2.3: Center of Mass and Torque
 Page ID
 89248
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Center of Mass
If you support a body at its center of mass (in a uniform gravitational field) it balances perfectly. That's the definition of the center of mass of the body.
If the body consists of a finite number of masses \(m_1\text{,}\) \(\cdots\text{,}\) \(m_n\) attached to an infinitely strong, weightless (idealized) rod with mass number \(i\) attached at position \(x_i\text{,}\) then the center of mass is at the (weighted) average value of \(x\text{:}\)
\[ \bar x =\frac{\sum_{i=1}^n m_ix_i}{\sum_{i=1}^n m_i} \nonumber \]
The denominator \(m=\sum_{i=1}^n m_i\) is the total mass of the body.
This formula for the center of mass is derived in the following (optional) section. See equation (2.3.14).
For many (but certainly not all) purposes an (extended rigid) body acts like a point particle located at its center of mass. For example it is very common to treat the Earth as a point particle. Here is a more detailed example in which we think of a body as being made up of a number of component parts and compute the center of mass of the body as a whole by using the center of masses of the component parts. Suppose that we have a dumbbell which consists of
 a left end made up of particles of masses \(m_{l,1}\text{,}\) \(\cdots\text{,}\) \(m_{l,3}\) located at \(x_{l,1}\text{,}\) \(\cdots\text{,}\) \(x_{l,3}\) and
 a right end made up of particles of masses \(m_{r,1}\text{,}\) \(\cdots\text{,}\) \(m_{r,4}\) located at \(x_{r,1}\text{,}\) \(\cdots\text{,}\) \(x_{r,4}\) and
 an infinitely strong, weightless (idealized) rod joining all of the particles.
Then the mass and center of mass of the left end are
\[ M_l=m_{l,1}+\cdots +m_{l,3}\qquad \bar X_l = \frac{m_{l,1}x_{l,1}+\cdots +m_{l,3}x_{l,3}}{M_l} \nonumber \]
and the mass and center of mass of the right end are
\[ M_r=m_{r,1}+\cdots +m_{r,4}\qquad \bar X_r = \frac{m_{r,1}x_{r,1}+\cdots +m_{r,4}x_{r,4}}{M_r} \nonumber \]
The mass and center of mass of the entire dumbbell are
\begin{align*} M&= m_{l,1}+\cdots +m_{l,3}\ +\ m_{r,1}+\cdots +m_{r,4}\\ &= M_l+M_r\\ \bar x &=\frac{m_{l,1}x_{l,1}+\cdots +m_{l,3}x_{l,3}\ +\ m_{r,1}x_{r,1}+\cdots +m_{r,4}x_{r,4}}{M}\\ &=\frac{M_l \bar X_l + M_r \bar X_r}{M_r+M_l} \end{align*}
So we can compute the center of mass of the entire dumbbell by treating it as being made up of two point particles, one of mass \(M_l\) located at the center of mass of the left end, and one of mass \(M_r\) located at the center of mass of the right end.
Here is another example in which an extended body acts like a point particle located at its center of mass. Imagine that there are a finite number of masses \(m_1,\cdots,m_n\) arrayed along a (vertical) \(z\)axis with mass number \(i\) attached at height \(z_i\text{.}\) Note that the total mass of the array is \(M=\sum_{i=1}^n m_i\) and that the center of mass of the array is at height
\begin{gather*} \bar z =\frac{\sum_{i=1}^n m_iz_i}{\sum_{i=1}^n m_i} =\frac{1}{M} \sum_{i=1}^n m_iz_i \end{gather*}
Now suppose that we lift all of the masses, against gravity, to height \(Z\text{.}\) So after the lift there is a total mass \(M\) located at height \(Z\text{.}\) The \(i^{\rm th}\) mass is subject to a downward gravitational force of \(m_i g\text{.}\) So to lift the \(i^{\rm th}\) mass we need to apply a compensating upward force of \(m_ig\) through a distance of \(Zz_i\text{.}\) This takes work \(m_i g (Zz_i)\text{.}\) So the total work required to lift all \(n\) masses is
\begin{align*} \text{Work} &= \sum_{i=1}^n m_i g (Zz_i)\\ &= g Z \sum_{i=1}^n m_i g \sum_{i=1}^n m_i z_i\\ &= g Z M  g M \bar z\\ & =Mg(Z\bar z) \end{align*}
So the work required to lift the array of \(n\) particles is identical to the work required to lift a single particle, whose mass, \(M\text{,}\) is the total mass of the array, from height \(\bar z\text{,}\) the center of mass of the array, to height \(Z\text{.}\)
Imagine, as in Example 2.3.2, that there are a finite number of masses \(m_1,\cdots,m_n\) arrayed along a (vertical) \(z\)axis with mass number \(i\) attached at height \(z_i\text{.}\) Again, the total mass and center of mass of the array are
\[ M=\sum_{i=1}^n m_i \qquad \bar z =\frac{\sum_{i=1}^n m_iz_i}{\sum_{i=1}^n m_i} =\frac{1}{M} \sum_{i=1}^n m_iz_i \nonumber \]
Now suppose that we lift, for each \(1\le i\le n\text{,}\) mass number \(i\text{,}\) against gravity, from its initial height \(z_i\) to a final height \(Z_i\text{.}\) So after the lift we have a new array of masses with total mass and center of mass
\[ M=\sum_{i=1}^n m_i \qquad \bar Z =\frac{\sum_{i=1}^n m_iZ_i}{\sum_{i=1}^n m_i} =\frac{1}{M} \sum_{i=1}^n m_iZ_i \nonumber \]
To lift the \(i^{\rm th}\) mass took work \(m_i g (Z_iz_i)\text{.}\) So the total work required to lift all \(n\) masses was
\begin{align*} \text{Work} &= \sum_{i=1}^n m_i g (Z_iz_i)\\ &= g \sum_{i=1}^n m_i Z_i g \sum_{i=1}^n m_i z_i\\ &= g M \bar Z  g M \bar z =Mg(\bar Z\bar z) \end{align*}
So the work required to lift the array of \(n\) particles is identical to the work required to lift a single particle, whose mass, \(M\text{,}\) is the total mass of the array, from height \(\bar z\text{,}\) the initial center of mass of the array, to height \(\bar Z\text{,}\) the final center of mass of the array.
