Centers of mass
The total mass of a rigid body is as follows:
Total mass of a rigid body.
\[ m = \iiint_{\mathcal{B}} \rho \, dV \]
The mass of an infinitesimally small element of the rigid body $dm$ assuming constant density is: \[ dm = \rho \, dV \] Therefore, integrating over the entire volume of the body gives us the expression above.
The center of mass of a rigid body can be calculated as follows:
Center of mass $C$ of a rigid body.
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\[\begin{gathered} \vec{r}_C = \frac{1}{m}\iiint_{\mathcal{B}} \rho \vec{r} \, dV \\ \end{gathered}\] |
$\vec{r}$ is the vector from a reference origin $O$ to $dV$.
The integral term $\iiint_{\mathcal{B}} \rho \vec{r} \, dV$ is known as the "first moment of mass".
The center of mass of a collection of particles is given by the finite sum: \[ \vec{r}_C = \frac{1}{m} \sum_{i=1}^{N} m_i \, \vec{r}_i \]
Therefore, for a continuous mass distribution and infinite number of particles, the sum becomes an integral and the masses become the mass of an infinitesimally small element on the rigid body: \[ \vec{r}_C = \frac{1}{m}\iiint_{\mathcal{B}} \vec{r} dm \]
Where $dm = \rho \, dV$ from the differential of #rcm-tm. Substituting in: \[\begin{aligned} \vec{r}_C &= \frac{1}{m}\iiint_{\mathcal{B}} \vec{r} dm \\ &= \frac{1}{m}\iiint_{\mathcal{B}} \vec{r} \rho \, dV \\ &= \frac{1}{m}\iiint_{\mathcal{B}} \rho \vec{r} \, dV \\ \end{aligned}\]
Example Problem: Center of mass of a right-triangular plate.
A solid uniform right-triangular plate with mass $m$ has width $w$, height $h$. Where is the center of mass $\vec{r}_C$ located?
In order to figure out where the center of mass is located, we will need to choose an origin. The origin chosen in this instant is where the base and height begin, as shown. Using the center of mass formula will allow us to calculate the vector shown below (i.e we are calculating $\vec{r}_{OC}$).
Using the formula #rcm-cm: \[\begin{aligned} \vec{r}_C &= \frac{1}{m} \iiint_{\mathcal{B}} \rho \vec{r} \, dV \\ &= \frac{1}{m} \iint_{A} \rho \vec{r} \, tdA \\ \end{aligned}\]
Where $A$ represents the surface area of the plate, and $t$ is the thickness. We will assume the thickness is 1, as we are only interested in the center of mass in the $x$-$y$ plane. Our triple integral with $dV$ instead becomes a double integral with $dV = t dA$. We therefore have: \[ \vec{r}_C = \frac{1}{m} \int_{x_1}^{x_2} \int_{y_1}^{y_2} \rho \vec{r} \, dA \]
We will redraw the figure and show the $dA$ element. We will also find the $x$-coordinate and $y$-coordinate ranges for the body.
