2 Finding the centre of mass of a plane uniform lamina
In the previous Section we calculated the centre of mass of several individual point masses. We are now interested in finding the centre of mass of a thin sheet of material, such as a plane sheet of metal, called a lamina . The mass is not now located at individual points. Rather, it is distributed continuously over the lamina. In what follows we assume that the mass is distributed uniformly over the lamina and you will see how integration as the limit of a sum is used to find the centre of mass.
Figure 6 shows a lamina where the centre of mass has been marked at point $G$ with coordinates $\left(\stackrel{\u0304}{x},\u0233\right)$ . If the total mass of the lamina is $M$ then the moments about the $y$  and $x$ axes are respectively $M\stackrel{\u0304}{x}$ and $M\u0233$ . Our approach to locating the position of $G$ , i.e. finding $\stackrel{\u0304}{x}$ and $\u0233$ , is to divide the lamina into many small pieces, find the mass of each piece, and calculate the moment of each piece about the axes. The sum of the moments of the individual pieces about the $y$ axis must then be equal to $M\stackrel{\u0304}{x}$ and the sum of the moments of the individual pieces about the $x$ axis must equal $M\u0233$ .
Figure 6 :
There are no formulae which can be memorized for finding the centre of mass of a lamina because of the wide variety of possible shapes. Instead you should be familiar with the general technique for deriving the centre of mass.
An important preliminary concept is ‘mass per unit area’ which we now introduce.
2.1 Mass per unit area
Suppose we have a uniform lamina and select a piece of the lamina which has area equal to one unit . Let $\rho $ stand for the mass of such a piece. Then $\rho $ is called the mass per unit area . The mass of any other piece can be expressed in terms of $\rho $ . For example, an area of 2 units must have mass $2\rho $ , an area of 3 units must have mass $3\rho $ , and so on. Any portion of the lamina which has area $A$ has mass $\rho A$ .
Key Point 4
If a lamina has mass per unit area, $\rho $ , then the mass of part of the lamina having area $A$ is $A\rho $ .
We will investigate the calculation of centre of mass through the following Tasks.
Task!
Consider the plane sheet, or lamina, shown below. Find the location of its centre of mass $\left(\stackrel{\u0304}{x},\u0233\right)$ . (By symmetry the centre of mass of this lamina lies on the $x$ axis.)

First inspect the figure and note the symmetry of the lamina. Purely from the symmetry, what must be the
$y$
coordinate,
$\u0233$
, of the centre of mass ?
$\u0233=0$ since the centre of mass must lie on the $x$ axis

Let
$\rho $
stand for the mass per unit area of the lamina. The total area is 3 units. The total mass is therefore 3
$\rho $
. Its moment about the
$y$
axis is
$3\rho \stackrel{\u0304}{x}$
.
To find $\stackrel{\u0304}{x}$ first divide the lamina into a large number of thin vertical slices. In the figure below a typical slice has been highlighted. Note that the slice has been drawn from the point $P$ on the line $y=3x$ . The point $P$ has coordinates $\left(x,y\right)$ . The thickness of the slice is $\delta x$ .
Assuming that the slice is rectangular in shape, write down its area:
$2y\delta x$

Writing
$\rho $
as the mass per unit area, write down the mass of the slice:
$\left(2y\delta x\right)\rho $

The centre of mass of this slice lies on the
$x$
axis. So the slice can be assumed to be a point mass,
$2y\rho \delta x$
, located a distance
$x$
from
$O$
.
Write down the moment of the mass of the slice about the $y$ axis:
$\left(2y\delta x\right)\rho x$

By adding up contributions from all such slices in the lamina we obtain the sum of the moments of the individual masses:
$\phantom{\rule{2em}{0ex}}{\sum}_{x=0}^{x=1}2\rho xy\delta x$
The limits on the sum are chosen so that all slices are included.
Write down the integral defined by letting $\delta x\to 0$ :
${\int}_{x=0}^{x=1}2\rho xy\phantom{\rule{0.3em}{0ex}}dx$