Now we'll extend the above ideas to cover more general classes of bodies. If the body consists of mass distributed continuously along a straight line, say with mass density \(\rho(x)\)kg/m and with \(x\) running from \(a\) to \(b\text{,}\) rather than consisting of a finite number of point masses, the formula for the center of mass becomes
\[ \bar x = \frac{\int_a^b x\ \rho(x)\,\, d{x}}{\int_a^b \rho(x)\,\, d{x}} \nonumber \]
Think of \(\rho(x)\,\, d{x}\) as the mass of the “almost point particle” between \(x\) and \(x+\, d{x}\text{.}\)
If the body is a two dimensional object, like a metal plate, lying in the \(xy\)plane, its center of mass is a point \((\bar x,\bar y)\) with \(\bar x\) being the (weighted) average value of the \(x\)coordinate over the body and \(\bar y\) being the (weighted) average value of the \(y\)coordinate over the body. To be concrete, suppose the body fills the region
\[ \big\{\ (x,y)\ \big\ a\le x\le b,\ B(x)\le y\le T(x)\ \big\} \nonumber \]
in the \(xy\)plane. For simplicity, we will assume that the density of the body is a constant, say \(\rho\text{.}\) When the density is constant, the center of mass is also called the centroid and is thought of as the geometric center of the body.
To find the centroid of the body, we use our standard “slicing” strategy. We slice the body into thin vertical strips, as illustrated in the figure below.
Here is a detailed description of a generic strip.
 The strip has width \(\, d{x}\text{.}\)
 Each point of the strip has essentially the same \(x\)coordinate. Call it \(x\text{.}\)
 The top of the strip is at \(y=T(x)\) and the bottom of the strip is at \(y=B(x)\text{.}\)
 So the strip has
 height \(T(x)B(x)\)
 area \([T(x)B(x)]\,\, d{x}\)
 mass \(\rho[T(x)B(x)]\,\, d{x}\)
 centroid, i.e. middle point, \(\big(x\,,\,\frac{B(x)+T(x)}{2}\big)\text{.}\)
In computing the centroid of the entire body, we may treat each strip as a single particle of mass \(\rho[T(x)B(x)]\,\, d{x}\) located at \(\big(x\,,\,\frac{B(x)+T(x)}{2}\big)\text{.}\) So:
The mass of the entire body bounded by curves \(T(x)\) above and \(B(x)\) below is
\[\begin{align*} M&= \rho\int_a^b [T(x)B(x)]\,\, d{x} =\rho A \tag{a}\\ \end{align*}\]where \(A=\int_a^b [T(x)B(x)]\,\, d{x}\) is the area of the region. The coordinates of the centroid are
\begin{align*} \bar x &= \frac{\int_a^b x\ \overbrace{\rho[T(x)B(x)]\,\, d{x}} ^{\mathrm{mass\ of\ slice}}} {M } &&= \frac{\int_a^b x [T(x)B(x)]\,\, d{x}}{A} \tag{b}\\ \bar y &= \frac{\int_a^b\!\!\!\!\overbrace{\tfrac{B(x)+T(x)}{2}} ^{\mathrm{average}\ y\ \mathrm{on\ slice}}\!\! \overbrace{\rho[T(x)B(x)]\,\, d{x}} ^{\mathrm{mass\ of\ slice}}} {M }\ &&= \frac{\int_a^b\, [T(x)^2B(x)^2]\,\, d{x}}{2A} \tag{c} \end{align*}
We can of course also slice up the body using horizontal slices.
If the body has constant density \(\rho\) and fills the region
\[ \big\{\ (x,y)\ \big\ L(y)\le x\le R(y),\ c\le y\le d\ \big\} \nonumber \]
then the same computation as above gives:
The mass of the entire body bounded by curves \(L(y)\) to the left and \(R(y)\) to the right is
\[\begin{align*} M&= \rho\int_c^d [R(y)L(y)]\,\, d{y} =\rho A \tag{a}\\ \end{align*}\]where \(A=\int_c^d [R(y)L(y)]\,\, d{y}\) is the area of the region, and gives the coordinates of the centroid to be
\begin{align*} \bar x &= \frac{\int_c^d\!\!\!\!\overbrace{\tfrac{R(y)+L(y)}{2}} ^{\mathrm{average}\ x\ \mathrm{on\ slice}}\!\! \overbrace{\rho[R(y)L(y)]\,\, d{y}} ^{\mathrm{mass\ of\ slice}}} {M } \ &&= \frac{\int_c^d\, [R(y)^2L(y)^2]\,\, d{y}}{2A} \tag{b}\\ \bar y &= \frac{\int_c^d y\ \overbrace{\rho[R(y)L(y)]\,\, d{y}} ^{\mathrm{mass\ of\ slice}}} {M } &&= \frac{\int_c^d y [R(y)L(y)]\,\, d{y}}{A} \tag{c} \end{align*}
Find the \(x\)coordinate of the centroid (center of gravity) of the plane region \(R\) that lies in the first quadrant \(x\ge 0, \ y\ge 0\) and inside the ellipse \(4x^2+9y^2=36\text{.}\) (The area bounded by the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(\pi ab\) square units.)