Examining the figure above, we can see that the $x$-coordinate range is $0$ to $w$. Our $y$-coordinate range has a dependence on the $x$-value, and the equation of the hypotenuse is $y = h - \frac{h}{w}x$, using the point-slope formula. Therefore, the $y$-coordinate range is $0$ to $h - \frac{h}{w}x$. As a result of this, we will have to integrate with respect to $y$ first. All in all, our center of mass equation becomes: \[ \vec{r}_C = \frac{1}{m} \int_{0}^{w} \int_{0}^{h - \frac{h}{w}x} \rho \vec{r} \, dydx \]
where $\vec{r}$ is the distance to the $dA$ element. We can see the coordinates of $dA$ are $[x,y]$. Thus, $\vec{r} = x\hat{\imath} + y\hat{\jmath}$. Therefore: \[\begin{aligned} \vec{r}_C &= \frac{1}{m} \int_{0}^{w} \int_{0}^{h - \frac{h}{w}x} \rho \left(x\hat{\imath} + y\hat{\jmath}\right) \, dydx \\ &= \frac{\rho}{m} \int_{0}^{w} \int_{0}^{h - \frac{h}{w}x} \left(x\hat{\imath} + y\hat{\jmath}\right) \, dydx \\ &= \frac{\rho}{m} \left[\int_{0}^{w} \int_{0}^{h - \frac{h}{w}x} \left(x\hat{\imath} \, dy\right)dx + \int_{0}^{w} \int_{0}^{h - \frac{h}{w}x} \left(y\hat{\jmath} \, dy\right)dx\right] \\ &= \frac{\rho}{m} \left[\int_{0}^{w} x\left(h - \frac{h}{w}x\right)\hat{\imath}dx + \int_{0}^{w} \frac{1}{2}\left(h - \frac{h}{w}x\right)^2\hat{\jmath}dx\right] \\ &= \frac{\rho}{m} \left[\frac{w^2 h}{6}\hat{\imath} + \frac{wh^2}{6}\hat{\jmath}\right] \end{aligned}\]
where $\frac{\rho}{m} = \frac{1}{A} = \frac{1}{\frac{1}{2}wh} = \frac{2}{wh}$. Substituting that in yields: \[ \vec{r}_C = \frac{1}{3}w \hat{\imath} + \frac{1}{3}h \hat{\jmath} \]
Meaning the center of mass for a right-triangular plate is located at 1/3rd of the width, and 1/3rd of the height. This formula has been used to calculate the center of mass of different common shapes.
Finding the center of mass allows us to treat complex shapes as point-masses with all their mass at the center of mass.
To find the center of mass of a body made up of composite shapes, we simply do the weighted average of each body by using the fact above.
Center of mass $C$ of composite bodies.
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\[ \vec{r}_C = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \] |
$\vec{r}_{i}$ is the center of mass of the $i$th body. The whole body $\mathcal{B}$ is composed of sub-bodies $\mathcal{B}_1$ and $\mathcal{B}_2$
Finding the center of mass of the whole body follows directly from the fact that the integral over the whole body is the sum of the integrals over the sub-bodies. That is, for $\mathcal{B} = \mathcal{B}_1 \cup \mathcal{B}_2$ (and $\mathcal{B}_1$ not overlapping with $\mathcal{B}_2$), we have: \[\begin{aligned} \vec{r}_C &= \frac{1}{M}\iiint_\mathcal{B} \rho \vec{r} \, dV \\ &= \frac{1}{M}\iiint_{\mathcal{B}_1 \cup \mathcal{B}_2} \rho \vec{r} \, dV \\ &= \frac{1}{M}\left[\iiint_{\mathcal{B}_1} \rho_1 \vec{r}_1 \, dV + \iiint_{\mathcal{B}_2} \rho_2 \vec{r}_2 \, dV \right] \\ &= \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{M} \\ &= \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \end{aligned}\]
Example Problem: Center of mass of an L-shaped plate.
A solid uniform L-shaped plate has mass $m$ and dimensions as shown. What is the location of the center of mass?
In order to figure out where the center of mass is located, we will need to choose an origin. The origin chosen here is the corner shown.
