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)