A cylinder with moment of inertia I 1 rotates about a vertical, frictionless axl

Question

A cylinder with moment of inertia I1 rotates about a vertical, frictionless axle with angular velocity i. 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 f.

(a) Calculate f. (Use I1 for I1, I2 for I2, and omega_i for i, as necessary.)
f =  

(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 ofi. (Use omega_i for i, I1 for I1, and I2 for I2.)

Explanation / Answer

Here,

let the final angular speed is wf

a) as there is no external torque acting between the disks

the total angular mometum of the system will be conserved

I1 * wi = (I1 + I2) * wf

wf = I1 * wi/(I1 + I2)

the angular speed wf is I1 * wi/(I1 + I2)

part b)

initial kinetic energy = 0.5 * I1 * wi^2

final kinetic energy = 0.5 *(I1 + I2) * wf^2

final kinetic energy = 0.5 *(I1 + I2) * (I1 * wi/(I1 + I2))^2

final kinetic energy = 0.5 * I1^2 * wi^2/(I1 + I2 )

now, for the ratio

final kinetic energy/initial kinetic energy = ( 0.5 * I1^2 * wi^2/(I1 + I2 ))/( 0.5 * I1 * wi^2 )

final kinetic energy/initial kinetic energy = I1/(I1 + I2)

the ratio of final to initial kinetic energy is I1/(I1 + I2)

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