Solution: In standard form \(4x^2+9y^2=36\) is \(\frac{x^2}{9}+\frac{y^2}{4}=1\text{.}\) So, on \(R\text{,}\) \(x\) runs from \(0\) to \(3\) and \(R\) has area \(A=\frac{1}{4}\pi\times 3\times 2=\frac{3}{2}\pi\text{.}\) For each fixed \(x\text{,}\) between \(0\) and \(3\text{,}\) \(y\) runs from \(0\) to \(2\sqrt{1\frac{x^2}{9}}\text{.}\) So, applying (2.3.5.b) with \(a=0\text{,}\) \(b=3\text{,}\) \(T(x)=2\sqrt{1\frac{x^2}{9}}\) and \(B(x)=0\text{,}\)
\[ \bar x =\frac{1}{A}\int_0^3 x\,T(x)\,\, d{x} =\frac{1}{A}\int_0^3 x\,2\sqrt{1\frac{x^2}{9}}\,\, d{x} =\frac{4}{3\pi}\int_0^3 x\sqrt{1\frac{x^2}{9}}\,\, d{x} \nonumber \]
Sub in \(u=1\frac{x^2}{9}\text{,}\) \(du=\frac{2}{9}x\,\, d{x}\text{.}\)
\[ \bar x =\frac{9}{2}\frac{4}{3\pi}\int_1^0 \sqrt{u}\,du =\frac{9}{2}\frac{4}{3\pi}\Big[\frac{u^{3/2}}{3/2}\Big]_1^0 =\frac{9}{2}\frac{4}{3\pi}\Big[\frac{2}{3}\Big] =\frac{4}{\pi} \nonumber \]
Find the centroid of the quarter circular disk \(x\ge 0\text{,}\) \(y\ge 0\text{,}\) \(x^2+y^2\le r^2\text{.}\)
Solution: By symmetry, \(\bar x=\bar y\text{.}\) The area of the quarter disk is \(A=\frac{1}{4}\pi r^2\text{.}\) By (2.3.5.b) with \(a=0\text{,}\) \(b=r\text{,}\) \(T(x)=\sqrt{r^2x^2}\) and \(B(x)=0\text{,}\)
\[ \bar x = \frac{1}{A}\int_0^r x\sqrt{r^2x^2}\,\, d{x} \nonumber \]
To evaluate the integral, sub in \(u=r^2x^2\text{,}\) \(du=2x\,\, d{x}\text{.}\)
\[ \int_0^r x\sqrt{r^2x^2}\,\, d{x} = \int_{r^2}^0 \sqrt{u}\,\frac{du}{2} = \frac{1}{2}\Big[\frac{u^{3/2}}{3/2}\Big]^0_{r^2} = \frac{r^3}{3} \tag{$*$} \nonumber \]
So
\[ \bar x = \frac{4}{\pi r^2}\Big[\frac{r^3}{3}\Big] =\frac{4r}{3\pi} \nonumber \]
As we observed above, we should have \(\bar x=\bar y\text{.}\) But, just for practice, let's compute \(\bar y\) by the integral formula (2.3.5.c), again with \(a=0\text{,}\) \(b=r\text{,}\) \(T(x)=\sqrt{r^2x^2}\) and \(B(x)=0\text{,}\)
\begin{align*} \bar y & = \frac{1}{2A}\int_0^r \big(\sqrt{r^2x^2}\big)^2\,\, d{x}\ &&= \frac{2}{\pi r^2}\int_0^r \big(r^2x^2\big)\,\, d{x}\\ &= \frac{2}{\pi r^2}\Big[r^2x\frac{x^3}{3}\Big]_0^r &&= \frac{2}{\pi r^2}\frac{2r^3}{3} \\ &=\frac{4r}{3\pi} \end{align*}
as expected.
Find the centroid of the half circular disk \(y\ge 0\text{,}\) \(x^2+y^2\le r^2\text{.}\)
Solution: Once again, we have a symmetry — namely the half disk is symmetric about the \(y\)axis. So the centroid lies on the \(y\)axis and \(\bar x=0\text{.}\) The area of the half disk is \(A=\frac{1}{2}\pi r^2\text{.}\) By (2.3.5.c), with \(a=r\text{,}\) \(b=r\text{,}\) \(T(x)=\sqrt{r^2x^2}\) and \(B(x)=0\text{,}\)
\begin{align*} \bar y & = \frac{1}{2A}\int_{r}^r \big(\sqrt{r^2x^2}\big)^2\,\, d{x}\ &= \frac{1}{\pi r^2}\int_{r}^r \big(r^2x^2\big)\,\, d{x} \\ &= \frac{2}{\pi r^2}\int_0^r \big(r^2x^2\big)\,\, d{x} & \text{since the integrand is even}\\ &= \frac{2}{\pi r^2}\Big[r^2x\frac{x^3}{3}\Big]_0^r \\ &=\frac{4r}{3\pi} \end{align*}
Find the centroid of the region \(R\) in the diagram.
Solution: By symmetry, \(\bar x=\bar y\text{.}\) The region \(R\) is a \(2\times 2\) square with one quarter of a circle of radius \(1\) removed and so has area \(2\times 2\frac{1}{4}\pi=\frac{16\pi}{4}\text{.}\) The top of \(R\) is \(y=T(x)=2\text{.}\) The bottom is \(y=B(x)\) with \(B(x)\!=\!\sqrt{1x^2}\) when \(0\le x\le 1\) and \(B(x)\!=\!0\) when \(1\le x\le 2\text{.}\) So
\[\begin{align*} \bar y = \bar x &=\frac{1}{A}\bigg[\int_0^1x[2\sqrt{1x^2}]\,\, d{x} +\int_1^2x[20]\,\, d{x}\bigg]\cr &=\frac{4}{16\pi}\bigg[x^2\big_0^1 +x^2\big_1^2\int_0^1x\sqrt{1x^2}\,\, d{x}\bigg]\\ \end{align*}\]Now we can make use of the starred equation in Example 2.3.8 with \(r=1\) to obtain
\begin{align*} &=\frac{4}{16\pi}\Big[4\frac{1}{3}\Big]\\ &=\frac{44}{483\pi} \end{align*}
Prove that the centroid of any triangle is located at the point of intersection of the medians. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
Solution: Choose a coordinate system so that the vertices of the triangle are located at \((a,0)\text{,}\) \((0,b)\) and \((c,0)\text{.}\) (In the figure below, \(a\) is negative.)
The line joining \((a,0)\) and \((0,b)\) has equation \(bx+ay=ab\text{.}\) (Check that \((a,0)\) and \((0,b)\) both really are on this line.) The line joining \((c,0)\) and \((0,b)\) has equation \(bx+cy=bc\text{.}\) (Check that \((c,0)\) and \((0,b)\) both really are on this line.) Hence for each fixed \(y\) between \(0\) and \(b\text{,}\) \(x\) runs from \(a\frac{a}{b}y\) to \(c\frac{c}{b}y\text{.}\)
We'll use horizontal strips to compute \(\bar x\) and \(\bar y\text{.}\) We could just apply equation (2.3.6) with \(c=0\text{,}\) \(d=b\text{,}\) \(R(y)= \frac{c}{b}(by)\) (which is gotten by solving \(bx+cy=bc\) for \(x\)) and \(L(y)= \frac{a}{b}(by)\) (which is gotten by solving \(bx+ay=ab\) for \(x\)).
But rather than memorizing or looking up those formulae, we'll derive them for this example. So consider a thin strip at height \(y\) as illustrated in the figure above.