Then we must realize that the L-shape can be decomposed into four square plates, as shown. Each square plate has the same mass, so: \[ m_i = \frac{1}{4}m \]
Using the addition formula #rcm-cb: \[ \vec{r}_C = \frac{1}{m} \sum_{i=1}^{4} m_i \vec{r}_i \]
where: \[\begin{aligned} \vec{r}_1 &= d\hat{\imath} + 3d\hat{\jmath} \\ \vec{r}_2 &= d\hat{\imath} + d\hat{\jmath} \\ \vec{r}_3 &= 3d\hat{\imath} + d\hat{\jmath} \\ \vec{r}_4 &= 5d\hat{\imath} + d\hat{\jmath} \end{aligned}\]
Substituting in: \[\begin{aligned} x_C = \frac{\frac{1}{4}md + \frac{1}{4}md + \frac{3}{4}md + \frac{5}{4}md}{m} = \frac{5}{2}d \\ y_C = \frac{\frac{3}{4}md + \frac{1}{4}md + \frac{1}{4}md + \frac{1}{4}md}{m} = \frac{3}{2}d \end{aligned}\]
Thus, the center of mass of the L-shape plate: \[ \vec{r}_C = \frac{5}{2}d \hat{\imath} + \frac{3}{2}d \hat{\jmath} \]
And its location on the plate is shown below.
Basic shapes
The centers of mass listed below are all computed directly from the integral #rcm-cm. Note the center of mass provided is the vector from point $O$ (the reference origin).
Rectangular plate: center of mass
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\[ \vec{r}_C = \frac{\ell_x}{2}\hat{\imath} + \frac{\ell_y}{2}\hat{\jmath} \] |
\[\begin{aligned} \vec{r}_C &= \frac{1}{m}\iiint_\mathcal{B} \rho \vec{r} \, dV \\ &= \frac{1}{m}\iint_{A} \rho \vec{r} \, dA \\ &= \frac{\rho}{m}\int_{0}^{\ell_x}\int_{0}^{\ell_y} \left[x\hat{\imath} + y\hat{\jmath}\right] \, dydx \\ &= \frac{\rho}{m}\int_{0}^{\ell_x} \left[xy\hat{\imath} + \frac{y^2}{2}\hat{\jmath}\right]_{y = 0}^{y = \ell_y} dx \\ & = \frac{\rho}{m}\int_{0}^{\ell_x} \left[x\ell_y \hat{\imath} + \frac{\ell^2_y}{2}\hat{\jmath}\right]dx \\ & = \frac{\rho}{m} \left[\frac{x^2}{2}\ell_y \hat{\imath} + \frac{\ell^2_y}{2}x\right]_{x = 0}^{x = \ell_x} \\ & = \frac{\rho}{m} \left[\frac{\ell_y \, \ell^2_x}{2} \hat{\imath} + \frac{\ell^2_y \, \ell_x}{2} \hat{\jmath}\right] \\ \end{aligned}\]
The total mass of the plate is $m = \rho A = \rho \ell_x \ell_y$. Thus we have: \[ \vec{r}_C = \frac{\ell_x}{2}\hat{\imath} + \frac{\ell_y}{2}\hat{\jmath} \]
Triangular plate: center of mass
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\[ \vec{r}_C = \frac{x_P + x_Q}{3}\hat{\imath} + \frac{y_Q}{3}\hat{\jmath} \] |
\[\begin{aligned} \vec{r}_C &= \frac{1}{m}\iiint_\mathcal{B} \rho \vec{r} \, dV \\ &= \frac{1}{m}\iint_{A} \rho \vec{r} \, dA \\ \end{aligned}\]
It is convenient to do a linear transformation from $x$-$y$ plane to the $u$-$v$ plane, to transform the triangle into a right-triangle with $b = 1$, $h = 1$: \[\begin{aligned} x &= au + bv \\ y &= cu + dv \\ \end{aligned}\]
Or solve the following system of equations: \[\begin{aligned} x_P &= a(1) + b(0) \\ x_Q &= a(0) + b(1) \\ \\ y_P &= c(1) + d(0) \\ y_Q &= c(0) + d(1) \\ \end{aligned}\]
