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Subsection 2.3.1 Centre 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 \(x\text{:}\)

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

\begin{equation*} 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} \end{equation*}

and the mass and center of mass of the right end are

\begin{equation*} 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} \end{equation*}

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 centre 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 centre 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 centre 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 \(Z-z_i\text{.}\) This takes work \(m_i g (Z-z_i)\text{.}\) So the total work required to lift all \(n\) masses is

\begin{align*} \text{Work} &= \sum_{i=1}^n m_i g (Z-z_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 centre 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 centre of mass of the array are

\begin{equation*} 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 \end{equation*}

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 centre of mass

\begin{equation*} 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 \end{equation*}

To lift the \(i^{\rm th}\) mass took work \(m_i g (Z_i-z_i)\text{.}\) So the total work required to lift all \(n\) masses was

\begin{align*} \text{Work} &= \sum_{i=1}^n m_i g (Z_i-z_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 centre of mass of the array, to height \(\bar Z\text{,}\) the final centre 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

Think of \(\rho(x)\,\dee{x}\) as the mass of the “almost point particle” between \(x\) and \(x+\dee{x}\text{.}\)

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

\begin{equation*} \big\{\ (x,y)\ \big|\ a\le x\le b,\ B(x)\le y\le T(x)\ \big\} \end{equation*}

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 \(\dee{x}\text{.}\)
  • 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 \([T(x)-B(x)]\,\dee{x}\)
    • mass \(\rho[T(x)-B(x)]\,\dee{x}\)
    • 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 \(\rho[T(x)-B(x)]\,\dee{x}\) located at \(\big(x\,,\,\frac{B(x)+T(x)}{2}\big)\text{.}\) So:

We can of course also slice up the body using horizontal slices.

If the body has constant density \(\rho\) and fills the region

\begin{equation*} \big\{\ (x,y)\ \big|\ L(y)\le x\le R(y),\ c\le y\le d\ \big\} \end{equation*}

then the same computation as above gives:

Find the \(x\)-coordinate of the centroid (centre 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 \(B(x)=0\text{,}\)

\begin{equation*} \bar x =\frac{1}{A}\int_0^3 x\,T(x)\,\dee{x} =\frac{1}{A}\int_0^3 x\,2\sqrt{1-\frac{x^2}{9}}\,\dee{x} =\frac{4}{3\pi}\int_0^3 x\sqrt{1-\frac{x^2}{9}}\,\dee{x} \end{equation*}

Sub in \(u=1-\frac{x^2}{9}\text{,}\) \(\dee{u}=-\frac{2}{9}x\,\dee{x}\text{.}\)

\begin{equation*} \bar x =-\frac{9}{2}\frac{4}{3\pi}\int_1^0 \sqrt{u}\,\dee{u} =-\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} \end{equation*}

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 \(a=0\text{,}\) \(b=r\text{,}\) \(T(x)=\sqrt{r^2-x^2}\) and \(B(x)=0\text{,}\)

\begin{equation*} \bar x = \frac{1}{A}\int_0^r x\sqrt{r^2-x^2}\,\dee{x} \end{equation*}

To evaluate the integral, sub in \(u=r^2-x^2\text{,}\) \(\dee{u}=-2x\,\dee{x}\text{.}\)

\begin{equation*} \int_0^r x\sqrt{r^2-x^2}\,\dee{x} = \int_{r^2}^0 \sqrt{u}\,\frac{\dee{u}}{-2} = -\frac{1}{2}\Big[\frac{u^{3/2}}{3/2}\Big]^0_{r^2} = \frac{r^3}{3} \tag{$*$} \end{equation*}

So

\begin{equation*} \bar x = \frac{4}{\pi r^2}\Big[\frac{r^3}{3}\Big] =\frac{4r}{3\pi} \end{equation*}