 The strip has length
\[ \ell(y)=\Big[\frac{c}{b}(by)\frac{a}{b}(by)\Big]=\frac{ca}{b}(by) \nonumber \]
 The strip has width \(\, d{y}\text{.}\)
 On this strip, \(y\) has average value \(y\text{.}\)
 On this strip, \(x\) has average value \(\frac{1}{2}\big[\frac{a}{b}(by)+\frac{c}{b}(by)\big]=\frac{a+c}{2b}(by)\text{.}\)
As the area of the triangle is \(A=\frac{1}{2} (ca)b\text{,}\)
\begin{align*} \bar y&=\frac{1}{A}\int_0^b y\ \ell(y)\, d{y} =\frac{2}{(ca)b}\int_0^b y\frac{ca}{b}(by)\, d{y}\\ &=\frac{2}{b^2}\int_0^b (byy^2)\, d{y} =\frac{2}{b^2}\Big(b\frac{b^2}{2}\frac{b^3}{3}\Big)\\ &=\frac{2}{b^2}\frac{b^3}{6}=\frac{b}{3}\\ \bar x&=\frac{1}{A}\int_0^b \frac{a+c}{2b}(by)\ \ell(y)\, d{y}\\ & =\frac{2}{(ca)b}\int_0^b \frac{a+c}{2b}(by)\frac{ca}{b}(by)\, d{y}\\ & =\frac{a+c}{b^3}\int_0^b (yb)^2\, d{y}\\ &=\frac{a+c}{b^3}\Big[\frac{1}{3}(yb)^3\Big]_0^b =\frac{a+c}{b^3}\ \frac{b^3}{3} =\frac{a+c}{3} \end{align*}
We have found that the centroid of the triangle is at \((\bar x,\bar y)=\big(\frac{a+c}{3},\frac{b}{3}\big)\text{.}\) We shall now show that this point lies on all three medians.
 One vertex is at \((a,0)\text{.}\) The opposite side runs from \((0,b)\) and \((c,0)\) and so has midpoint \(\frac{1}{2}(c,b)\text{.}\) The line from \((a,0)\) to \(\frac{1}{2}(c,b)\) has slope \(\frac{b/2}{c/2a}=\frac{b}{c2a}\) and so has equation \(y=\frac{b}{c2a}(xa)\text{.}\) As \(\frac{b}{c2a}(\bar xa) =\frac{b}{c2a}\big(\frac{a+c}{3}a\big) =\frac{1}{3}\frac{b}{c2a}(c+a3a) =\frac{b}{3} =\bar y\text{,}\) the centroid does indeed lie on this median. In this computation we have implicitly assumed that \(c\ne 2a\) so that the denominator \(c2a\ne 0\text{.}\) In the event that \(c=2a\text{,}\) the median runs from \((a,0)\) to \(\big(a,\frac{b}{2}\big)\) and so has equation \(x=a\text{.}\) When \(c=2a\) we also have \(\bar x=\frac{a+c}{3}=a\text{,}\) so that the centroid still lies on the median.
 Another vertex is at \((c,0)\text{.}\) The opposite side runs from \((a,0)\) and \((0,b)\) and so has midpoint \(\frac{1}{2}(a,b)\text{.}\) The line from \((c,0)\) to \(\frac{1}{2}(a,b)\) has slope \(\frac{b/2}{a/2c}=\frac{b}{a2c}\) and so has equation \(y=\frac{b}{a2c}(xc)\text{.}\) As \(\frac{b}{a2c}(\bar xc) =\frac{b}{a2c}\big(\frac{a+c}{3}c\big) =\frac{1}{3}\frac{b}{a2c}(a+c3c) =\frac{b}{3} =\bar y\text{,}\) the centroid does indeed lie on this median. In this computation we have implicitly assumed that \(a\ne 2c\) so that the denominator \(a2c\ne 0\text{.}\) In the event that \(a=2c\text{,}\) the median runs from \((c,0)\) to \(\big(c,\frac{b}{2}\big)\) and so has equation \(x=c\text{.}\) When \(a=2c\) we also have \(\bar x=\frac{a+c}{3}=c\text{,}\) so that the centroid still lies on the median.
 The third vertex is at \((0,b)\text{.}\) The opposite side runs from \((a,0)\) and \((c,0)\) and so has midpoint \(\big(\frac{a+c}{2},0\big)\text{.}\) The line from \((0,b)\) to \(\big(\frac{a+c}{2},0\big)\) has slope \(\frac{b}{(a+c)/2}=\frac{2b}{a+c}\) and so has equation \(y=b\frac{2b}{a+c}x\text{.}\) As \(b\frac{2b}{a+c}\bar x =b\frac{2b}{a+c}\frac{a+c}{3} =\frac{b}{3} =\bar y\text{,}\) the centroid does indeed lie on this median. This time, we have implicitly assumed that \(a+c\ne 0\text{.}\) In the event that \(a+c=0\text{,}\) the median runs from \((0,b)\) to \((0,0)\) and so has equation \(x=0\text{.}\) When \(a+c=0\) we also have \(\bar x=\frac{a+c}{3}=0\text{,}\) so that the centroid still lies on the median.
Optional — Torque
Newton's law of motion says that the position \(x(t)\) of a single particle moving under the influence of a force \(F\) obeys \(mx''(t)=F\text{.}\) Similarly, the positions \(x_i(t)\text{,}\) \(1\le i\le n\text{,}\) of a set of particles moving under the influence of forces \(F_i\) obey \(mx_i''(t)=F_i\text{,}\) \(1\le i\le n\text{.}\) Often systems of interest consist of some small number of rigid bodies. Suppose that we are interested in the motion of a single rigid body, say a piece of wood. The piece of wood is made up of a huge number of atoms. So the system of equations determining the motion of all of the individual atoms in the piece of wood is huge. On the other hand, because the piece of wood is rigid, its configuration is completely determined by the position of, for example, its center of mass and its orientation. (Rather than get into what is precisely meant by “orientation”, let's just say that it is certainly determined by, for example, the positions of a few of the corners of the piece of wood). It is possible to extract from the huge system of equations that determine the motion of all of the individual atoms, a small system of equations that determine the motion of the center of mass and the orientation. We can avoid some vector analysis, that is beyond the scope of this course, by assuming that our rigid body is moving in two rather than three dimensions.
