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