An object with a mass of # 2 kg# is traveling in a circular path of a radius of #4 m#. If the object's angular velocity changes from # 4 Hz# to # 8 Hz# in # 3 s#, what torque was applied to the object?

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Answer 1 · 2017-01-22T07:39:23

The torque was #=268.1Nm#

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

The moment of inertia of the object is #I=mr^2#

#=2*4^2= 32 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(8-4)/3*2pi#

#=((8pi)/3) rads^(-2)#

So the torque is #tau=32*(8pi)/3 Nm=256/3piNm=268.1Nm#