W} =\dfrac{10-\ln(11)\,\text{ft-lb}}{\ln(11))\,\text{lb}}\approx 3.170\,\text{ft} \]

We check the center of mass \(\bar{x}\approx 3.170\,\text{ft}\) is within the \(10\,\text{ft}\) bar.

You can practice computing Mass and Center of Mass of a Bar by using the following Maplet (requires Maple on the computer where this is executed):

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