Another center of mass problem

• Jul 7th 2009, 02:29 AM
Twig
Another center of mass problem
Hi!

Attached picture below.

Problem: Calcuate center of mass of the slender rod.

Solution attempt:

Since the density varies, I did: With $\displaystyle dm = \rho \cdot A \cdot dx$ , because I guess the area is the same throughout the rod.

$\displaystyle \bar{x}=\frac{\int_{0}^{1}x\cdot \rho_{0}(1-\frac{x}{2})A\; dx}{\int_{0}^{1} \rho_{0}(1-\frac{x}{2})A\; dx} = \frac{4}{9}$

• Jul 7th 2009, 02:44 AM
alexmahone
Quote:

Originally Posted by Twig
Hi!

Attached picture below.

Problem: Calcuate center of mass of the slender rod.

Solution attempt:

Since the density varies, I did: With $\displaystyle dm = \rho \cdot A \cdot dx$ , because I guess the area is the same throughout the rod.

$\displaystyle \bar{x}=\frac{\int_{0}^{1}x\cdot \rho_{0}(1-\frac{x}{2})A\; dx}{\int_{0}^{1} \rho_{0}(1-\frac{x}{2})A\; dx} = \frac{4}{9}$

I confirmed that $\displaystyle \frac{\int_{0}^{1}x\cdot \rho_{0}(1-\frac{x}{2})A\; dx}{\int_{0}^{1} \rho_{0}(1-\frac{x}{2})A\; dx} = \frac{4}{9}$

but I really don't know if that's the correct formula for the centre of mass of a rod.
• Jul 7th 2009, 02:50 AM
Random Variable
It seems correct except for the constant A. The rod is a 1-dimensional object.
• Jul 7th 2009, 02:52 AM
Twig
I was thinking about the cross sectional area of the rod.

So you mean I should just take $\displaystyle dm=\rho \cdot dx$ ?
• Jul 7th 2009, 02:54 AM
Random Variable
Quote:

Originally Posted by Twig
I was thinking about the cross sectional area of the rod.

So you mean I should just take $\displaystyle dm=\rho \cdot dx$ ?

Yes. The word "slender" in this problem is telling you to treat the rod as 1-dimensional.
• Jul 7th 2009, 03:01 AM
Twig
Quote:

Originally Posted by Random Variable
Yes. The word "slender" in this problem is telling you that can treat the rod as 1-dimensional.

Ah, I see. Really didnīt know that word in english =)

Note: But the constant A didnīt affect the result? $\displaystyle \frac{4}{9}$ is correct?
• Jul 7th 2009, 03:25 AM
malaygoel
Quote:

Originally Posted by Twig
Ah, I see. Really didnīt know that word in english =)

Note: But the constant A didnīt affect the result? $\displaystyle \frac{4}{9}$ is correct?

A doesn't affect the result because it is assumed to be constant.

A two-dimensional figure is a as good as a one-dimensional as long as one dimension remains unchanged.

For example, if we are dealing with all rectangles with the same same, then we have to consider only their relative length..hence problem comes in the realm of one-dimension.