Rotational Kinetic Energy discrepancy

AI Thread Summary
The discussion centers on the comparison of rotational kinetic energy (Kr) between a rolling disk and a rolling hoop, highlighting that the hoop has a greater moment of inertia (I) which leads to a smaller angular acceleration (alpha) and angular velocity (omega). This raises the question of why the hoop's Kr isn't less than that of the disk, given that omega is squared in the kinetic energy formula. Participants note that while the hoop has a higher I, the torque remains constant, suggesting that the disk's torque may increase proportionally with I. The conversation emphasizes the need for clarity on the relationship between torque, moment of inertia, and angular acceleration in this context. Overall, the rolling disk consistently outperforms the hoop due to its more favorable distribution of mass.
FlyDoc
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I was going over the rolling disk versus rolling hoop problem, in which the hoop has more Kr due to greater I and therefore smaller Kt and v. I know this can be algebraically proved with two unique expressions for V that don't involve omega. The question in class that came up concerns torque. If torque=I*alpha and the torques are the same (Fr), wouldn't the hoop produce a smaller alpha, therefore a smaller omega, which would make for a smaller Kr than the disk since omega is squared? I can only seem to resolve this discrepancy if I assume that the torque on the disk increases proportionally with I. Can somebody clarify for me. Thank you
 
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FlyDoc said:
Summary:: Why isn't Rotational Kinetic Energy of a hoop less than a disk? Should angular acceleration and therefore angular velocity decrease due to an increasing in moment of inertia?

I was going over the rolling disk versus rolling hoop problem, in which the hoop has more Kr due to greater I and therefore smaller Kt and v. I know this can be algebraically proved with two unique expressions for V that don't involve omega. The question in class that came up concerns torque. If torque=I*alpha and the torques are the same (Fr), wouldn't the hoop produce a smaller alpha, therefore a smaller omega, which would make for a smaller Kr than the disk since omega is squared? I can only seem to resolve this discrepancy if I assume that the torque on the disk increases proportionally with I. Can somebody clarify for me. Thank you

Why don't you show us the equations properly? Is this rolling down an incline?

A uniform disk (or solid cylinder) beats a hoop (or hollow cylinder) down an incline, right?
 
yes, disk beats hoop.
 
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FlyDoc said:
yes, disk beats hoop.
... and that's because ...
 
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