34.1: Discrete Masses
- Page ID
- 92262
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)For a collection of discrete point masses in one dimension, the center of mass \(x_{\mathrm{cm}}\) is defined to be
\[x_{\mathrm{cm}}=\frac{\sum_{i} m_{i} x_{i}}{\sum_{i} m_{i}}\]
where the summations are over all of the point masses. This is just the weighted average of the positions of the masses, where the "weights" are the masses. Note that the denominator is the total mass of all the point masses.
Suppose there is a mass of \(3 \mathrm{~kg}\) at \(x=1 \mathrm{~m}\), a mass of \(2 \mathrm{~kg}\) at the origin, and a mass of \(4 \mathrm{~kg}\) at \(x=2 \mathrm{~m}\). Where is the center of mass?
Solution
Let's put the data in a table:
| \(i\) | \(m_{i}(\mathrm{~kg})\) | \(x_{i}(\mathrm{~m})\) |
|---|---|---|
| 1 | 3.0 | 1.0 |
| 2 | 2.0 | 0.0 |
| 3 | 4.0 | 2.0 |
Then by Eq. \(\PageIndex{1}\),
\[x_{\mathrm{cm}} =\frac{(3 \mathrm{~kg})(1 \mathrm{~m})+(2 \mathrm{~kg})(0 \mathrm{~m})+(4 \mathrm{~kg})(2 \mathrm{~m})}{3
\mathrm{~kg}+2 \mathrm{~kg}+4 \mathrm{~kg}} \]
\[ =1.222 \mathrm{~m} .\]
In two or three dimensions, the \(x, y\), and \(z\) coordinates of the center of mass are calculated independently:
\[ x_{\mathrm{cm}}=\frac{\sum_{i} m_{i} x_{i}}{\sum_{i} m_{i}} \]
\[ y_{\mathrm{cm}}=\frac{\sum_{i} m_{i} y_{i}}{\sum_{i} m_{i}} \]
\[ z_{\mathrm{cm}}=\frac{\sum_{i} m_{i} z_{i}}{\sum_{i} m_{i}}\]
In two dimensions: suppose there is a mass of \(3 \mathrm{~kg}\) at \((x, y)=(1,3) \mathrm{m}\), a mass of \(2 \mathrm{~kg}\) at the origin, and a mass of \(4 \mathrm{~kg}\) at \((x, y)=(5,-1) \mathrm{m}\). Where is the center of mass?
Solution
Let's put the data in a table:
| \(i\) | \(m_{i}(\mathrm{~kg})\) | \(x_{i}(\mathrm{~m})\) | \(y_{i}(\mathrm{~m})\) |
|---|---|---|---|
| 1 | 3.0 | 3.0 | 3.0 |
| 2 | 2.0 | 0.0 | 0.0 |
| 3 | 4.0 | 5.0 | -1.0 |
Then by Eqs. \(\PageIndex{4}\) and \(\PageIndex{5}\),
\[x_{\mathrm{cm}} =\frac{(3 \mathrm{~kg})(1 \mathrm{~m})+(2 \mathrm{~kg})(0 \mathrm{~m})+(4 \mathrm{~kg})(5 \mathrm{~m})}{3 \mathrm{~kg}+2 \mathrm{~kg}+4 \mathrm{~kg}} \]
\[ =2.556 \mathrm{~m} .\]
and
\[y_{\mathrm{cm}} =\frac{(3 \mathrm{~kg})(3 \mathrm{~m})+(2 \mathrm{~kg})(0 \mathrm{~m})+(4 \mathrm{~kg})(-1 \mathrm{~m})}{3 \mathrm{~kg}+2 \mathrm{~kg}+4 \mathrm{~kg}} \]
\[ =0.556 \mathrm{~m} .\]
The center of mass is at \(\left(x_{\mathrm{cm}}, y_{\mathrm{cm}}\right)=(2.556,0.556) \mathrm{m}\).