Another center of mass problem

**Twig** · July 7th 2009, 02:29 AM

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.

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

Thx for your help!

**alexmahone** · July 7th 2009, 02:44 AM

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.

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

Thx for your help!

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.

**Random Variable** · July 7th 2009, 02:50 AM

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

**Twig** · July 7th 2009, 02:52 AM

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

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

**Random Variable** · July 7th 2009, 02:54 AM

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.

**Twig** · July 7th 2009, 03:01 AM

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?

**malaygoel** · July 7th 2009, 03:25 AM

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?

your answer 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.

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