So, imagine a piece of wood moving in the \(xy\)plane.
Furthermore, imagine that the piece of wood consists of a huge number of particles joined by a huge number of weightless but very strong steel rods. The steel rod joining particle number one to particle number two just represents a force acting between particles number one and two. Suppose that
 there are \(n\) particles, with particle number \(i\) having mass \(m_i\)
 at time \(t\text{,}\) particle number \(i\) has \(x\)coordinate \(x_i(t)\) and \(y\)coordinate \(y_i(t)\)
 at time \(t\text{,}\) the external force (gravity and the like) acting on particle number \(i\) has \(x\)coordinate \(H_i(t)\) and \(y\)coordinate \(V_i(t)\text{.}\) Here \(H\) stands for horizontal and \(V\) stands for vertical.
 at time \(t\text{,}\) the force acting on particle number \(i\text{,}\) due to the steel rod joining particle number \(i\) to particle number \(j\) has \(x\)coordinate \(H_{i,j}(t)\) and \(y\)coordinate \(V_{i,j}(t)\text{.}\) If there is no steel rod joining particles number \(i\) and \(j\text{,}\) just set \(H_{i,j}(t)=V_{i,j}(t)=0\text{.}\) In particular, \(H_{i,i}(t)=V_{i,i}(t)=0\text{.}\)
The only assumptions that we shall make about the steel rod forces are
 (A1)

for each \(i\ne j\text{,}\) \(H_{i,j}(t)=H_{j,i}(t)\) and \(V_{i,j}(t)=V_{j,i}(t)\text{.}\) In words, the steel rod joining particles \(i\) and \(j\) applies equal and opposite forces to particles \(i\) and \(j\text{.}\)
 (A2)

for each \(i\ne j\text{,}\) there is a function \(M_{i,j}(t)\) such that \(H_{i,j}(t)=M_{i,j}(t)\big[x_i(t)x_j(t)\big]\) and \(V_{i,j}(t)=M_{i,j}(t)\big[y_i(t)y_j(t)\big]\text{.}\) In words, the force due to the rod joining particles \(i\) and \(j\) acts parallel to the line joining particles \(i\) and \(j\text{.}\) For (A1) to be true, we need \(M_{i,j}(t)=M_{j,i}(t)\text{.}\)
Newton's law of motion, applied to particle number \(i\text{,}\) now tells us that
\begin{align*} m_i x''_i(t)&= H_i(t)+\sum_{j=1}^n H_{i,j}(t)\tag{$X_i$}\\ m_i y''_i(t)&= V_i(t)+\sum_{j=1}^n V_{i,j}(t)\tag{$Y_i$} \end{align*}
Adding up all of the equations \((X_i)\text{,}\) for \(i=1,\ 2,\ 3,\ \cdots,\ n\) and adding up all of the equations \((Y_i)\text{,}\) for \(i=1,\ 2,\ 3,\ \cdots,\ n\) gives
\begin{align*} \sum_{i=1}^n m_i x''_i(t)&= \sum_{i=1}^n H_i(t)+\sum_{1\le i,j\le n} H_{i,j}(t) \tag{$\Sigma_iX_i$}\\ \sum_{i=1}^n m_i y''_i(t)&= \sum_{i=1}^n V_i(t)+\sum_{1\le i,j\le n} V_{i,j}(t) \tag{$\Sigma_iY_i$} \end{align*}
The sum \(\sum_{1\le i,j\le n} H_{i,j}(t)\) contains \(H_{1,2}(t)\) exactly once and it also contains \(H_{2,1}(t)\) exactly once and these two terms cancel exactly, by assumption (A1). In this way, all terms in \(\sum_{1\le i,j\le n} H_{i,j}(t)\) with \(i\ne j\) exactly cancel. All terms with \(i=j\) are assumed to be zero. So \(\sum_{1\le i,j\le n} H_{i,j}(t)=0\text{.}\) Similarly, \(\sum_{1\le i,j\le n} V_{i,j}(t)=0\text{,}\) so the equations \((\Sigma_iX_i)\) and \((\Sigma_iY_i)\) simplify to
\begin{align*} \sum_{i=1}^n m_i x''_i(t)&= \sum_{i=1}^n H_i(t) \tag{$\Sigma_iX_i$}\\ \sum_{i=1}^n m_i y''_i(t)&= \sum_{i=1}^n V_i(t) \tag{$\Sigma_iY_i$} \end{align*}
Denote by
\[ M=\sum\limits_{i=1}^n m_i \nonumber \]
the total mass of the system, by
\[ X(t)=\frac{1}{M}\sum\limits_{i=1}^n m_ix_i(t)\qquad \text{and}\qquad Y(t)=\frac{1}{M}\sum\limits_{i=1}^n m_iy_i(t) \nonumber \]
the \(x\) and \(y\)coordinates of the center of mass of the system at time \(t\) and by
\[ H(t)=\sum\limits_{i=1}^n H_i(t) \qquad\text{ and }\qquad V(t)=\sum\limits_{i=1}^n V_i(t) \nonumber \]
the \(x\) and \(y\)coordinates of the total external force acting on the system at time \(t\text{.}\) In this notation, the equations \((\Sigma_iX_i)\) and \((\Sigma_iY_i)\) are
\[ MX''(t)=H(t)\qquad MY''(t)=V(t) \nonumber \]
So the center of mass of the system moves just like a single particle of mass \(M\) subject to the total external force.
