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The kinetic energy of a mass M rotating at distance R from an axis with angular velocity w , is K = (1/2)Iw^2 where I is the mass's moment of inertia, which is just MR^2 if its concentrated at distance R. So for part (a) K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2 =(1/2)[M1 + M2]R^2w^2 = (1/2)[4.65 + 3](.5)^2(2.2)^2 = 4.63 J In part (b) you have to add the moment of inertia of the rod which is I = (1/12)ML^2 = (1/12)(2.05)(1)^2 = .171K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2 + (1/2)Iw^2 = 4.63 + (1/2)(.171)(2.2)^2 = 5.04 J

hi this is mjoody

sorry not me the computer

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