A cylinder with moment of inertia I1 rotates with an angular velocity Wo about a vertical frictionless axle. Asecond cylinder with moment of inertia I2 initially not rotating drops on to the first cylinder since the surfaces are rough the two eventually reach the same angular velocity then find: a) finall angular velocity? b) show that kinetic energy is lost in this situation. c) the ratio of the final to the initial kinetic energy
a) Conservation of angular momentum:
I1Wo = (I1 + I2)Wf
Wf = I1Wo / (I1 + I2)
b) Since the surface is rough, some energy is lost due to friction between the two cylinders. This energy is converted to heat and sound, but does not contribute to the final kinetic energy of the system. Therefore, there is a loss of kinetic energy in this situation.
c) The initial kinetic energy is given by:
KEi = 0.5 * I1 * Wo^2
The final kinetic energy is given by:
KEf = 0.5 * (I1 + I2) * Wf^2
Using the result from part (a), we can substitute for Wf:
KEf = 0.5 * (I1 + I2) * (I1Wo / (I1 + I2))^2
KEf = 0.5 * I1 * Wo^2 * (I1 / (I1 + I2))^2
Taking the ratio of KEf to KEi:
KEf / KEi = 0.5 * I1 * (I1 / (I1 + I2))^2 / I1
KEf / KEi = (I1 / (I1 + I2))^2
Therefore, the ratio of final to initial kinetic energy is (I1 / (I1 + I2))^2, which is less than 1 due to energy loss.