An object with a mass of # 3 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 3 Hz# to # 5 Hz# in # 5 s#, what torque was applied to the object?

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Answer 1 · 2017-03-11T14:40:23

The torque was #=30.2Nm#

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*2^2= 12 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(5-3)/5*2pi#

#=(4/5pi) rads^(-2)#

So the torque is #tau=12*(4/5pi) Nm=48/5piNm=30.2Nm#