Energy in rotation

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Suppose you push a stick by a distance of s like on the picture. It will start to rotate as you along. My question is: Will the rotational energy that it gains be the distance that the point of applicance covers on the circumference of the circle of rotation, while the translational energy will be the one gained from the horizontal displacement of the point of applicance?
 

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Doc Al
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Suppose you push a stick by a distance of s like on the picture. It will start to rotate as you along. My question is: Will the rotational energy that it gains be the distance that the point of applicance covers on the circumference of the circle of rotation, while the translational energy will be the one gained from the horizontal displacement of the point of applicance?
Not exactly. Assuming the force is always horizontal, that force times the horizontal displacement of the point of application represents the total work done. That includes both rotational and translational KE. That force times the horizontal displacement of the center of mass will give the translational KE.

(I assume that the given force is the only force acting.)
 
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right, and since the point will in general rotate and thus cover extra distance compared to the center of mass, then this energy will be the one gained as rotational energy right? I do realize that in general the force will not be perpendicular to this displacement and am thus not saying that the work is just F*s where s is the distance moved on the circle of rotation.
But now my question is:
How do you see that this extra energy MUST go to rotation and not to a translation of center of mass? Is it because the rotation is caused by internal forces and these can do no work on the COM?
 

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