Center of Mas of a planar Lamina

[hagobarcos]

Homework Statement

Find the center of mass of a planar lamina, in the form of a triangle with vertices (0,0),(0,a),(a,a),
if ρ=k

Homework Equations

m = ∫∫f dA

xbar = My/m

ybar = Mx/m

The Attempt at a Solution

mass = ka²/2

Mx = ∫∫yk dy dx

My = ∫∫xk dy dx


**Side Question, how do we use math type or some other type of symbolic language on physics forums? ***

Any thoughts would be appreciated.
Below is my photo:

 

[Simon Bridge]

You can use LaTeX to type maths.
i.e. $$M_x = \iint ky\;dy\;dx$$

Did you have a question?

 

[HallsofIvy]

If the density is a constant, the "center of mass" is the "centroid", the geometric center of the figure. And for a triangle that is particularly simple to find. If you really want to use the double integral, x goes from 0 to a and, for each x, y goes from 0 to x:

m= \int_0^a\int_0^x k dydx
M_x= \int_0^a\int_0^x kx dydx
M_y= \int_0^a\int_0^x ky dydx

 

[hagobarcos]

Aw sweet, turns out I had the wrong variable in my Mx & My formulas, kept getting the thing wrong:)And for symbols, my screen just shows "quick symbols", none of the fancy LaTex y'all are using ^^

 

[hagobarcos]

Okay actually I do have a question, in my book the formula for Mx is double integral of ky dA

And for My is double integral kx dA Which is correct?

 

[HallsofIvy]

What do you mean "which" is correct? M_x is the moment of inertia about the x-axis and is \int\int ky dA. The y coordinate of the center of mass is \overline{y}= M_x/m= \int\int ky dA/m and the x coordinate of the center of mass is \overline{x}= M_y/m= \int\int kx dA/m.

Personally, I find the "M_x", "M_y" notation "non-intuitive" and prefer the \overline{x}= \int\int kx dA, \overline{y}= \int\int ky dA notation.

 

[hagobarcos]

oh! Yes :D Excellent, thank you for clearing that up.