Attached to each end of a thin steel rod of length 1.20 m and mass6.40 kg is a s

Question

Attached to each end of a thin steel rod of length 1.20 m and mass6.40 kg is a small ball of mass 1.06 kg. The rod is constrained torotate in a horizontal plane about a vertical axis through itsmidpoint. At a certain instant, it is rotating at 39.0 rev/s.Because of friction, it slows to a stop in 32.0 s. Assuming aconstant retarding torque due to friction, compute(a) the angular acceleration, (b)the retarding torque, (c) the total energytransferred from mechanical energy to thermal energy by friction,and (d) the number of revolutions rotated duringthe 32.0 s.

Explanation / Answer

Given that
intial angular acceleration 0 final angular accleration = which isconstant = angular accleration by using kinematic eq = 0 = 0+t                                        = = -0/t                                              = - 39/32 = - 1.21 rev/s                                     = -1.21 *2 = - 7.66 rad/s2 (B) the moment of inertia of rod = I =Ml2/12 the contribution of each ball = M(l/2)2 total moment of inertia = I = Ml2/12 +2 ml2/4         plug valuesand                                   do caliculations for I (c) since the system comes to rest the mechnical energyis converted to thermal energy which is simply intial kineticenergy       Ki = 1/2I02 plug values docaliculations forKi----J (d) applying kinematic eq =             = 0t + 1/2t2                2 ( 39) (32) +1/2 (-7.66) (32)2         docaliculations andthat value gives angular displacement (e) only mechnial energy is converted into thermal energy canstill be computed without any additional information         docaliculations andthat value gives angular displacement (e) only mechnial energy is converted into thermal energy canstill be computed without any additional information

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