We find that the linear transformation is: \[\begin{aligned} x &= x_P u + x_Q v \\ y &= y_P u + y_Q v \\ \end{aligned}\]
The Jacobian of this coordinate transformation is: \[\begin{aligned} J &= \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} x_P & x_Q \\ y_P & y_Q \end{vmatrix} = x_P y_Q - x_Q y_P \end{aligned}\]
Starting with the $x$-coordinate: \[\begin{aligned} x_C &= \frac{\rho}{m}\int_{0}^{1}\int_{0}^{1 - u} (x_P u + x_Q v) \, J \, dvdu \\ &= \frac{\rho}{m}\int_{0}^{1}\int_{0}^{1 - u} (x_P u + x_Q v) \, (x_P y_Q - x_Q y_P) \, dvdu \\ &= \frac{\rho}{m}(x_P y_Q - x_Q y_P)\int_{0}^{1}\int_{0}^{1 - u} (x_P u + x_Q v) \, dvdu \\ &= \frac{\rho}{m}(x_P y_Q - x_Q y_P)\int_{0}^{1}x_P u\left[v\right]_{v = 0}^{v = 1-u} + \frac{x_Q}{2}\left[v^2\right]_{v = 0}^{v = 1-u}du \\ &= \frac{\rho}{m}(x_P y_Q - x_Q y_P)\int_{0}^{1} x_P u(1 - u) + \frac{x_Q}{2}(1 - u)^2 du \\ &= \frac{\rho}{m}(x_P y_Q - x_Q y_P) \left(\frac{x_P}{2} \left[u^2\right]_{0}^{1} - \frac{x_P}{3}\left[u^3\right]_{0}^{1} + \frac{x_Q}{2}\left[u\right]_{0}^{1} - \frac{x_Q}{2}\left[u^2\right]_{0}^{1} + \frac{x_Q}{6}\left[u^3\right]_{0}^{1}\right) \\ &= \frac{\rho}{m}(x_P y_Q - x_Q y_P) \left(\frac{x_P}{2} - \frac{x_P}{3} + \frac{x_Q}{2} - \frac{x_Q}{2} + \frac{x_Q}{6}\right) \\ &= \frac{\rho}{m}(x_P y_Q - x_Q y_P) \left(\frac{x_P + x_Q}{6}\right) \end{aligned}\]
The total mass of the plate is $m = \rho A = \frac{1}{2}\rho x_P y_Q$, and with the chosen configuration $y_P = 0$. Thus: \[ x_C = \frac{x_P + x_Q}{3} \]
Elliptical sector: center of mass
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\[ \vec{r}_C = \frac{2a\sin\theta}{3\theta}\hat{\imath} + \frac{2b\left(1 - \cos\theta\right)}{3\theta}\hat{\jmath} \] |
\[\begin{aligned} \vec{r}_C &= \frac{1}{m}\iiint_\mathcal{B} \rho \vec{r} \, dV \\ &= \frac{1}{m}\iint_{A} \rho \vec{r} \, dA \\ \end{aligned}\]
It is convenient to switch to cylindrical coordinates: \[\begin{aligned} x &= a \, r \cos\theta \\ y &= b \, r \sin\theta \\ \end{aligned}\]
The Jacobian of this coordinate transformation is: \[\begin{aligned} J &= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} a\cos\theta & -a r \sin\theta \\ b \sin\theta & b r\cos\theta \end{vmatrix} = a b r \cos^2\theta + a b r\sin^2\theta = a b r \end{aligned}\]
Therefore: \[\begin{aligned} \vec{r}_C &= \frac{1}{m}\iint_{A} \rho \vec{r} \, dx \, dy \\ &= \frac{\rho}{m}\int_{0}^{\theta} \int_{0}^{1} \left[ar\cos\theta\hat{\imath} + br\sin\theta\hat{\jmath}\right] \, J \, dr \, d\theta \\ &= \frac{\rho ab}{m}\int_{0}^{\theta} \int_{0}^{1} \left[ar^2\cos\theta\hat{\imath} + br^2\sin\theta\hat{\jmath}\right] \, dr \, d\theta \\ &= \frac{\rho ab}{m}\int_{0}^{\theta} \left[\frac{r^3}{3}\right]_{0}^{1}a\cos\theta\hat{\imath} + \left[\frac{r^3}{3}\right]_{0}^{1}b\sin\theta\hat{\jmath} \, d\theta \\ &= \frac{\rho ab}{m}\int_{0}^{\theta} \left[\frac{1}{3}a\cos\theta\hat{\imath} + \frac{1}{3}b\sin\theta\hat{\jmath}\right] \, d\theta \\ &= \frac{\rho ab}{m}\left(\left[\sin\theta\right]_{0}^{\theta}\frac{1}{3}a\hat{\imath} + \left[-\cos\theta\right]_{0}^{\theta}\frac{1}{3}b\hat{\jmath}\right) \\ &= \frac{\rho ab}{m}\left[\frac{1}{3}a\sin\theta\hat{\imath} + \frac{1}{3}b\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ \end{aligned}\]
The total mass of the sector is $m = \rho A = \frac{\theta}{2}\rho ab$. Thus: \[\begin{aligned} \vec{r}_C &= \frac{\rho ab}{m}\left[\frac{1}{3}a\sin\theta\hat{\imath} + \frac{1}{3}b\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ &= \frac{\rho ab}{\frac{\theta}{2}\rho ab}\left[\frac{1}{3}a\sin\theta\hat{\imath} + \frac{1}{3}b\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ &= \frac{2}{\theta}\left[\frac{1}{3}a\sin\theta\hat{\imath} + \frac{1}{3}b\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ &= \frac{2a\sin\theta}{3\theta}\hat{\imath} + \frac{2b\left(1 - \cos\theta\right)}{3\theta}\hat{\jmath} \end{aligned}\]
Simplified shapes
The centers of mass listed below are all special cases of the basic shapes given in Section #rcm-bs. Other special cases can be easily obtained by similar methods, or directly computing the integral.
Right triangular plate: center of mass
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\[ \vec{r}_C = \frac{b}{3}\hat{\imath} + \frac{h}{3}\hat{\jmath} \] |
See example problem on how to derive it by directly computing the integrals.
The other perhaps simpler approach is to let $x_Q = 0$ in #rcm-et, which forms a right triangle if the configuration is the same as the one shown in the figure. Thus: \[\begin{aligned} \vec{r}_C &= \frac{x_P + x_Q}{3}\hat{\imath} + \frac{y_Q}{3}\hat{\jmath} \\ &= \frac{x_P}{3}\hat{\imath} + \frac{y_Q}{3}\hat{\jmath} \\ &= \frac{b}{3}\hat{\imath} + \frac{h}{3}\hat{\jmath} \end{aligned}\]
Isosceles triangular plate: center of mass
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\[ \vec{r}_C = \frac{b}{2}\hat{\imath} + \frac{h}{3}\hat{\jmath} \] |
Using #rcm-et, we let $x_Q = \frac{x_P}{2}$, which forms an isosceles triangle if the configuration is the same as the one shown in the figure. Thus: \[\begin{aligned} \vec{r}_C &= \frac{x_P + x_Q}{3}\hat{\imath} + \frac{y_Q}{3}\hat{\jmath} \\ &= \frac{x_P + \frac{x_P}{2}}{3}\hat{\imath} + \frac{y_Q}{3}\hat{\jmath} \\ &= \frac{\frac{3}{2}x_P}{3}\hat{\imath} + \frac{y_Q}{3}\hat{\jmath} \\ &= \frac{b}{2}\hat{\imath} + \frac{h}{3}\hat{\jmath} \end{aligned}\]
Circular sector: center of mass
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\[ \vec{r}_C = \frac{2r\sin\theta}{3\theta}\hat{\imath} + \frac{2r\left(1 - \cos\theta\right)}{3\theta}\hat{\jmath} \] |