Now multiply equation \((Y_i)\) by \(x_i(t)\text{,}\) subtract from it equation \((X_i)\) multiplied by \(y_i(t)\text{,}\) and sum over \(i\text{.}\) This gives the equation \({\sum_i\big[x_i(t)\,(Y_i)y_i(t)\,(X_i)\big]}\text{:}\)
\begin{align*} \sum_{i=1}^n m_i\big[x_i(t)y''_i(t)y_i(t)x''_i(t)\big] & = \sum_{i=1}^n \big[x_i(t)V_i(t)y_i(t)H_i(t)\big]\\ & \qquad +\sum_{1\le i,j\le n} \big[x_i(t)V_{i,j}(t)y_i(t)H_{i,j}(t)\big] \end{align*}
By the assumption (A2)
\begin{align*} x_1(t)V_{1,2}(t)y_1(t)H_{1,2}(t) &=x_1(t)M_{1,2}(t)\big[y_1(t)y_2(t)\big] y_1(t)M_{1,2}(t)\big[x_1(t)x_2(t)\big]\\ &=M_{1,2}(t)\big[y_1(t)x_2(t)x_1(t)y_2(t)\big]\\ x_2(t)V_{2,1}(t)y_2(t)H_{2,1}(t) &=x_2(t)M_{2,1}(t)\big[y_2(t)y_1(t)\big] y_2(t)M_{2,1}(t)\big[x_2(t)x_1(t)\big]\\ &=M_{2,1}(t)\big[y_1(t)x_2(t)+x_1(t)y_2(t)\big]\\ &=M_{1,2}(t)\big[y_1(t)x_2(t)+x_1(t)y_2(t)\big] \end{align*}
So the \(i=1\text{,}\) \(j=2\) term in \(\sum_{1\le i,j\le n} \big[x_i(t)V_{i,j}(t)y_i(t)H_{i,j}(t)\big]\) exactly cancels the \(i=2\text{,}\) \(j=1\) term. In this way all of the terms in \(\sum_{1\le i,j\le n} \big[x_i(t)V_{i,j}(t)y_i(t)H_{i,j}(t)\big]\) with \(i\ne j\) cancel. Each term with \(i=j\) is exactly zero. So \(\sum_{1\le i,j\le n} \big[x_i(t)V_{i,j}(t)y_i(t)H_{i,j}(t)\big]=0\) and
\[ \sum_{i=1}^n m_i\big[x_i(t)y''_i(t)y_i(t)x''_i(t)\big] = \sum_{i=1}^n \big[x_i(t)V_i(t)y_i(t)H_i(t)\big] \nonumber \]
Define
\begin{align*} L(t)&= \sum_{i=1}^n m_i\big[x_i(t)y'_i(t)y_i(t)x'_i(t)\big]\\ T(t)&=\sum_{i=1}^n \big[x_i(t)V_i(t)y_i(t)H_i(t)\big] \end{align*}
In this notation
\[ \dfrac{d}{dt} L(t)=T(t) \nonumber \]
 Equation (2.3.13) plays the role of Newton's law of motion for rotational motion.
 \(T(t)\) is called the torque and plays the role of “rotational force”.
 \(L(t)\) is called the angular momentum (about the origin) and is a measure of the rate at which the piece of wood is rotating.
 For example, if a particle of mass \(m\) is traveling in a circle of radius \(r\text{,}\) centerd on the origin, at \(\omega\) radians per unit time, then \(x(t)=r\cos(\omega t)\text{,}\) \(y(t)=r\sin(\omega t)\) and
\begin{align*} m\big[x(t)y'(t)y(t)x'(t)\big] &= m \big[r\cos(\omega t)\ r\omega\cos(\omega t) r\sin(\omega t)\ \big(r\omega\sin(\omega t)\big)\big]\\ &=m r^2\ \omega \end{align*}
is proportional to \(\omega\text{,}\) which is the rate of rotation about the origin.
 For example, if a particle of mass \(m\) is traveling in a circle of radius \(r\text{,}\) centerd on the origin, at \(\omega\) radians per unit time, then \(x(t)=r\cos(\omega t)\text{,}\) \(y(t)=r\sin(\omega t)\) and
In any event, in order for the piece of wood to remain stationary, that is to have \(x_i(t)\) and \(y_i(t)\) be constant for all \(1\le i\le n\text{,}\) we need to have
\[ X''(y)=Y''(t)=L(t)=0 \nonumber \]
and then equations (2.3.12) and (2.3.13) force
\[ H(t)=V(t)=T(t)=0 \nonumber \]
Now suppose that the piece of wood is a seesaw that is long and thin and is lying on the \(x\)axis, supported on a fulcrum at \(x=p\text{.}\) Then every \(y_i=0\) and the torque simplifies to \(T(t)=\sum_{i=1}^n x_i(t)V_i(t)\text{.}\) The forces consist of
 gravity, \(m_ig\text{,}\) acting downwards on particle number \(i\text{,}\) for each \(1\le i\le n\) and the
 force \(F\) imposed by the fulcrum that is pushing straight up on the particle at \(x=p\text{.}\)
So
 The net vertical force is \(V(t)=F\sum\limits_{i=1}^n m_ig =FMg\text{.}\) If the seesaw is to remain stationary, this must be zero so that \(F=Mg\text{.}\)
 The total torque (about the origin) is
\[ T=Fp\sum_{i=1}^n m_ig x_i =Mgp\sum_{i=1}^n m_ig x_i \nonumber \]
If the seesaw is to remain stationary, this must also be zero and the fulcrum must be placed at\[ p=\frac{1}{M}\sum_{i=1}^n m_i x_i \nonumber \]
Exercises
Stage 1
In Questions 8 through 10, you will derive the formulas for the center of mass of a rod of variable density, and the centroid of a twodimensional region using vertical slices (Equations 2.3.4 and 2.3.5 in the text). Knowing the equations by heart will allow you to answer many questions in this section; understanding where they came from will you allow to generalize their ideas to answer even more questions.
Using symmetry, find the centroid of the finite region between the curves \(y=(x1)^2\) and \(y=x^2+2x+1\text{.}\)
Using symmetry, find the centroid of the region inside the unit circle, and outside a rectangle centerd at the origin with width 1 and height 0.5.
A long, straight, thin rod has a number of weights attached along it. True or false: if it balances at position \(x\text{,}\) then the mass to the right of \(x\) is the same as the mass to the left of \(x\text{.}\)
A straight rod with negligible mass has the following weights attached to it:
 A weight of mass 1 kg, 1m from the left end,
 a weight of mass 2 kg, 3m from the left end,
 a weight of mass 2 kg, 4m from the left end, and
 a weight of mass 1 kg, 6m from the left end.
Where is the center of mass of the weighted rod?
For each picture below, determine whether the centroid is to the left of, to the right of, or along the line \(x=a\text{,}\) or whether there is not enough information to tell. The shading of a region indicates density: darker shading corresponds to a denser area.
