6 -44 11.P030 WI of inertia I1 rotates about a frictionless, vertical axle with

Question

6 -44 11.P030 WI of inertia I1 rotates about a frictionless, vertical axle with angular speed wi. A second disk, this one having moment of inertia 12 and initially not rotating, drops onto the first disk (figure). Because of friction between the surfaces, the two eventually reach the same angular speed us. o, Before After (a) Calculate of. (Use the following as necessary: I1, 12, w) (b) Calculate the ratio of the final to the initial rotational energy. (Use the following as necessary: h, 12, a ) Need Help? Read it

Explanation / Answer

(a) Solve this problem using conservation of angular momentum.

Means, initial angular momentum = final angular momentum

=> I1 wi = (I1+I2) wf

=> wf = [I1 /( I1+I2)] * wi

(b) Initial Kinetic energy Ki = (1/2)*I1*wi^2

Final kinetic energy Kf = (1/2)*(I1 + I2)wf^2 = (1/2)*I1^2 / (I1 + I2)*wi^2

The kinetic energy ratio = Kf / Ki = 1/2 (I1+I2) wf^2 / ( 1/2 I1 wi^2 )

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

= [(I1 + I2) / I1]*[wf / wi]^2

= [(I1 + I2) / I1]*[I1 / (I1 + I2)]^2 = I1 / (I1 + I2)

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