Just after a motorcycle rides off the end of a ramp and launches into the air, i

Question

Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 8000 rev/min. The motorcycle rider forgets to throttle back, so the engine's angular speed increases to 11300 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 4.10 rev/min. Calculate the ratio Ie/Im of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.

Explanation / Answer

In rotational motion, Torque acting on an object () is defined as the rate of change of Angular momntum (L).                                                 = dL/dt              But Angular momentum L = I                                                               where I = Moment of inertia                                                                         = Angular velocity.                        Hence torque = dL/dt Notation 1 stands for Engine case.                2 stands for Man case. Given that Angular speeds:              1 = 2(11300 rev/min) = 2(11300/60   rev/s) = 1183.33 rad/s               2 = 2(4.10 rev/min) = 2(4.10/60 rev/s) = 0.4293 rad/s But External torque acting on the bike is zero.So                                             dL/dt = 0                                                   L = Constant                                                  I = Constant                                             I1 1 = I2 2        Ratio of moment of inertias I1/I2 = 2/1                                                          = (0.4293 rad/s)/(1183.33 rad/s)                                                           = 3.628*10^-4                       

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