Noting that
$y=3x$
, express the integrand in terms of
$x$
and evaluate it:
${\int}_{0}^{1}6\rho {x}^{2}\phantom{\rule{0.3em}{0ex}}dx={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}2\rho {x}^{3}\right]}_{0}^{1}=2\rho $

Calculate
$\stackrel{\u0304}{x}$
and hence find the centre of mass of the lamina:
This must equal the moment of the total mass acting at the centre of mass so $3\rho \stackrel{\u0304}{x}=2\rho $ giving $\stackrel{\u0304}{x}=\frac{2}{3}$ . Now the coordinates of the centre of mass are thus $\left(\frac{2}{3},0\right)$ .
Task!
Find the centre of mass of the plane lamina shown below.
The coordinates of $\stackrel{\u0304}{x}$ and $\u0233$ must be calculated separately.
Stage 1: To calculate $\stackrel{\u0304}{x}$

Let
$\rho $
equal the mass per unit area. Write down the total area, the total mass, and its moment about the
$y$
axis:
$2,2\rho ,2\rho \stackrel{\u0304}{x}$

To calculate
$\stackrel{\u0304}{x}$
the lamina is divided into thin slices; a typical slice is shown in the figure above. We assume that the shaded slice is rectangular, which is a reasonable approximation.
Write down the height of the typical strip shown in the figure, its area, and its mass:
$y$ , $y\delta x$ , $\left(y\delta x\right)\rho $

Write down the moment about the
$y$
axis of the typical strip:
$\left(y\delta x\right)\rho x$

The sum of the moments of all strips is
$\phantom{\rule{2em}{0ex}}{\sum}_{x=0}^{x=2}\rho xy\delta x$
Write down the integral which follows as $\delta x\to 0$ :
${\int}_{0}^{2}\rho xy\phantom{\rule{0.3em}{0ex}}dx$

In this example,
$y=x$
because the line
$y=x$
defines the upper limit of each strip (and hence its height). Substitute this value for
$y$
in the integral, and evaluate it:
${\int}_{0}^{2}\rho {x}^{2}\phantom{\rule{0.3em}{0ex}}dx=\frac{8}{3}\rho $

Equating the sum of individual moments and the total moment gives
$2\rho \stackrel{\u0304}{x}=\frac{8}{3}\rho $
. Deduce
$\stackrel{\u0304}{x}$
:
$\stackrel{\u0304}{x}=\frac{4}{3}$
We will illustrate two alternative ways of calculating $\u0233$ .
Stage 2 : To calculate $\u0233$ using vertical strips

Referring to the figure again, which we repeat here, the centre of mass of the slice must lie half way along its length, that is its
$y$
coordinate is
$\frac{y}{2}$
. Assume that all the mass of the slice,
$y\rho \delta x$
, acts at this point. Then its moment about the
$x$
axis is
$y\rho \delta x\phantom{\rule{0.3em}{0ex}}\frac{y}{2}$
. Adding contributions from all slices gives the sum
$\phantom{\rule{2em}{0ex}}{\sum}_{x=0}^{x=2}\frac{{y}^{2}\rho}{2}\delta x$

Write down the integral which is defined as
$\delta x\to 0$
:
${\int}_{x=0}^{2}\frac{\rho {y}^{2}}{2}\phantom{\rule{0.3em}{0ex}}dx$

We can write the above as
$\phantom{\rule{2em}{0ex}}\rho {\int}_{x=0}^{2}\frac{{y}^{2}}{2}\phantom{\rule{0.3em}{0ex}}dx$ and in this example $y=x$ , so the integral becomes
$\phantom{\rule{2em}{0ex}}\rho {\int}_{x=0}^{2}\frac{{x}^{2}}{2}\phantom{\rule{0.3em}{0ex}}dx$
Evaluate this.
$\frac{4\rho}{3}$

This is the sum of the individual moments about the
$x$
axis and must equal the moment of the total mass about the
$x$
axis which has already been found as
$2\rho \u0233$
. Therefore
$\phantom{\rule{2em}{0ex}}2\rho \u0233=\frac{4\rho}{3}\phantom{\rule{2em}{0ex}}\text{fromwhich}\phantom{\rule{1em}{0ex}}\u0233=\frac{2}{3}$

