Rotational Kinetic Energy discrepancy

In summary, the disk has more Rotational Kinetic Energy (Kr) because it has greater moment of inertia (I).
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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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  • #2
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?
 
  • #3
yes, disk beats hoop.
 
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  • #4
FlyDoc said:
yes, disk beats hoop.
... and that's because ...
 
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