An object with a mass of # 3 kg# is traveling in a circular path of a radius of #12 m#. If the object's angular velocity changes from # 5 Hz# to # 4 Hz# in # 5 s#, what torque was applied to the object?

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Answer 1 · 2017-01-12T08:31:31

The torque is #=542.9Nm#

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#

#=3*12^2= 432 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(5-4)/5*2pi#

#=((2pi)/5) rads^(-2)#

So the torque is #tau=432*(2pi)/5 Nm=864/5piNm=542.9Nm#