- #1
MasterNewbie
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Homework Statement
Find the moment of inertia of a circular wire hoop about a point on the hoop.
mass = m
radius = r
Homework Equations
Izz = Iyy + Ixx
The Attempt at a Solution
Well, using the above equation I first tried to solve for Ixx. Since we are looking for the moment of inertia about a point on the hoop I picked an arbitrary point on the hoop to be the origin for the calculation. With my chosen system the hoop would appear suspended from the origin, x-axis running from left to right, y-axis bottom to top and z-axis behind to front.
Ixx = m[tex]\int[/tex][tex]\int[/tex](y2+z2)dydx
z is 0 so it drops from the equation, leaving m[tex]\int[/tex][tex]\int[/tex](y2)dydx
My issue is trying to find some function to model the curve. At first I attempted to use polar coordinates but I couldn't find an equation that would give the shape of the hoop in my coordinate system. Then I used parametric equations for each x and y. x = cos(t) and y = 1 - sin(t), 0<t<2[tex]\pi[/tex] for my limits. dy = -cos(t) and dx = -sin(t). Making those substitutions in the integral and evaluating it I found that one of the integral steps requires a u-substitution, and changing the limits of integration to reflect the substitution would make both limits the same (integral = 0) which doesn't make much sense to me. After that I stopped trying to get Ixx
Trying to find Iyy For this, because shifting the hoop vertically won't affect the distance each point is away from the y-axis, I made the center of the hoop the origin in this calculation. This let me write x = [tex]\sqrt{r^2 - y^2}[/tex]. Setting up the integral, m[tex]\int[/tex][tex]\int[/tex]x2dxdy, with the inner limit between 0 and the above equation for x, and for the y limits, -r/2 and r/2. This would give one side of the loop so I multiplied by 2 to include both sides. Evaluating the inner integral, I get 2m[tex]\int[/tex][tex]\sqrt{R^2-y^2}[/tex]3dy which is evaluated with a trig-sub. However when I make the substitution, because the root is being cubed and an extra r arising from the dy=rcos[tex]\vartheta[/tex]d[tex]\vartheta[/tex], r4 can be pulled out, making it dimensionally inconsistent with the moment of inertia. I spotted a mistake I made while writing this so I will continue again, but I don't think it will work out with an extra r2 outside.
I want to apologize for my obvious terribleness with the equation writing feature here on these forums.