This cam is a circular disk rotating on a shaft that does not pass through the center of the disk. It is manufactured by first making the cam with radius R, then drilling an off-center hole, radius R/2, parallel to the axis of the cylinder and centered at a point R/2 from the center of the cam.(adsbygoogle = window.adsbygoogle || []).push({});

The cam, of mass M is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed w about the axis of the shaft?

This is giving me real problems, so I would appreciate any suggestions.

I used Icm = MR^2/2 for the moment of inertia of a solid, rotating cylinder around its center of mass, and the parallel axis theorem to get:

I = MR^2/2 + MD^2 = MR^2/2 + M(R/2)^2 = 3MR^2/4

From which the rotational kinetic energy should just be:

Kw = (3MR^2/4)w^2(1/2) = (3MR^2/8)w^2

But the answer in the book is (23/48)MR^2w^2

So, wow, am I off.

The only thing I can think of is that the shaft might be made of a different material, so I would have to change the mass calculation somehow, but the problem doesn't say anything about that. Anyway, wouldn't they make these things out of the same stuff?

Thanks for any help,

Dorothy

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# Homework Help: Moment of Inertia of a Rotating Cam

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