Finally deduce
$\u0233$
and state the coordinates of the centre of mass:
$\u0233=\frac{2}{3}$ and the coordinates of the centre of mass are $\left(\frac{4}{3},\frac{2}{3}\right)$
Stage 3 : To calculate $\u0233$ using horizontal strips

This time the lamina is divided into a number of horizontal slices; a typical slice is shown below.
The length of the typical slice shown is $2x$ .
Write down its area, its mass and its moment about the $x$ axis :
$\left(2x\right)\delta y$ , $\rho \left(2x\right)\delta y$ , $\rho \left(2x\right)y\delta y$

Write down the expression for the sum of all such moments and the corresponding integral as
$\delta y\to 0$
.
${\sum}_{y=0}^{y=2}\rho \left(2x\right)y\delta y$ , ${\int}_{0}^{2}\rho \left(2x\right)y\phantom{\rule{0.3em}{0ex}}dy$

Now, since
$y=x$
the integral can be written entirely in terms of
$y$
as
$\phantom{\rule{2em}{0ex}}{\int}_{0}^{2}\rho \left(2y\right)y\phantom{\rule{0.3em}{0ex}}dy$
Evaluate the integral and hence find $\u0233$ :
$\frac{4\rho}{3}$ ; As before the total mass is $2\rho $ , and its moment about the $x$ axis is $2\rho \u0233$ . Hence
$\phantom{\rule{2em}{0ex}}2\rho \u0233=\frac{4\rho}{3}\phantom{\rule{2em}{0ex}}\text{fromwhich}\phantom{\rule{1em}{0ex}}\u0233=\frac{2}{3}\phantom{\rule{2em}{0ex}}\text{whichwastheresultobtainedbeforeinStage2.}$
Task!
Find the position of the centre of mass of a uniform semicircular lamina of radius $a$ , shown below.
The equation of a circle centre the origin, and of radius $a$ is ${x}^{2}+{y}^{2}={a}^{2}$ .
By symmetry $\stackrel{\u0304}{x}=0$ . However it is necessary to calculate $\stackrel{\u0304}{y}$ .

The lamina is divided into a number of horizontal strips and a typical strip is shown. Assume that each strip is rectangular. Writing the mass per unit area as
$\rho $
, state the area and the mass of the strip:
$2x\delta y$ , $2x\rho \delta y$

Write down the moment of the mass about the
$x$
axis:
$2x\rho y\delta y$

Write down the expression representing the sum of the moments of all strips and the corresponding integral obtained as
$\delta y\to 0$
:
${\sum}_{y=0}^{y=a}2x\rho y\delta y$ , ${\int}_{0}^{a}2x\rho y\phantom{\rule{0.3em}{0ex}}dy$

Now since
${x}^{2}+{y}^{2}={a}^{2}$
we have
$x=\sqrt{{a}^{2}{y}^{2}}$
and the integral becomes:
$\phantom{\rule{2em}{0ex}}{\int}_{0}^{a}2\rho y\sqrt{{a}^{2}{y}^{2}}\phantom{\rule{0.3em}{0ex}}dy$
Evaluate this integral by making the substitution ${u}^{2}={a}^{2}{y}^{2}$ to obtain the total moment.
$\frac{2\rho {a}^{3}}{3}$

The total area is half that of a circle of radius
$a$
, that is
$\frac{1}{2}\pi {a}^{2}$
. The total mass is
$\frac{1}{2}\pi {a}^{2}\rho $
and its moment is
$\frac{1}{2}\pi {a}^{2}\rho \u0233$
.
Deduce $\u0233$ :
$\phantom{\rule{2em}{0ex}}\frac{1}{2}\pi {a}^{2}\rho \u0233=\frac{2\rho {a}^{3}}{3}\phantom{\rule{2em}{0ex}}\text{fromwhich}\phantom{\rule{1em}{0ex}}\u0233=\frac{4a}{3\pi}$