# Another center of mass problem

• Jul 7th 2009, 03:29 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 $dm = \rho \cdot A \cdot dx$ , because I guess the area is the same throughout the rod.

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

• Jul 7th 2009, 03:44 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 $dm = \rho \cdot A \cdot dx$ , because I guess the area is the same throughout the rod.

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

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

but I really don't know if that's the correct formula for the centre of mass of a rod.
• Jul 7th 2009, 03:50 AM
Random Variable
It seems correct except for the constant A. The rod is a 1-dimensional object.
• Jul 7th 2009, 03:52 AM
Twig
I was thinking about the cross sectional area of the rod.

So you mean I should just take $dm=\rho \cdot dx$ ?
• Jul 7th 2009, 03:54 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 $dm=\rho \cdot dx$ ?

Yes. The word "slender" in this problem is telling you to treat the rod as 1-dimensional.
• Jul 7th 2009, 04:01 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? $\frac{4}{9}$ is correct?
• Jul 7th 2009, 04:25 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? $\frac{4}{9}$ is correct?