2.3 Continuous Extended Objects

In the case where the number of particles in the system becomes so large that the distinction between the individual particles becomes smooth, we stop computing the centre of mass quantities as discrete sums and switch to integrals. The intuition for this is as follows. As the number of particles $N$ tends to infinity, the mass $m_{i}$ becomes a small mass element $\operatorname{\mathrm{d}}\!m$ , which is what we integrate over.

M=\int\operatorname{\mathrm{d}}\!m,\quad\boldsymbol{r}_{\text{COM}}=\frac{1}{M% }\int\boldsymbol{r}\operatorname{\mathrm{d}}\!m.

(2.20)

To evaluate this integral, we will need to convert the mass element $\operatorname{\mathrm{d}}\!m$ into a spatial element, for example a length, area, or volume element. This is done using a linear, surface, or volume density.

\operatorname{\mathrm{d}}\!m=\lambda(x)\operatorname{\mathrm{d}}\!x,\quad% \operatorname{\mathrm{d}}\!m=\sigma(\boldsymbol{r})\operatorname{\mathrm{d}}\!% A,\quad\operatorname{\mathrm{d}}\!m=\rho(\boldsymbol{r})\operatorname{\mathrm{% d}}\!V.

(2.21)

We will look at some examples here in the form of uniform lamina, which are flat (two-dimensional) extended objects with uniform surface density.

Example 2.2.

Find the centre of mass of a right triangle with both small side lengths $a$ .

We will find each component of the centre of mass position separately. In fact, by the symmetry of the shape we only need to find the average $x$ position, because the average $y$ position will be the same. We will break the triangle up into thin vertical slices of area $y\operatorname{\mathrm{d}}\!x$ . Note that $y=a-x$ (equation of a straight line), so we have

\bar{x}=\frac{1}{M}\int x\operatorname{\mathrm{d}}\!m=\frac{1}{M}\int_{0}^{a}% \sigma xy\operatorname{\mathrm{d}}\!x=\frac{1}{M}\int_{0}^{a}\sigma x(a-x)% \operatorname{\mathrm{d}}\!x.

(2.22)

The total mass of the uniform lamina is $M=\frac{1}{2}\sigma a^{2}$ , so the integral becomes

\bar{x}=\frac{2}{a^{2}}\int_{0}^{a}(ax-x^{2})\operatorname{\mathrm{d}}\!x=% \frac{2}{a^{2}}\left[\frac{ax^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{a}=\frac{a}{% 3}.

(2.23)

Thus the centre of mass position is $\boldsymbol{r}_{\text{COM}}=\left(\frac{a}{3},\frac{a}{3}\right)$ .

Example 2.3.

Find the centre of mass of a hemicircular uniform lamina of radius $r$ .

By symmetry, we can see that $\bar{x}$ must be zero. Then, dividing the lamina into horizontal slices of area $2x\operatorname{\mathrm{d}}\!y$ (remember the factor of two because the slice goes from negative $x$ to positive $x$ !), and using the equation of a circle to write $x=\sqrt{r^{2}-y^{2}}$ , we get

\bar{y}=\frac{1}{M}\int y\operatorname{\mathrm{d}}\!m=\frac{1}{M}\int_{0}^{r}2% \sigma xy\operatorname{\mathrm{d}}\!y=\frac{1}{M}\int_{0}^{r}2\sigma y\sqrt{r^% {2}-y^{2}}\operatorname{\mathrm{d}}\!y.

(2.24)

The total mass of the hemicircle is $M=\frac{1}{2}\sigma\pi r^{2}$ , so

\bar{y}=\frac{4}{\pi r^{2}}\int_{0}^{r}y\sqrt{r^{2}-y^{2}}\operatorname{% \mathrm{d}}\!y=\frac{4r}{3\pi}.

(2.25)

The center of mass is therefore $\boldsymbol{r}_{\text{COM}}=\left(0,\frac{4r}{3\pi}\right)$ .

Example 2.4.

Sometimes we can solve the problem using symmetries without having to do any integrals at all. Consider the following uniform lamina.

Using the techniques we have built up in this chapter, we can find the centre of mass without doing any integrals.

Firstly, we can consider the whole object in two separate subsections. We will call the left section $A$ and the right section $B$ . Then the centre of mass of the whole lamina will be the average of the centres of mass of each subsection.

\boldsymbol{r}_{\text{COM}}=\left(\frac{m_{A}x_{\text{COM},A}+m_{B}x_{\text{% COM},B}}{m_{A}+m_{B}},\frac{m_{A}y_{\text{COM},A}+m_{B}y_{\text{COM},B}}{m_{A}% +m_{B}}\right)

(2.26)

Since both subsections are rectangles, by symmetry the centre of mass must be the geometric centre. Thus, the centre of mass of section $A$ is $(1,4)$ and for section $B$ it is $(5,2)$ . Now, we need to know the mass of each subsection. Note that since the surface density is uniform, the mass is proportional to area. So the mass of section $A$ is proportional to the area of section $A$ , likewise for section $B$ , and the total mass is proportional to the total area.

m_{A}\propto A_{A},\quad m_{B}\propto A_{B},\quad m_{A}+m_{B}\propto A_{A}+A_{% B}.

(2.27)

The area of section $A$ is $16$ square units, and the area of section $B$ is $24$ square units, so now we can calculate the ratios of mass of subsection to total mass:

\frac{m_{A}}{m_{A}+m_{B}}=\frac{A_{A}}{A_{A}+A_{B}}=\frac{16}{40}=\frac{2}{5},% \quad\frac{m_{B}}{m_{A}+m_{B}}=\frac{A_{B}}{A_{A}+A_{B}}=\frac{24}{40}=\frac{3% }{5}.

(2.28)

Therefore, the centre of mass of the whole lamina is

\boldsymbol{r}_{\text{COM}}=\left(\frac{2}{5}1+\frac{3}{5}5,\frac{2}{5}4+\frac% {3}{5}2\right)=(3.4,2.8).

(2.29)