(part 1 of 2) A cylinder with moment of inertia 27.7 kg m2 rotates with angular

Question

(part 1 of 2) A cylinder with moment of inertia 27.7 kg m2 rotates with angular velocity 4.79 rad/s on a frictionless vertical axle. A second cylinder, with moment of inertia 32.2 kg m2 , initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity.

Calculate the final angular velocity. Answer in units of rad/s.

(part 2 of 2) Show that energy is lost in this situation by calculating the ratio of the final to the initial kinetic energy.

Explanation / Answer

I1 = 27.7 kg.m2, I2 = 32.2 kg.m^2, wi = 4.79 rad/s

(1) from conservation of angular momentum

Iiwi= Ifwf

I1*wi = (I1+I2)wf

(27.7*4.79) = (27.7 +32.2)*wf

wf = 2.22 rad/s

(2) Lost KE = KEf - KEi

KEi = 0.5I1wi^2

KEf = 0.5Ifinalwf^2

= 0.5(I1+I2)(wi*I1/(I1+T2))^2

= 0.5I1wi^2(I1/(I1+I2))

KEf = KEi(I1/(I1+I2))

KEf/KEi = I1/(I1+I2)

= 27.7/(27.7+32.2)

= 0.46

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