# Math Help - Another center of mass problem

1. ## 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.

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

2. 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.

3. It seems correct except for the constant A. The rod is a 1-dimensional object.

4. I was thinking about the cross sectional area of the rod.

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

5. 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.

6. 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 didn´t know that word in english =)

Note: But the constant A didn´t affect the result? $\frac{4}{9}$ is correct?

7. Originally Posted by Twig
Ah, I see. Really didn´t know that word in english =)

Note: But the constant A didn´t affect the result? $\frac{4}{9}$ is correct?