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 any variable or symbol stated above 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 of i. (Use any variable or symbol stated above as necessary.)

Explanation / Answer

anglar momentum of this system is always conserved,
so I1 i + I2 X 0 = (I1 + I2 )f

f = I1 i / (I1 + I2 ) 

b)  KEi = I2 /2 = I1 i2 / 2  

KEf = (I1 + I2 )(I1 i / (I1 + I2 ) )2 / 2  =     I12 i2  / 2(I1 + I2 )

KEf / KEi  = I1 / (I1 + I2 ) 

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