Do you consider both regular and rotational kinetic energy?

AI Thread Summary
In analyzing the energy of a ball moving in a circular path, both translational kinetic energy (1/2)mv^2 and rotational kinetic energy (1/2)Iomega^2 can be considered. However, if the radius of the ball is significantly smaller than the radius of the circular path, the rotational kinetic energy becomes negligible compared to the translational kinetic energy. In such cases, the ball can be treated as a "point particle," simplifying calculations. This principle also applies to other objects, like cars, moving in circular motion. Therefore, while both forms of energy exist, only translational kinetic energy is typically relevant in practical scenarios.
fangrz
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Homework Statement


If I have a ball moving in a circular path (ball is connected to a string), as shown in this picture:
http://w3.shorecrest.org/~Lisa_Peck.../circularmotion/Images/cent_force_on_ball.gif

should I say that the energy of the ball is both its kinetic energy (1/2)mv^2 and its rotational kinetic energy = (1/2)Iomega^2?

Homework Equations


kinetic energy (1/2)mv^2
rotational kinetic energy = (1/2)Iomega^2

The Attempt at a Solution


I was just thinking about this. I mean, what if you have a car moving in a circle--do you consider both kinetic energy and rotational kinetic energy?
 
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Hello fangrz.

In principle, the total kinetic energy would include rotational KE as well as translational KE. However, if the radius R of the ball is much less than the radius r of the circle, then the rotational KE is very small compared to the translational KE. You can easily check this. So, the rotational KE is often neglected in this type of situation. Then the ball is effectively treated as a "point particle".
 
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