A cylinder with moment of inertia 38.7 kg m^2 rotates with angular velocity 9.12

Question

A cylinder with moment of inertia 38.7 kg m^2 rotates with angular velocity 9.12 rad/s on a frictionless vertical axle. A second cylinder, with moment of inertia 31.5 kgm^2, 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. Show that energy is lost in this situation by calculating the ratio of the final to the initial kinetic energy.

Explanation / Answer

Here ,
I1 = 38.7 Kg.m^2

w1 = 9.12 rad/s

I2 = 31.5 Kg.m^2

let the final angular velocity is wf

Using conservation of angular momentum

( I1 * w1) = (I1 + I2) * wf

(38.7 * 9.12) = (38.7 + 31.5) * wf

solving for wf

wf = 5.03 rad/s

the final angular speed is 5.03 rad/s

2)

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

final kinetic energy = 0.5 * (38.7 + 31.5) * 5.03^2

final kinetic energy = 887.3 J

initial kinetic energy = 0.5 * I1 * w^2

initial kinetic energy = 0.5 * 38.7 * 9.12^2

initial kinetic energy = 1609.4 J

as the initial kinetic energy is more than the final kinetic energy , energy is lost during this situation

ratio of final energy to initial kinetic energy = 887.3/1609.4

ratio of final energy to initial kinetic energy = 0.551

the ratio of final energy to initial kinetic energy is 0.551

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