Now, let’s get some practice on calculating centre of mass of objects.
An object of mass $M$ is in the shape of a right-angle triangle whose dimensions are shown in the figure. Locate the coordinates of the centre of mass, assuming that the object has a uniform mass per unit area.
Recall that the equations for centre of mass:
$$\begin{aligned} x_{CM} &= \frac{1}{M} \int x \, dm \\ y_{CM} &= \frac{1}{M} \int y \, dm \end{aligned}$$
First, in order to find $x_{CM}$, we shall slice the triangle into thin slices with mass $dm$, height y and thickness $dx$, as shown in the figure below. This is such that every point in $dm$ has the same x value (same distance from y-axis).
Look at the figure above, from theory of similar triangles, we can obtain this relation:
$$\begin{aligned} \frac{y}{x} &= \frac{b}{a} \\ y &= \frac{b}{a} x \end{aligned}$$
We will use this relation in the calculation for $dm$ below.
The mass per unit area, $\rho$ is given by:
$$\begin{aligned} M &= \rho \left( \frac{1}{2} a b \right) \\ &= \frac{2M}{ab} \end{aligned}$$
Hence, $dm$ will be given by:
$$\begin{aligned} dm &= \rho \left( y \, dx \right) \\ &= \rho \left( \frac{b}{a} x \right) \, dx \end{aligned}$$
Using $dm$ in the equation for centre of mass:
$$\begin{aligned} x_{CM} &= \frac{1}{M} \int x \, dm \\ &= \frac{\rho}{M} \int\limits_{0}^{a} x \left( \frac{b}{a} x \right) \, dx \\ &= \frac{1}{M} \frac{2M}{ab} \left[ \frac{b}{a} \frac{x^{3}}{3} \right]_{0}^{a} \\ &= \frac{2}{ab} \left( \frac{b}{a} \frac{a^{3}}{3} \right) \\ &= \frac{2}{3} a \end{aligned}$$
Now, to find $y_{CM}$, we will choose $dm$ such that every point in $dm$ has the same y value.
From the above figure, using the theory of similar triangles, we can arrive at a relation:
$$\begin{aligned} \frac{y}{x} &= \frac{b}{a} \\ x &= \frac{a}{b} y \end{aligned}$$
Hence, $dm$ is given by:
$$\begin{aligned} dm &= \rho \left( a-x \right) dy \\ &= \rho \left( a-\frac{a}{b} y \right) dy \end{aligned}$$
Using $dm$ in the equation for centre of mass:
$$\begin{aligned} y_{CM} &= \frac{1}{M} \int y \, dm \\ &= \frac{\rho}{M} \int\limits_{0}^{b} y \left( a-\frac{a}{b} y \right) \, dy \\ &= \frac{1}{M} \frac{2M}{ab} \left[ \frac{ay^{2}}{2}-\frac{ay^{3}}{3b} \right]_{0}^{b} \\ &= \frac{2}{ab} \left( \frac{ab^{2}}{2}-\frac{ab^{3}}{3b} \right) \\ &= \frac{1}{3} b \end{aligned}$$
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