Center of Mass of Thin Plates

We’ve examined point masses on a line and in a plane. Now, we look at systems where mass is distributed continuously across a thin sheet, called a lamina. We assume the lamina’s density is constant.

Laminas are often two-dimensional regions in a plane, with the geometric center called its centroid. The center of mass of a lamina depends only on the shape, not the density. For a rectangular lamina, the center of mass is where the diagonals intersect, which follows the symmetry principle.

Let’s examine general laminas. Suppose we have a lamina bounded above by the graph of a continuous function [latex]f(x),[/latex] below by the [latex]x[/latex]-axis, and on the left and right by the lines [latex]x=a[/latex] and [latex]x=b,[/latex].

To find the center of mass, we need the total mass of the lamina. We divide the lamina into thin vertical strips, approximating each strip’s mass using the density [latex]ρ[/latex]. The mass of the strip is given by [latex]\rho f({x}_{i}^{*})\text{Δ}x.[/latex]

To get the approximate mass of the lamina, we add the masses of all the rectangles to get

This is a Riemann sum. Taking the limit as [latex]n\to \infty[/latex] gives the exact mass of the lamina:

Next, we calculate the moment of the lamina with respect to the x-axis. For each rectangle, the center of mass is at [latex]x_{i}^{*}[/latex]. The moment with respect to the [latex]x[/latex]-axis is given by:

Similarly, the moment with respect to the [latex]y[/latex]-axis is:

The coordinates of the center of mass are:

[latex]\overline{x}=\frac{{M}_{y}}{m} \text{ and }\overline{y}=\frac{{M}_{x}}{m}[/latex]

We summarize these findings in the following theorem.

In the next example, we use this theorem to find the center of mass of a lamina.

Let R be the region bounded above by the graph of the function [latex]f(x)=\sqrt{x}[/latex] and below by the [latex]x[/latex]-axis over the interval [latex]\left[0,4\right].[/latex] Find the centroid of the region.

Show Solution

The region is depicted in the following figure.

Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant [latex]\rho[/latex] cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume [latex]\rho =1.[/latex]

First, we need to calculate the total mass:

[latex]\begin{array}{cc}\hfill m& =\rho {\displaystyle\int }_{a}^{b}f(x)dx={\displaystyle\int }_{0}^{4}\sqrt{x}dx\hfill \\ & ={\frac{2}{3}{x}^{3\text{/}2}|}_{0}^{4}=\frac{2}{3}\left[8-0\right]=\frac{16}{3}.\hfill \end{array}[/latex]

Next, we compute the moments:

[latex]\begin{array}{cc}\hfill {M}_{x}& =\rho {\displaystyle\int }_{a}^{b}\frac{{\left[f(x)\right]}^{2}}{2}dx\hfill \\ & ={\displaystyle\int }_{0}^{4}\frac{x}{2}dx={\frac{1}{4}{x}^{2}|}_{0}^{4}=4\hfill \end{array}[/latex]

and

[latex]\begin{array}{cc}\hfill {M}_{y}& =\rho {\displaystyle\int }_{a}^{b}xf(x)dx\hfill \\ & ={\displaystyle\int }_{0}^{4}x\sqrt{x}dx={\displaystyle\int }_{0}^{4}{x}^{3\text{/}2}dx\hfill \\ & ={\frac{2}{5}{x}^{5\text{/}2}|}_{0}^{4}=\frac{2}{5}\left[32-0\right]=\frac{64}{5}.\hfill \end{array}[/latex]

Thus, we have

[latex]\overline{x}=\frac{{M}_{y}}{m}=\frac{64\text{/}5}{16\text{/}3}=\frac{64}{5}·\frac{3}{16}=\frac{12}{5}\text{ and }\overline{y}=\frac{{M}_{x}}{y}=\frac{4}{16\text{/}3}=4·\frac{3}{16}=\frac{3}{4}.[/latex]

The centroid of the region is [latex](12\text{/}5,3\text{/}4).[/latex]