A cylinder with moment of inertia I 1 rotates with angular velocity 0 about afri

Question

A cylinder with moment of inertia I1rotates with angular velocity 0 about africtionless vertical axle. A second cylinder, with moment ofinertia I2, initially not rotating, drops ontothe first cylinder. Because the surfaces are rough, the twoeventually reach the same angular velocity, . (a) Calculate . (b) Show that energy is lost in this situation andcalculate the ratio of the final to the initial kineticenergy. i just solve for the equation but my ans is still wrong A cylinder with moment of inertia I1rotates with angular velocity 0 about africtionless vertical axle. A second cylinder, with moment ofinertia I2, initially not rotating, drops ontothe first cylinder. Because the surfaces are rough, the twoeventually reach the same angular velocity, . (a) Calculate . (b) Show that energy is lost in this situation andcalculate the ratio of the final to the initial kineticenergy. i just solve for the equation but my ans is still wrong

Explanation / Answer

             Given that the moment of inertia of the cylinder is I1              initial angular velocity0             The moment of inertia of cylinder is I2               initialangular velocity of the second cylinder is 0              Final angualr velocity is   ---------------------------------------------------------------------------------------        since there is noexternal torque on the system then the angular momentum isconserved                               initial angualr momentum = final angular momentum                                        I1*o+ I2*0 = (I1+I2)                                                          = I1*o / ( I1 + I2) --------(1)                                                                      from the aboveequation the < o then the final kineticenergy lost in this situation       The ratio of the final kineticenergy to the initila kinetic energy                                                                                                                                       Kf/Ki   = (1/2)( I1 + I2)*2 / (1/2)I1*02                                                                  =  ( I1 + I2)*2 / I1*02     substitude the value of from theequation (1) we get                                                        Kf/ Ki = I1 / ( I1+ I2)                                                                                                                                        =  ( I1 + I2)*2 / I1*02     substitude the value of from theequation (1) we get                                                        Kf/ Ki = I1 / ( I1+ I2)                                                                                            

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