Rolling Disk Inertia: Comparing Friction and Slip on Inclines

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SUMMARY

The discussion focuses on the comparison of two identical disks, A and B, as they roll up inclines with different frictional properties. Disk A, which rolls up a frictional incline, reaches a height of 12 cm due to its ability to convert both translational and rotational energy into potential energy. In contrast, Disk B rolls up a frictionless incline and cannot convert its rotational energy, resulting in a lower height. The key takeaway is that the presence of friction allows for the conversion of energy forms, impacting the maximum height achieved by the disks.

PREREQUISITES
  • Understanding of rotational inertia, specifically I (= 1/2 MR²)
  • Knowledge of energy conservation principles in physics
  • Familiarity with the concepts of translational and rotational energy
  • Basic understanding of friction and its effects on motion
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  • Explore the principles of energy conservation in rolling motion
  • Study the effects of friction on rotational dynamics
  • Learn about the equations governing translational and rotational kinetic energy
  • Investigate the differences between rolling with and without slipping
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Physics students, educators, and anyone interested in the mechanics of rolling motion and energy conversion in physical systems.

ctv1337
Two identical disks, with rotational inertia I (= 1/2 MR2), roll without slipping across a horizontal floor and then up inclines. Disk A rolls up its incline without slipping. On the other hand, disk B rolls up a frictionless incline. Otherwise the inclines are identical. Disk A reaches a height 12 cm above the floor before rolling down again. Disk B reaches a height above the floor of?
I've tried a couple of things but i don't think I'm going in the correct direction...can anyone help me?
 
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You can readily see

that the disc on the frictionless incline can't convert it's rotational energy into potential energy, whereas the disc on the frictional incline can covert both its translational and rotational energy into potential energy. Solving the problem is just a matter of finding the ratios of translational and rotational energies and plugging them into the formulas for total and translational only.
 

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