Rolling Disk Inertia: Comparing Friction and Slip on Inclines

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Two identical disks, A and B, are analyzed as they roll up inclines, with disk A rolling on a frictional incline and disk B on a frictionless incline. Disk A reaches a height of 12 cm due to its ability to convert both translational and rotational energy into potential energy. In contrast, disk B cannot convert its rotational energy, limiting its height gain. The discussion emphasizes the importance of understanding energy conversion in rolling motion. Ultimately, the problem requires calculating the ratios of translational and rotational energies to determine the height reached by disk B.
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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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