A uniform steel rod of length 1.20 m and mass 6.40 kg has attached to each end a

Question

A uniform steel rod of length 1.20 m and mass 6.40 kg has attached to each end a small ball of mass 1.06 kg. The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant it is observed to be rotating with an angular speed of 39 rev/s. Because of axle friction it becomes to rest 32.0 s later. Compute, assuming a constant frictional torque, (a) the angular acceleration, (b) the retarding torque exerting by axle friction, (c) the total work done by the axle friction and (d) the number of revolutions executed during the 32.0 s.

Explanation / Answer

a)

=0-T

=0,=0/T=39/32=1.21

b)

TORQUE =I

=[6.40*(1.20)2/12 +1.60*(1.20)2]*1.21

=3.717

c)

WORK DONE BY TORQUE =CHANGE IN KE=0.5I02

=0.5*[6.40*(1.20)2/12 +1.60*(1.20)2]*392

=2336.25J

d)

=0T-0.5T2

=0.5*0T

=624 RADIANS

NO OF REVOLUTIONS =624/ =198.72

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