Moment of inertia of a hoop about a point

[MasterNewbie]

Homework Statement

Find the moment of inertia of a circular wire hoop about a point on the hoop.
mass = m
radius = r

Homework Equations

I_zz = I_yy + I_xx

The Attempt at a Solution

Well, using the above equation I first tried to solve for I_xx. 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.

I_xx = m\int\int(y²+z²)dydx
z is 0 so it drops from the equation, leaving m\int\int(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\pi 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 I_xx

Trying to find I_yy 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 = \sqrt{r^2 - y^2}. Setting up the integral, m\int\intx²dxdy, 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\int\sqrt{R^2-y^2}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\varthetad\vartheta, 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.

 

[MasterNewbie]

I dug through my mechanics book, and there is an equation you can use to calculate the components of the inertia tensor about any point, if you know the components of the tensor about the center or mass. Keeping in mind the point is displaced from the cm by (x,y,z) (so for mine x=z=0, y=-r)

Izz = Izz(cm) + M(x^2+y^2), since x^2+y^2=r^2 regardless of any point on the hoop, and the moment of inertia of a thin hoop is mr^2, Izz about any point on the hoop is 2mr^2.