A cylinder with moment of inertia I1 rotates about a vertical, frictionless axle

Question

A cylinder with moment of inertia I1 rotates about a vertical, frictionless axle with angular velocity omega j. A second cylinder; this one having a moment of inertia of I2 and initially not rotating, drops onto the first cylinder. Because of friction between the surfaces, the two eventually reach the same angular speed omega f. Calculate omega f. (Use any variable or symbol stated above as necessary.) omega f = Show that the kinetic energy of the system decreases in this interaction by calculating the ratio of the final to initial rotational energy. Express your answer in terms of omega j. (Use any variable or symbol stated above as necessary.) KEj = KEf = KEf/ KEj =

Explanation / Answer

as there is no external torque the angular momentum will be conserved I1Wi=(I1+I2)Wf Wf =I1Wi/(I1+I2) K.E i = 0.5I1Wi^2 K.E f =0.5(I1Wi)^2/(I1+I2) K.Ef/K.E = (1+I2/I1)(I1/(I1+I2)^2 cheers :)

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