A man stands on a platform that is rotating (without friction) withan angular sp

Question

A man stands on a platform that is rotating (without friction) withan angular speed of 1.94 rev/s; his arms are outstretched and heholds a brick in each hand. The rotational inertia of the systemconsisting of the man, bricks, and platform about the central axisis 5.60 kg·m2. If by moving the bricks the mandecreases the rotational inertia of the system to 2.33kg·m2, (a) what is theresulting angular speed of the platformand (b) what is the ratio of the newkinetic energy of the system to the original kinetic energy?

Explanation / Answer

   Given that the initial moment of inertia I1 = 5.60 kg.m2 and final moment of inertia is I2 = 2.33kg.m2 initial angular velocity of first wheel is i = 1.94 rev /s =12.18 rad/s final combined angular velocity of two wheels is f = ? ------------------------------------------------------------------ Since there is no external force on the system then theangular momentum is conserved                        I1*i = I2 * f                             f = I1*i / I2    = --------rad/s    Total initial kinetic energy is KEi =(1/2)*I1*i2    Total final kinetic energy is KEf = (1/2)*(I2 )*f2          =(1/2)*(I2 )*{ I1*i / I2 }2                                                     = (1/2)*I12*i2/ (I2 )    and KEf / KEi = I1 / ( I2)

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