Tank \(A\) is spherical, of radius 1 metre, and filled completely with water. The bottom of tank \(A\) is three metres above the ground, where Tank \(B\) sits. Tank \(B\) is tall and rectangular, with base dimensions 2 metres by 1 metre, and empty. Calculate the work done by gravity to drain all the water from Tank \(A\) to Tank \(B\) by modelling the situation as a point mass, of the same mass as the water, being moved from the height of the center of mass of \(A\) to the height of the center of mass of \(B\text{.}\)
You may use \(1000\) kg/m\(^3\) for the density of water, and \(g=9.8\) m/sec\(^2\) for the acceleration due to gravity.
Let \(S\) be the region bounded above by \(y=\frac{1}{x}\) and and below by the \(x\)axis, \(1 \le x \le 3\text{.}\) Let \(R\) be a rod with density \(\rho(x)=\frac{1}{x}\) at position \(x\text{,}\) \(1 \le x \le 3\text{.}\)
 What is the area of a thin slice of \(S\) at position \(x\) with width \(\, d{x}\text{?}\)
 What is the mass of a small piece of \(R\) at position \(x\) with length \(\, d{x}\text{?}\)
 What is the total area of \(S\text{?}\)
 What is the total mass of \(R\text{?}\)
 What is the \(x\)coordinate of the centroid of \(S\text{?}\)
 What is the center of mass of \(R\text{?}\)
Suppose \(R\) is a straight, thin rod with density \(\rho(x)\) at a position \(x\text{.}\) Let the left endpoint of \(R\) lie at \(x=a\text{,}\) and the right endpoint lie at \(x=b\text{.}\)
 To approximate the center of mass of \(R\text{,}\) imagine chopping it into \(n\) pieces of equal length, and approximating the mass of each piece using the density at its midpoint. Give your approximation for the center of mass in sigma notation.
 Take the limit as \(n\) goes to infinity of your approximation in part (a), and express the result using a definite integral.
Suppose \(S\) is a twodimensional object and at (horizontal) position \(x\) its height is \(T(x)B(x)\text{.}\) Its leftmost point is at position \(x=a\text{,}\) and its rightmost point is at position \(x=b\text{.}\)
To approximate the \(x\)coordinate of the centroid of \(S\text{,}\) we imagine it as a straight, thin rod \(R\text{,}\) where the mass of \(R\) from \(a \le x \le b\) is equal to the area of \(S\) from \(a \leq x \leq b\text{.}\)
 If \(S\) is the sheet shown below, sketch \(R\) as a rod with the same horizontal length, shaded darker when \(R\) is denser, and lighter when \(R\) is less dense.
 If we cut \(S\) into strips of very small width \(\, d{x}\text{,}\) what is the area of the strip at position \(x\text{?}\)
 Using your answer from (b), what is the density \(\rho(x)\) of \(R\) at position \(x\text{?}\)
 Using your result from Question 8(b), give the \(x\)coordinate of the centroid of \(S\text{.}\) Your answer will be in terms of \(a\text{,}\) \(b\text{,}\) \(T(X)\text{,}\) and \(B(x)\text{.}\)
Suppose \(S\) is flat sheet with uniform density, and at (horizontal) position \(x\) its height is \(T(x)B(x)\text{.}\) Its leftmost point is at position \(x=a\text{,}\) and its rightmost point is at position \(x=b\text{.}\)
To approximate the \(y\)coordinate of the centroid of \(S\text{,}\) we imagine it as a straight, thin, vertical rod \(R\text{.}\) We slice \(S\) into thin, vertical strips, and model these as weights on \(R\) with:
 position \(y\) on \(R\text{,}\) where \(y\) is the center of mass of the strip, and
 mass in \(R\) equal to the area of the strip in \(S\text{.}\)
 If \(S\) is the sheet shown below, slice it into a number of vertical pieces of equal length, approximated by rectangles. For each rectangle, mark its center of mass. Sketch \(R\) as a rod with the same vertical height, with weights corresponding to the slices you made of \(S\text{.}\)
 Imagine a thin strip of \(S\) at position \(x\text{,}\) with thickness \(\, d{x}\text{.}\) What is the area of the strip? What is the \(y\)value of its center of mass?
 Recall the center of mass of a rod with \(n\) weights of mass \(M_i\) at position \(y_i\) is given by
\[ \frac{\sum\limits_{i=1}^n (M_i\times y_i) }{\sum\limits_{i=1}^n M_i} \nonumber \]
Considering the limit of this formula as \(n\) goes to infinity, give the \(y\)coordinate of the center of mass of \(S\text{.}\)
Express the \(x\)coordinate of the centroid of the triangle with vertices \((1,{3})\text{,}\) \((1,{3})\text{,}\) and \((0,0)\) in terms of a definite integral. Do not evaluate the integral.
Stage 2
Use Equations 2.3.4 and 2.3.5 to find centroids and centers of mass in Questions 12 through 23.
A long, thin rod extends from \(x=0\) to \(x=7\) metres, and its density at position \(x\) is given by \(\rho(x) = x\) kg/m. Where is the center of mass of the rod?
A long, thin rod extends from \(x=3\) to \(x=10\) metres, and its density at position \(x\) is given by \(\rho(x) = \frac{1}{1+x^2}\) kg/m. Where is the center of mass of the rod?
Find the \(y\)coordinate of the centroid of the region bounded by the curves \(y=1\text{,}\) \(y=e^x\text{,}\) \(x=0\) and \(x=1\text{.}\) You may use the fact that the area of this region equals \(e\text{.}\)
Consider the region bounded by \(y=\frac{1}{\sqrt{16x^2}}\text{,}\) \(y=0\text{,}\) \(x=0\) and \(x=2\text{.}\)
 Sketch this region.
 Find the \(y\)coordinate of the centroid of this region.
Find the centroid of the finite region bounded by \(y = \sin(x)\text{,}\) \(y = \cos(x)\text{,}\) \(x = 0\text{,}\) and \(x = \pi/4\text{.}\)
Let \(A\) denote the area of the plane region bounded by \(x=0\text{,}\) \(x=1\text{,}\) \(y=0\) and \(y=\dfrac{k}{\sqrt{1+x^2}}\text{,}\) where \(k\) is a positive constant.
