A cylinder with moment of inertia i_1 rotates about a vertical, frictionless axl

Question

A cylinder with moment of inertia i_1 rotates about a vertical, frictionless axle with angular velocity a)j. A second cylinder; this one having a moment of inertia of I_2 and initially not rotating, drops onto the first cylinder. Because of friction between the surfaces, the two eventually reach the same angular speed Omega. (a) Calculate omegaf. (Use any variable or symbol stated above as necessary.) (b) 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 omegai. (Use any variable or symbol stated above as necessary.)

Explanation / Answer

A.

Using the angular momentum conservation

Li = Lf

I1*w1 = (I1 + I2)*wf

wf = I1*w1/(I1 + I2)

I'm getting same answer as you, It should be correct.

B.

Rotational KE is given by

KE = 0.5*Iw^2

KEi = 0.5*I1*wi^2

KEf = 0.5*(I1 + I2)*wf^2

ratio

KEf/KEi = 0.5*(I1 + I2)*wf^2/(0.5*I1*wi^2)

using wf from A part

KEf/KEi = I1/(I1 + I2)

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