Center of mass of a slender rod with variable density

[jonjacson]

Hi to everybody

Homework Statement

I´ll show the problem with a picture:

Homework Equations

L[itex]\overline{x}[/itex]=∫xc ρ dl

The Attempt at a Solution

Well the total length of the rod is 1 feet, I only need to calculate the integral.

The moment xc of a differential element of mass of the rod is the distance x to the y axis, and the density is known so:

∫x * (1-x/2) dx , from 0 to 1, the result for me is:

1 * [itex]\overline{x}[/itex] = 1/3

So [itex]\overline{x}[/itex]= 1/3;

Unfortunately the result is 4/9, I can´t see where are my mistakes, maybe I´m not using the proper arm length or something.

The problem doesn´t give you any coordinate system, only the x axis, I assume that the y-axis is orthogonal, and the equation for the moments around that axis gives you the x center of mass.

I suppose that nothing would change if I had used another coordinate system, Am I right?

I assumed too that the constant ρ0 at the equation for the density would not change anything, but maybe that is a mistake.

Where are my mistakes?

 

[Doc Al]

Homework Equations

L[itex]\overline{x}[/itex]=∫xc ρ dl

Instead of L*x_cm, the left hand side should be M*x_cm. Where M is the mass of the rod.

I assumed too that the constant ρ0 at the equation for the density would not change anything, but maybe that is a mistake.

You can't just drop a constant, like you did when you integrated.

 

[jonjacson]

Instead of L*x_cm, the left hand side should be M*x_cm. Where M is the mass of the rod.


You can't just drop a constant, like you did when you integrated.

-Ups, you are right I was used to use the equation of the centroids with constant density and I wrote the wrong equation.

-I see, they cancel after integrating on both parts of the equation.

Now I get the right result.

Thank you very much!