 Find the coordinates of the centroid of this region in terms of \(k\) and \(A\text{.}\)
 For what value of \(k\) is the centroid on the line \(y=x\text{?}\)
The region \(R\) is the portion of the plane which is above the curve \(y=x^23x\) and below the curve \(y=xx^2\text{.}\)
 Sketch the region \(R\)
 Find the area of \(R\text{.}\)
 Find the \(x\) coordinate of the centroid of \(R\text{.}\)
Let \(R\) be the region where \(0\le x\le 1\) and \(0\le y\le\frac{1}{1+x^2}\text{.}\) Find the \(x\)coordinate of the centroid of \(R\text{.}\)
Find the centroid of the region below, which consists of a semicircle of radius \(3\) on top of a rectangle of width \(6\) and height \(2\text{.}\)
Let \(D\) be the region below the graph of the curve \(y=\sqrt{94x^2}\) and above the \(x\)axis.
 Using an appropriate integral, find the area of the region \(D\text{;}\) simplify your answer completely.
 Find the center of mass of the region \(D\text{;}\) simplify your answer completely. (Assume it has constant density \(\rho\text{.}\))
The finite region \(S\) is bounded by the lines \(y=\arcsin x\text{,}\) \(y=\arcsin(2x)\text{,}\) and \(y=\frac{\pi}{2}\text{.}\) Find the centroid of \(S\text{.}\)
Calculate the centroid of the figure bounded by the curves \(y=e^x\text{,}\) \(y=3(x1)\text{,}\) \(y=0\text{,}\) \(x=0\text{,}\) and \(x=2\text{.}\)
Stage 3
Find the \(y\)coordinate of the center of mass of the (infinite) region lying to the right of the line \(x=1\text{,}\) above the \(x\)axis, and below the graph of \(y=8/x^3\text{.}\)
Let \(A\) be the region to the right of the \(y\)axis that is bounded by the graphs of \(y=x^2\) and \(y = 6x\text{.}\)
 Find the centroid of \(A\text{,}\) assuming it has constant density \(\rho=1\text{.}\) The area of \(A\) is \(\dfrac{22}{3}\) (you don't have to show this).
 Write down an expression, using horizontal slices (disks), for the volume obtained when the region \(A\) is rotated around the \(y\)axis. Do not evaluate any integrals; simply write down an expression for the volume.
(a) Find the \(y\)coordinate of the centroid of the region bounded by \(y = e^x\text{,}\) \(x = 0\text{,}\) \(x = 1\text{,}\) and \(y = 1\text{.}\)
(b) Calculate the volume of the solid generated by rotating the region from part (a) about the line \(y = 1\text{.}\)
Suppose a rectangle has width 4 m, height 3 m, and its density \(x\) metres from its left edge is \(x^2\) kg/m\(^2\text{.}\) Find the center of mass of the rectangle.
Suppose a circle of radius 3 m has density \((2+y)\) kg/m\(^2\) at any point \(y\) metres above its bottom. Find the center of mass of the circle.
A right circular cone of uniform density has base radius \(r\) m and height \(h\) m. We want to find its center of mass. By symmetry, we know that the center of mass will occur somewhere along the straight vertical line through the tip of the cone and the center of its base. The only question is the height of the center of mass.
We will model the cone as a rod \(R\) with height \(h\text{,}\) such that the mass of the section of the rod from position \(a\) to position \(b\) is the same as the volume of the cone from height \(a\) to height \(b\text{.}\) (You can imagine that the cone is an umbrella, and we've closed it up to look like a cane.^{2} )
 Using this model, calculate how high above the base of the cone its center of mass is.
 If we cut off the top \(hk\) metres of the cone (leaving an object of height \(k\)), how high above the base is the new center of mass?
An hourglass is shaped like two identical truncated cones attached together. Their base radius is 5 cm, the height of the entire hourglass is 18 cm, and the radius at the thinnest point is \(.5\) cm. The hourglass contains sand that fills up the bottom 6 cm when it's settled, with mass \(600\) grams and uniform density. We want to know the work done flipping the hourglass smoothly, so the sand settles into a truncated, invertedcone shape before it starts to fall down.
Using the methods of Section 2.1 to calculate the work done would be quite tedious. Instead, we will model the sand as a point of mass \(0.6\) kg, being lifted from the center of mass of its original position to the center of mass of its upturned position. Using the results of Question 29, how much work was done on the sand?
To simplify your calculation, you may assume that the height of the upturned sand (that is, the distance from the skinniest part of the hourglass to the top of the sand) is 8.8 cm. (Actually, it's \(\sqrt[3]{937}1\approx 8.7854\) cm.) So, the top 0.2 cm of the hourglass is empty.
Tank \(A\) is in the shape of half a sphere of radius 1 metre, with its flat face resting on the ground, and is completely filled with water. Tank \(B\) is empty and rectangular, with a square base of side length 1 m and a height of 3 m.
 To pump the water from Tank \(A\) to Tank \(B\text{,}\) we need to pump all the water from Tank \(A\) to a height of 3 m. How much work is done to pump all the water from Tank \(A\) to a height of 3 m? You may model the water as a point mass, originally situated at the center of mass of the full Tank \(A\text{.}\)
 Suppose we could move the water from Tank \(A\) directly to its final position in Tank \(B\) without going over the top of Tank \(B\text{.}\) (For example, maybe tank \(A\) is elastic, and Tank \(B\) is just Tank \(A\) after being smooshed into a different form.) How much work is done pumping the water? (That is, how much work is done moving a point mass from the center of mass of Tank \(A\) to the center of mass of Tank \(B\text{?}\))
 What percentage of work from part (a) was “wasted” by pumping the water over the top of Tank \(B\text{,}\) instead of moving it directly to its final position?
You may assume that the only work done is against the acceleration due to gravity, \(g=9.8\) m/sec\(^2\text{,}\) and that the density of water is 1000 kg/m\(^3\text{.}\)
Remark: the answer from (b) is what you might think of as the net work involved in pumping the water from Tank \(A\) to Tank \(B\text{.}\) When work gets “wasted,” the pump does some work pumping water up, then gravity does equal and opposite work bringing the water back down.
Let \(R\) be the region bounded above by \(y=2x\sin (x^2)\) and below by the \(x\)axis, \(0 \le x \le \sqrt{\frac{\pi}{2}}\text{.}\) Give an approximation of the \(x\)value of the centroid of \(R\) with error no more than \(\frac{1}{100}\text{.}\)
You may assume without proof that \(\left\frac{d^{4}}{dx^{4}}\left\{2x^2\sin(x^2)\right\} \right \leq 415\) over the interval \(\left[0,\sqrt{\frac{\pi}{2}}\right]\text{.}\)