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There is a disk of radius $R$, that can rotate and some guy decides to glue on a point mass to the rim and see what the angular speed of the rotating disk is when the point mass is immediately below the $X$ axis. My approach to compute the speed was to use the conservation of mechanical energy and set the initial potential potential energy equal to the final kinetic energy. I set potential energy zero at the bottom of x axis.

I was wondering if setting up the equation requires me to include the linear kinetic energy of particle or do I just include the rotational kinetic energy, i.e. is the correct equation which represents the energy conversion $mgR = \frac{1}{2}Iw^2 + \frac{1}{2}mv^2$ or $mgR = \frac{1}{2}(Iw^2)$?

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  • $\begingroup$ Consider the conservation of angular momentum. $\endgroup$
    – ytlu
    Commented Mar 11, 2021 at 1:38
  • $\begingroup$ there's gravity doing work here. angular momentum is not conserved $\endgroup$
    – user256872
    Commented Mar 11, 2021 at 3:21

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You need to include the kinetic energy of the particle, otherwise you're omitting energy. Also, assuming the disk rotates about its CoM, there wouldn't be any accelerated rotation due to gravity without the point mass.

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