How Is the Center of Gravity Calculated for a Non-Uniform Rod?

[decathlonist]

Hello,

I was studying for the test and I ran into the problem that I can't solve.

This problem involves the use of Calculus (in particular - integration). Although I do have a solution for the problem, I don't have any explanation for it. Can someone please tell me how we arrived to the final answer?

"A rod of length L and linear mass density l(x)=Ax kg/m, where A is a constant and x is measured in meters from one end of the rod. Assuming the rod is in a uniform gravitational field, determine the location of the center of the rod."

( center of gravity = 2L/3 )

Thank you

 

[bogdan]

Are you sure that the answer is 2L/3 ?
Because I think it's L*sqrt(2)/2 (the same as L/sqrt(2))
sqrt -> square root;
[?]

 

[HallsofIvy]

As "sir-pinski" was pointing out, you can't do a problem involving center of gravity until you know what that means! (It's remarkable how many people don't seem to notice that you can't do a problem if you don't know the definitions!)
The "center of gravity" is the point at which the rod would balance: the torque (twisting force) on one side about that point is equal to (and opposite) the torque on the other. The torque about a point is itself defined as the force applied times the distance from the point.

Imagine dividing the rod into small lengths, each of length dx, and treat the weight as constant on each small length (this technique should be familiar to you). Take the left end of the bar as x=0 and let "a" be the distance to the center of gravity (I would prefer x with a bar over it but..). For each point on the bar, at distance x from the left end, the distance from the center of gravity is |x-a|. However, if we just use x-a (rather than the absolute value) that will also allow us to treat left and right of it as positive and negative. The mass of each little "piece" of the rod is its density times its length: l(x)dx. The torque about the center of gravity of a little piece of the rod is (x-a)l(x)dx. The total torque would be [sigma](x-a)l(x)dx and in order to be balanced must be 0: [sigma](x-a)l(x)dx = 0 which is the same as [sigma]xl(x)dx= a[sigma]l(x)dx.

These are Riemann sums and in the limit become the integral:

[integral]xl(x)dx= a [integral]l(x)dx.
(The integral on the right is, of course, the total weight of the rod.)

In this particular case l(x)= Ax so [integral]l(x)dx= [integral]Axdx= (1/2)Ax² evaluated between 0 and L= (1/2)Al².

[integral]xl(x)dx= [integral]Ax²dx= (1/3)Ax³
evaluated between 0 and L= (1/3)AL³.

We must have (1/3)A L³= a(1/2)A L² so

a= (2/3)L as advertised.