Attached to each end of a thin steel rod of length 1.10 m and mass 9.40 kg is a

Question

Attached to each end of a thin steel rod of length 1.10 m and mass 9.40 kg is a small ball of mass 1.07 kg. The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at 42.0 rev/s. Because of friction, it slows to a stop in 37.0 s. Assuming a constant retarding torque due to friction, compute (a) the angular acceleration, (b) the retarding torque, (c) the total energy transferred from mechanical energy to thermal energy by friction, and (d) the number of revolutions rotated during the 37.0 s.

Explanation / Answer

a)angular acceleration=(final angular speed -initial angular speed) /time=0-(2*42)/37=-2*3.14*42/37

=-7.13rad/s^2

b)retarding torque = total inertia*angular acceleration=((9.4*1.1^2)/12+2*1.07*(1.1/2)^2)*7.13=11.37Nm

c)the total energy transferred from mechanical energy to thermal energy by friction,=.5*total inertia*angular speed^2=.5*((9.4*1.1^2)/12+2*1.07*(1.1/2)^2)*(2*3.14*42)^2=55487.92j

d)v^2-u^2=2as

0-(2*3.14*42)^2=-2*7.13s

s=777 revolutions

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