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I am asked to find the moment about the $x$ axis for a thin wire of constant density. This thin wire lies along the curve $y=\sqrt{x}$ and the limits for integration are $x=0$ and $x=2$.

I know from my textbook that the moment about the $x$ axis is: $M_y = \int \tilde{y} dm$

Because this is a thin wire, I know that I need to subdivide the wire into small segments for integration. I have the following for relevant data for each segment:

Length: $dl = \sqrt{x}dx$

mass: $dm = \delta dl = \delta \sqrt{x}dx$

It's the part about the distance of the center of mass to the $x$ axis that I think I'm missing. I have the following:

$\tilde{y} = \sqrt{x}$

Therefore, my final integral is:

$$ M_y = \int \tilde{y} dm = \int_0^2 \delta \sqrt{x} \sqrt{x} dx = \delta \int_0^2 x dx = \delta \left.\frac{1}{2}x^2\right|_0^2 = \delta 2 $$

This particular problem is an odd numbered problem and so I know that I've got it incorrect. Please help me to see where I'm going wrong.

Thanks, Andy

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up vote 2 down vote accepted

The center of mass for a wire of constant density making a curve $y=f(x)$ between $x=a$ and $x=b$ is

$$\bar{y} = \frac{\int_a^b dx \: y \sqrt{1+y'^2}}{\int_a^b dx \:\sqrt{1+y'^2}}$$

In the case you describe

$$\bar{y} = \frac{\displaystyle \int_0^2 dx \: \sqrt{x} \sqrt{1+\frac{1}{4 x}}}{\displaystyle \int_0^2 dx \: \sqrt{1+\frac{1}{4 x}}}$$

The top integral is relatively simple and is equal to

$$\int_0^2 dx \: \sqrt{x+\frac{1}{4}} = \frac{2}{3} \left ( \frac{27}{8} - \frac{1}{8} \right ) = \frac{13}{6}$$

The bottom integral may be evaluated by making the substitution $\sec{\theta} = \sqrt{1+\frac{1}{4 x}}$ to get

$$\int_0^2 dx \: \sqrt{1+\frac{1}{4 x}} = \frac{1}{2} \int_{\arccos{\sqrt{8}/3}}^{\pi/2} d\theta \: \csc^3{\theta} = \frac{1}{8} \left(12 \sqrt{2}+\log \left(17+12 \sqrt{2}\right)\right)$$

Therefore your center of mass is

$$\bar{y} = \frac{52/3}{12 \sqrt{2}+\log \left(17+12 \sqrt{2}\right)} \approx 0.846$$

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The element of arclength is $ds=\sqrt {1+(\frac {dy}{dx})^2}dx$ or $ds=\sqrt {1+(\frac {dx}{dy})^2}dy$ and $dm=\delta ds$ To find the center of mass in the $y$ direction you need to find the average $y$. Please check your text, the center of mass in $y$ should be $y_{CM}=\frac 1m\int y dm$, then the moment of inertia in the $y$ direction around the center of mass is $\int (y-y_{CM})^2dm$. The expression you give doesn't have the proper units for MOI, which should be mass*length^2

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The question/comment I have is really relevant to both responders so I'll put it here. This is apparently, one of the areas of misunderstanding that I had. The problem asked me to find the moment about the x-axis which I understood to be $\tilde{y}$. So this is what I was trying to find. Because the problem was worded thusly, I didn't think it was a question of "Find the Center of Mass." I thought that finding center of mass was to find $(\bar{x},\bar{y})$. At any rate, thank you both for you guidance. – Andrew Falanga Feb 22 '13 at 0:48

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