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:

enter image description here

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?


The first equation that you have is the kinetic energy of the entire rectangle and is the equation we need. After knowing the kinetic energy of the rectangle, you can find the angular velocity of every point in the rectangle since they have the same angular velocity. To find the linear velocity of the point located at the radius R from the axis of rotation, just multiply the angular velocity by R.

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