# Period of swinging incomplete hula-hoop

I was working on a problem where I had to calculate the period of a swinging incomplete hula-hoop given its center of mass and radius. It only swings with very small amplitude so I considered the motion as harmonic so therefore I attempted to use the equation $$T=2\pi (\sqrt{(\frac{I support}{mgK})}$$ Where K is the distance to the point of rotation to the center of mass. My problem with this equation is that I don't know the moment of inertia and I'm not exactly sure how to calculate it.

Is my approach to this problem correct?

Thanks

The moment of inertia, $I$, is, of course, given by

$$I = \sum m_i r_i^2$$

where the sum is over every bit, $i$, of the object with mass, $m_i$, and at a distance, $r_i$, from the axis of rotation.

You might need to calculate this by integration.

$$I = \int_x ~r(x)^2 \rho dx$$

where

$$m = \int_x \rho dx$$

You integrate along the pendulum in infinitessimal steps of $dx$ and $r(x)$ is the function that determines $r$ as a function of $x$ and $\rho$ is density so $\rho dx = dm$, where $dm$ is the infinitessimal bit of mass at $dx$.

From my point of this this approach looks fine.