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Moment of Inertia of a Rotating Cam

  • #1
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.

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
 

Answers and Replies

  • #2
Doc Al
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I think they want the KE of the cam, not the cam plus shaft.

A hint and a trick. Hint: M is the mass of the cam, not of a solid disk of radius R. Trick: A disk with a hole can be thought of as a solid disk of radius R plus another solid disk of radius R/2 but with negative mass. :wink:
 
  • #3
Well, that was easy. Thank you! Cool trick, that negative mass idea.

Does this actually have any kind of physical significance? Wouldn't the effective KE of the cam be different, because really, it seems to me, the shaft and the cam form a single object. Is there any point to this calculation besides the routine torture of physics students? :rolleyes:

Thanks again, Doc Al. I hope you had a great thanksgiving.

Dorothy
 
  • #4
Request for full solution

Can you show me the full solution of this question?

Thanks :-)
 
  • #5
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0
So just to follow up on what Doc Al posted. When he is talking about the disk of radius R being subtracted with the missing disk we are assuming that the original disk does not follow the inertia of a regular disk right? Since the axis it is being turned on is not at the center of the large disk we use MR^2 instead of 1/2MR^2 right? Just double checking.

Thanks Matt
 
  • #6
Doc Al
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When he is talking about the disk of radius R being subtracted with the missing disk we are assuming that the original disk does not follow the inertia of a regular disk right?
A disk with a hole in it has a different rotational inertia than a complete disk, if that's what you are asking.
Since the axis it is being turned on is not at the center of the large disk we use MR^2 instead of 1/2MR^2 right? Just double checking.
To find the rotational inertia of a disk about a point not at its center, we use the parallel axis theorem.
 

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