How to calculate kinetic energy of rotating piece of wood?

I'm trying to figure out how to calculate the kinetic energy of a rotating piece of wood. I have the following diagram I have drawn: In $A$, I have a side view to show the setup. There is just a piece of wood that is attached to the ceiling that will swing. When it is perpendicular to the ceiling, it will have its linear velocity measured. I want to calculate the kinetic energy at this point. $B$ shows the piece of wood. Just a normal rectangular piece of wood with some thickness.

My first thought was to just calculate it using $KE_{lin} = (1/2) m v^2$ but this is only for linear motion. Then for rotational it would be $KE_{rot} = (1/2) I \omega^2$. But I'm measuring the linear velocity and only want to know this at one point. Is there any way that I can do this calculation?

• linear velocity $v$ = radius $r$ $\times$ angular velocity $\omega$ – Farcher Aug 3 '17 at 17:22
• A few hints: (1) Treat the plank as a uniform rod and look up the moment of inertia, $I$ of a rod about one end. (2) $v=L \omega$: Suggest you check this out in a textbook! (3) Apply conservation of energy. Good luck! – Philip Wood Aug 3 '17 at 17:23

The block of wood has many velocities at any instant. The velocity is zero at the hinge and a maximum at the end. If you measure the velocity $v$ of the end of the block, then the KE is $\frac12 I\omega^2$. Here the moment of inertia $I$ is measured about the end of the block, not the middle. If the block has negligible thickness (ie looking sideways as in A) then $I$ is the same as for a slender rod. And $\omega=v/L$ where $L$ is the length of the block.