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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:

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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?

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    $\begingroup$ linear velocity $v$ = radius $r$ $\times$ angular velocity $\omega$ $\endgroup$ – Farcher Aug 3 '17 at 17:22
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    $\begingroup$ 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! $\endgroup$ – Philip Wood Aug 3 '17 at 17:23
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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.

Assuming that there is no friction as the block swings, the KE gained should equal the loss of gravitational PE of the block as it falls. This depends on the vertical height through which the CM of the block has fallen.

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