Rotational inertia of a square piece of wood?

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SUMMARY

The discussion focuses on calculating the rotational inertia of a 20inx20inx0.25in piece of plywood when using a hole saw. The moment of inertia (I) is defined by the equation I = (1/12)ma² + md², where m represents the mass of the plywood, a is the side length of the square, and d is the distance from the center of the square to the center of the hole being drilled. It is established that the rotational inertia is a geometric property that does not depend on the RPM of the saw, but rather on the dimensions and mass of the plywood and the location of the cut.

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I was in the shop the other day cutting holes out of a 20inx20inx.25in piece of plywood and I noticed that the hole saw was slipping on the board (meaning that it would begin to make an unpleasant sound as the saw hit the board). I had the board clamped down and I realized that a large portion of the board was hanging of the edge. Even if the board was perfectly centered for a cut through the middle of the board (discouraged in the shop, it wastes a large amount of resources), there was too much torque on the board (it would have to be cut).
Assuming that the board wasn't clamped down, I'm wondering what the rotational inertia would be on such a piece of plywood if I was using a hole saw. I know that it would depend on the size of the hole i was cutting, the rpm of the saw, the size of the board, and the location of the cut. I'm just wondering HOW they would relate.
Use any other necessary units (mass etc.)

Is it even possible to create an equation for this?
 
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The moment of inertia (rotational inertia) of the plywood is a geometric quantity and does not depend on the rpm of the saw. The equation you are looking for is $$I=\frac{1}{12}ma^2+md^2$$where
##m##= mass of plywood,
##a##=side of square,
##d##=distance from the center of the square to the center of the hole you are about to drill.

The formula is an application of the parallel axis theorem.
 

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