In Section 8.5 we calculated the center of mass by considering objects composed of a finite number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals

${\mathbf{x}}_{\mathbf{cm}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{M}}\mathbf{\int }\mathbf{xdm}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}{\mathbf{y}}_{\mathbf{cm}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{M}}\mathbf{\int }\mathbf{ydm}$

where x and y are the coordinates of the small piece of the object that has mass dm. The integration is over the whole of the object. Consider a thin rod of length L, mass M, and cross-sectional area A. Let the origin of the coordinates be at the left end of the rod and the positive x-axis lies along the rod. (a) If the density$\mathbf{\rho }\mathbf{=}\frac{\mathbf{M}}{\mathbf{V}}$ of the object is uniform, perform the integration described above to show that the x-coordinate of the center of mass of the rod is at its geometrical center. (b) If the density of the object varies linearly with x-that is, $\mathbf{\rho }\mathbf{=}\mathbf{\alpha x}$, where$\mathbf{\alpha }$ is a positive constant—calculate the x-coordinate of the rod’s center of mass.

The length of the thin rod is L.

The mass of the rod is M.

The cross-sectional area of the thin rod is A.

Density tells that how different materials act together when they are mixed together. It depends upon the mass and volume of object or substance.

The expression of density is given by,

$\mathit{p}\mathbf{=}\frac{\mathbf{M}}{\mathbf{V}}$

Consider a small piece of the rod that has massdm and length dx.

The density of the thin rod is given by,

$$\begin{gathered} p=\frac{M}{V} \\ =\frac{M}{AL} \end{gathered}$$

Mass of the small piece can be written as,

$$\begin{gathered} dm=pAdx \\ =\left(\frac{M}{AL}\right)Adx \\ =\left(\frac{M}{L}\right)dx \end{gathered}$$

The center of mass of the thin rod has only coordinate, as the rod lies along the x-axis, and it can be calculated as follows.

$$\begin{gathered} x_{cm}=\frac{1}{M}\int xdm \\ =\frac{1}{M}{\int }_0^L\left(\frac{M}{L}\right)xdx \\ =\frac{1}{L}{\left(\frac{1}{2}{x}^2\right)}_0^L \\ =\frac{L}{2} \end{gathered}$$

Therefore, the x-coordinate of the center of mass of the thin rod is $\frac{L}{2}$.

If the density of the object varies linearly with x that is given by,

$\rho =\alpha x$

Here,$\alpha$ is positive constant.

Mass of the small piece can be written as,

$$\begin{gathered} dm=\rho Adx \\ =A\alpha xdx \end{gathered}$$

The mass of the thin rod is given by,

$$\begin{gathered} M=\int dm \\ =\alpha A{\int }_0^{L}xdx \\ =\alpha A{\left(\frac{x^2}{2}\right)}_0^L \\ =\frac{\alpha A{L}^2}{2} \end{gathered}$$

Find the x-coordinate of the center of mass of the thin rod as follows.

$$\begin{gathered} x_{cm}=\frac{1}{M}xdm \\ =\frac{2}{\alpha A{L}^2}{\int }_0^L\alpha A{x}^{2}dx \\ =\frac{2}{L^2}{\left(\frac{1}{3}{x}^3\right)}_0^L \\ =\frac{2L}{3} \end{gathered}$$

Therefore, the x-coordinate of the center of mass of the thin rod is $\frac{2L}{3}$.