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 \(a=0\text{,}\) \(b=r\text{,}\) \(T(x)=\sqrt{r^2-x^2}\) and \(B(x)=0\text{,}\)

\begin{align*} \bar y & = \frac{1}{2A}\int_0^r \big(\sqrt{r^2-x^2}\big)^2\,\dee{x}\ &&= \frac{2}{\pi r^2}\int_0^r \big(r^2-x^2\big)\,\dee{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 \(a=-r\text{,}\) \(b=r\text{,}\) \(T(x)=\sqrt{r^2-x^2}\) and \(B(x)=0\text{,}\)

\begin{align*} \bar y & = \frac{1}{2A}\int_{-r}^r \big(\sqrt{r^2-x^2}\big)^2\,\dee{x}\ &= \frac{1}{\pi r^2}\int_{-r}^r \big(r^2-x^2\big)\,\dee{x} \\ &= \frac{2}{\pi r^2}\int_0^r \big(r^2-x^2\big)\,\dee{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 \(B(x)\!=\!\sqrt{1-x^2}\) 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{1-x^2}]\,\dee{x} +\int_1^2x[2-0]\,\dee{x}\bigg]\cr &=\frac{4}{16-\pi}\bigg[x^2\big|_0^1 +x^2\big|_1^2-\int_0^1x\sqrt{1-x^2}\,\dee{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}{48-3\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 \(c=0\text{,}\) \(d=b\text{,}\) \(R(y)= \frac{c}{b}(b-y)\) (which is gotten by solving \(bx+cy=bc\) for \(x\)) and \(L(y)= \frac{a}{b}(b-y)\) (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
    \begin{equation*} \ell(y)=\Big[\frac{c}{b}(b-y)-\frac{a}{b}(b-y)\Big]=\frac{c-a}{b}(b-y) \end{equation*}
  • The strip has width \(\dee{y}\text{.}\)
  • On this strip, \(y\) has average value \(y\text{.}\)
  • On this strip, \(x\) has average value \(\frac{1}{2}\big[\frac{a}{b}(b-y)+\frac{c}{b}(b-y)\big]=\frac{a+c}{2b}(b-y)\text{.}\)

As the area of the triangle is \(A=\half (c-a)b\text{,}\)

\begin{align*} \bar y&=\frac{1}{A}\int_0^b y\ \ell(y)\dee{y} =\frac{2}{(c-a)b}\int_0^b y\frac{c-a}{b}(b-y)\dee{y}\\ &=\frac{2}{b^2}\int_0^b (by-y^2)\dee{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}(b-y)\ \ell(y)\dee{y}\\ \amp=\frac{2}{(c-a)b}\int_0^b \frac{a+c}{2b}(b-y)\frac{c-a}{b}(b-y)\dee{y}\\ & =\frac{a+c}{b^3}\int_0^b (y-b)^2\dee{y}\\ &=\frac{a+c}{b^3}\Big[\frac{1}{3}(y-b)^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 \(\half(c,b)\text{.}\) The line from \((a,0)\) to \(\half(c,b)\) has slope \(\frac{b/2}{c/2-a}=\frac{b}{c-2a}\) and so has equation \(y=\frac{b}{c-2a}(x-a)\text{.}\) As \(\frac{b}{c-2a}(\bar x-a) =\frac{b}{c-2a}\big(\frac{a+c}{3}-a\big) =\frac{1}{3}\frac{b}{c-2a}(c+a-3a) =\frac{b}{3} =\bar y\text{,}\) the centroid does indeed lie on this median. In this computation we have implicitly assumed that \(c\ne 2a\) so that the denominator \(c-2a\ne 0\text{.}\) 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 \(\half(a,b)\text{.}\) The line from \((c,0)\) to \(\half(a,b)\) has slope \(\frac{b/2}{a/2-c}=\frac{b}{a-2c}\) and so has equation \(y=\frac{b}{a-2c}(x-c)\text{.}\) As \(\frac{b}{a-2c}(\bar x-c) =\frac{b}{a-2c}\big(\frac{a+c}{3}-c\big) =\frac{1}{3}\frac{b}{a-2c}(a+c-3c) =\frac{b}{3} =\bar y\text{,}\) the centroid does indeed lie on this median. In this computation we have implicitly assumed that \(a\ne 2c\) so that the denominator \(a-2c\ne 0\text{.}\) 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.
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