# Moment of inertia

• Jul 19th 2009, 01:51 PM
arbolis
Moment of inertia
I'd like to know where I went wrong :
A lattice with constant density $\rho (x,y)= \rho$ is on the region under the curve $y= \sin x$ from $x=0$ to $x=\pi$. Find the moments of inertia $I_x$ and $I_y$.
My attempt : $I_x$ :
I realize that the region is an upper semi circle with radius $r=1$.
$I_x=\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta=\frac{\pi}{10}$, which mean that $I_x$ does not depend on the mass of the lattice!
I'm guessing that my error comes from the first step, that is $\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta$. Should I replace $y$ by $\sin x$, so $y=\sin (\cos (\theta))$?
If so, then I don't understand why I can't take $y$ as $\rho \sin (\theta)$...
• Jul 19th 2009, 03:38 PM
Jester
Quote:

Originally Posted by arbolis
I'd like to know where I went wrong :
A lattice with constant density $\rho (x,y)= \rho$ is on the region under the curve $y= \sin x$ from $x=0$ to $x=\pi$. Find the moments of inertia $I_x$ and $I_y$.
My attempt : $I_x$ :
I realize that the region is an upper semi circle with radius $r=1$.
$I_x=\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta=\frac{\pi}{10}$, which mean that $I_x$ does not depend on the mass of the lattice!
I'm guessing that my error comes from the first step, that is $\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta$. Should I replace $y$ by $\sin x$, so $y=\sin (\cos (\theta))$?
If so, then I don't understand why I can't take $y$ as $\rho \sin (\theta)$...

You've got to be careful. You are using $\rho$ for two different things here. A density and a changing radius.