We could use the integral definition of the center of mass and compute it directly:
\[\begin{aligned} \vec{r}_C &= \frac{1}{m}\iiint_\mathcal{B} \rho \vec{r} \, dV \\ &= \frac{1}{m}\iint_{A} \rho \vec{r} \, dA \\ \end{aligned}\]
It is convenient to switch to cylindrical coordinates: \[\begin{aligned} x &= r \cos\theta \\ y &= r \sin\theta \\ \end{aligned}\]
The Jacobian of this coordinate transformation is: \[\begin{aligned} J &= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r \sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r \cos^2\theta + r \sin^2\theta = r \end{aligned}\]
Therefore: \[\begin{aligned} \vec{r}_C &= \frac{1}{m}\iint_{A} \rho \vec{r} \, dx \, dy \\ &= \frac{\rho}{m}\int_{0}^{\theta} \int_{0}^{r} \left[r\cos\theta\hat{\imath} + r\sin\theta\hat{\jmath}\right] \, J \, dr \, d\theta \\ &= \frac{\rho}{m}\int_{0}^{\theta} \int_{0}^{r} \left[r^2\cos\theta\hat{\imath} + r^2\sin\theta\hat{\jmath}\right] \, dr \, d\theta \\ &= \frac{\rho}{m}\int_{0}^{\theta} \left[\frac{r^3}{3}\right]_{0}^{r}\cos\theta\hat{\imath} + \left[\frac{r^3}{3}\right]_{0}^{r}\sin\theta\hat{\jmath} \, d\theta \\ &= \frac{\rho}{m}\int_{0}^{\theta} \left[\frac{r^3}{3}\cos\theta\hat{\imath} + \frac{r^3}{3}\sin\theta\hat{\jmath}\right] \, d\theta \\ &= \frac{\rho}{m}\left(\left[\sin\theta\right]_{0}^{\theta}\frac{r^3}{3}\hat{\imath} + \left[-\cos\theta\right]_{0}^{\theta}\frac{r^3}{3}\hat{\jmath}\right) \\ &= \frac{\rho}{m}\left[\frac{1}{3}r^3\sin\theta\hat{\imath} + \frac{1}{3}r^3\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ \end{aligned}\]
The total mass of the sector is $m = \rho A = \frac{\theta}{2}\rho r^2$. Thus: \[\begin{aligned} \vec{r}_C &= \frac{\rho}{m}\left[\frac{1}{3}r^3\sin\theta\hat{\imath} + \frac{1}{3}r^3\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ &= \frac{\rho}{\frac{\theta}{2}\rho r^2}\left[\frac{1}{3}r^3\sin\theta\hat{\imath} + \frac{1}{3}r^3\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ &= \frac{2}{\theta r^2}\left[\frac{1}{3}r^3\sin\theta\hat{\imath} + \frac{1}{3}r^3\left(1 - \cos\theta\right)\hat{\jmath}\right] \\ &= \frac{2r\sin\theta}{3\theta}\hat{\imath} + \frac{2r\left(1 - \cos\theta\right)}{3\theta}\hat{\jmath} \end{aligned}\]
Another way to reach this formula is to use #rcm-ee, and for a circular sector, let $a = b = r$.
Semi-circular sector: center of mass
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\[ \vec{r}_C = \frac{4r}{3\pi}\hat{\jmath} \] |
Using #rcm-ec, for a semi-circular sector, the angle $\theta = \pi$. Therefore: \[\begin{aligned} \vec{r}_C &= \frac{2r\sin\theta}{3\theta}\hat{\imath} + \frac{2r\left(1 - \cos\theta\right)}{3\theta}\hat{\jmath} \\ &= \frac{2r\sin(\pi)}{3\pi}\hat{\imath} + \frac{2r\left(1 - \cos(\pi)\right)}{3\pi}\hat{\jmath} \\ &= 0 + \frac{4r}{3\pi}\hat{\jmath} \\ &= \frac{4r}{3\pi}\hat{\jmath} \end{aligned}\]