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 1.60 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?
d) The number of revolutions executed during 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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