I Problem about uniform thin disk

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The discussion centers on the treatment of kinetic energy in the context of a uniform thin disk, questioning why only translational kinetic energy is considered without accounting for rotational kinetic energy. The equations presented illustrate the relationship between translational and rotational motion, highlighting that both forms of kinetic energy should be included in calculations. The final equation shows that the outcome remains consistent regardless of whether a factor of 3/4 or 1/2 is used, suggesting a lack of precision in the original solution. This indicates a potential oversight in the problem-solving approach regarding the energy contributions of a rolling disk. The conversation emphasizes the importance of incorporating both types of kinetic energy for accurate analysis.
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Why do they only think of kinetic energy of motion?
Why don't they think of both kinetic of motion and kinetic of rolling energy?

So. i think
## L = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2##
## L = \frac{1}{2}m(r \omega)^2 + \frac{1}{2}(\frac{1}{2}mr^2) \omega^2##
## L = \frac{3}{4} mr^2 \omega^2##
 
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Look at the last three equations in the answer. The result will be the same whether you multiply ##mr^2\omega^2## by ##\frac{3}{4}## or ##\frac{1}{2}##. It looks like whoever wrote the solution knew the bottom line and became sloppy.
 
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