# Is mass moment of inertia measured to the point of interest or the entire object?

I'm trying to calculate the amount of kinetic energy that a point of a rectangle of wood will have after it swings and hits something. I have the following diagram: where there is a rectangle attached to a wall. The rectangle has length $L$, height $H$, and thickness $t$. I want to measure the amount of kinetic energy that point $X$ has when it hits the wall, which is a distance $R$ on the rectangle. I will be able to measure the linear velocity the moment it hits the wall. I calculated the kinetic energy just using the normal equation $E = 1/2 I \omega^2 = 1/2 I v^2/r^2$ where $I$ is the mass moment of inertia, $\omega$ is the angular velocity, $v$ is the linear velocity, and $r$ is the distance to the point. I calculate $I$ from the general equation:

\begin{align} I &= \int_0^L r^2 dm \\ &= \int_0^L y^2 \rho H t dy \\ &= \frac{\rho H t L^3}{3} \\ &= \frac{M L^2}{3} \end{align}

where $\rho$ is the density and $M$ is the mass of the entire rectangle. I used the fact that $M = \rho H t L$. Then is the kinetic energy is $E = \frac{M L^2 v^2}{6 r^2}$ or $E = \frac{M v^2}{6}$? I am thinking it is the first since I should be measuring the mass moment of inertia for the entire rectangle, not just up to the point that I care about, but I'm not positive and a little confused on this point. Can someone clear this up for me?