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:

$$\begin{aligned} y_{CM} &= \frac{1}{M} \int y \, dm \\ &= \frac{\rho}{M} \int\limits_{0}^{b} y \left( a-\frac{a}{b} y \right) \, dy \\ &= \frac{1}{M} \frac{2M}{ab} \left[ \frac{ay^{2}}{2}-\frac{ay^{3}}{3b} \right]_{0}^{b} \\ &= \frac{2}{ab} \left( \frac{ab^{2}}{2}-\frac{ab^{3}}{3b} \right) \\ &= \frac{1}{3} b \end{aligned}$$

Next: Centre Of Mass Of A Cone

Previous: What is centre of mass?

May I suggest a different point of view to the upright/facing yellow triangle when analyzing the problem?

It resembles the square I use in my workshop, and my initial interpretation of its “center of mass” was the pivot point P on base ‘a’ where the triangle could stand in equilibrium while upright, facing me, as shown in the figure.

It resembles the square I use in my workshop, and my initial interpretation of its “center of mass”, was the pivot point P on base ‘a’ where the triangle could stand in equilibrium while upright, facing me, as shown in the figure.

In this case, P would be at SQRT(1/2) * a

0,707 * a, slightly more than 2/3 of a

Imagine splitting the yellow triangle vertically at point P.

Let’s call Q the top point where this perpendicular line intersects the hypotenuse.

This new (left) triangle OPQ, we want its mass/area to be half of the total mass M

Mass of triangle OPQ = 1/2 * base * height = 1/2 * Pa * Pb

Total mass M = 1/2 ab

Half of M = 1/4 ab

1/4 ab = 1/2 Pa Pb

1/2 ab = Pa Pb

1/2 = PP (it is independant of length a and b)

SQRT(1/2) = P

0.707

You can verify this by calculating the remaining trapezoidal area or using a tool like GeoGebra, as I did. Alternatively, you can physically cut and rectangle triangle and weigh both pieces to be certain, as I did.

Why is this important? In other textbooks and videos, they often use a bookshelf example where the books have a triangular shape and apply the 2/3 rule to the base. This is incorrect.

It suggests that one could balance the bookshelf at 2/3 of its base, which is wrong.

Let’s have a discussion if you like.

Regards,

Richard Nault