Question

A triangular lamina has vertices (0, 0), (0, 1) and (c, 0) for some positive constant c. Assuming constant mass density, show that the y-coordinate of the center of mass of the lamina is independent of the constant c.

Answer 1

The equation of the line through (0, 1) and (c, 0) is

y - 0 = (0 - 1)/(c - 0) (x - c)   ==>   y = 1 - x/c

Let L denote the given lamina,

L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}

Then the center of mass of L is the point $(\bar x,\bar y)$ with coordinates given by

$\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m$

where $M_x$ is the first moment of L about the x-axis, $M_y$ is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.

Let ρ be the mass density of L. Then L has a mass of

$\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2$

Now we compute the first moment about the y-axis:

$\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6$

Then

$\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3$

but this clearly isn't independent of c ...

Maybe the x-coordinate was intended? Because we would have had

$\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6$

and we get

$\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13$