Question

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


(a) Calculate ωf. (Use I1 for I1, I2 for I2, and omega_i for ωi, 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 ω_i. (Use omega_i for ω_i, I1 for I₁, and I2 for I₂.)

Answer 1

a) Angular momentum is conserved

Ii .Wi = (Ii + Iii ) . w

w = Ii . Wi / ( Ii + Iii )

b) Original kinetic energy = 1/2 . Ii . Wi^2

Final kinetic energy = 1/2 . (Ii + Iii ) . w^2

Substitute the "w"

Final KE = 1/2 . (Ii + Iii ) . Ii^2 . Wi^2 / ( Ii + Iii )^2

Final KE / Original KE = I (original ) / I ( final )