Good question! Rotational KE is actually the sum of the translational KE of all the particles in a rigid body that's rotating. In this case we're dealing with a "particle" so we can use either 1/2mv^2 or 1/2Iω^2. Either way, you would get the same value. You can convince this to yourself by substituting I=mr^2 and ω=v/r into KE=1/2Iω^2, and you will end up with KE=1/2mv^2. I've since removed that part of the video because I mistakenly set the KEi=KEf. However, since it's an inelastic collision (some of the energy was dissipated via heat), the final KE will be less than the initial KE. So KE is not conserved here. Thanks for watching!👍
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for the last example with energy, why is the initial only rotational kinetic energy and not both rotational and translational kinetic energy?
Good question! Rotational KE is actually the sum of the translational KE of all the particles in a rigid body that's rotating. In this case we're dealing with a "particle" so we can use either 1/2mv^2 or 1/2Iω^2. Either way, you would get the same value. You can convince this to yourself by substituting I=mr^2 and ω=v/r into KE=1/2Iω^2, and you will end up with KE=1/2mv^2.
I've since removed that part of the video because I mistakenly set the KEi=KEf. However, since it's an inelastic collision (some of the energy was dissipated via heat), the final KE will be less than the initial KE. So KE is not conserved here. Thanks for watching!👍