Attached to each end of a thin steel rod of length 1.60 m and mass 8.20 kg is a

Question

Attached to each end of a thin steel rod of length 1.60 m and mass 8.20 kg is a small ball   of mass 1.03 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 44.0 rev/s. Because of friction, it slows to a stop in   36.0 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 36.0 s.

Explanation / Answer

(a)angular accel =44*2*3.14/36=1.22*2*3.14=7.66 m/s^2

(b)torque = (moment)*(angular accel) = (1/12ml^2+2*m(l/2)^2)*7.66=3.06*7.66=23.34Nm

(c)energy transferred=0.5*Iw^2=3.06*2*3.14*44=845.53 J

(d)time=36s

angle=wt-0.5(angular accel)t^2=2*3.14*44*36-0.5*7.66*36^2=4983.84

revolution=4983.84/(2*3.14)=793.6=73 approx

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