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 # 2 Hz# to # 6 Hz# in # 5 s#, what torque was applied to the object?

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Answer 1 · 2017-07-19T06:31:12

The torque was #=60.3Nm#

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=(6-2)/5*2pi#

#=(8/5pi) rads^(-2)#

So the torque is #tau=12*(8/5pi) Nm=96/5piNm=60.3Nm#