A large wooden turntable in the shape of a flat uniform disk has aradius of 1.90

Question

A large wooden turntable in the shape of a flat uniform disk has aradius of 1.90 m and a total mass of130 kg. The turntable is initiallyrotating at 2.65 rad/s about a verticalaxis through its center. Suddenly, a 67.5-kg parachutist makes a soft landing on theturntable at a point near the outer edge. (a) Find the angular speed of the turntableafter the parachutist lands. (Assume that you can treat theparachutist as a particle.)
1 rad/s

(b) Compute the kinetic energy of the system before and after theparachutist lands. KEbefore = 2 J KEafter = 3 J
Why are these kinetic energies not equal?
4
(a) Find the angular speed of the turntableafter the parachutist lands. (Assume that you can treat theparachutist as a particle.)
1 rad/s

(b) Compute the kinetic energy of the system before and after theparachutist lands. KEbefore = 2 J KEafter = 3 J
Why are these kinetic energies not equal?
4 KEbefore = 2 J KEafter = 3 J

Explanation / Answer

         Given that themass of table is M = 130 kg           radius oftable is r = 1.90 m           initialangular velocity of the table is 1 = 2.65 rad/s    mass of parachutist is m =67.5 kg ---------------------------------------------------------------------------      The angular moementum L = I*      Initial moment of inertia of the table isI1 = (1/2)Mr2                                                                   = 234.65 kg.m/s2    final moment of inertia of the table +parachutist is I2 = (1/2)Mr2 + mr2                                                                      = 234.65 kg.m2 + 243.67 kg.m2                                                                                        = 478.32 kg.m2       since there is no external forceacting on the system angular momentum conserved                        initial angular momentum = final angular momentum                                      I1 * 1 = ( I2) *2                                                          2 = I1*1 / I2                                                                = ---------- rad/s       Kinetic energy before is K.E1 =(1/2)I1*12                                                 = ----------- J                           Kinetic energy after is K.E2= (1/2)I2*22                                                 = ----------- J     (c) since final angular velocity decreases.

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