Rotational Motion Clarification

AI Thread Summary
The discussion centers on the moment of inertia of a ball attached to a massless rod, which is being accelerated about the free end. The initial assumption is that the moment of inertia can be simplified to mr^2, similar to a free ball. However, it is clarified that the ball also rotates around its own axis, necessitating a greater force for the same acceleration. A more accurate calculation of the moment of inertia includes both the ball's inertia and the rod's, represented as I = I_{ball} + mL^2. This highlights the importance of considering all rotational aspects in physics calculations.
Moose352
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I got a little confused thinking about this. Consider a ball attatched to a (massless) rod. This system is to be accelerate about the free end of the rod. From what I know, the moment of inertia for this system would simply be the mr^2, since this is the same case as a free ball being accelerated at the same rate. However, it seems to me that the ball is also rotating around its own axis (since it is fixed on the rod), and thus wouldn't the force necessary for the same rate of acceleration be greater?
 
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Is this question too dumb for anyone to even answer? :wink:
 
Originally posted by Moose352
Is this question too dumb for anyone to even answer? :wink:
No, it's an excellent question.

Originally posted by Moose352
However, it seems to me that the ball is also rotating around its own axis (since it is fixed on the rod), and thus wouldn't the force necessary for the same rate of acceleration be greater?
I would say that the answer is yes. Using I = mL2 (L is length of stick) is just an approximation: it assumes the radius of the ball can be neglected.

A more realistic value for rotational inertia must include that of the ball as well:

I = I_{ball} + mL^2 = \frac{2}{5}mR^2 + mL^2
 
Thanks Doc. My physics instructor (and the book) was saying otherwise and wouldn't concede to my view.
 

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