An object with a mass of # 2 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 3 Hz# to # 6 Hz# in # 1 s#, what torque was applied to the object?

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Answer 1 · 2017-05-06T09:37:29

The torque was #=150.8Nm#

The torque is the rate of change of angular momentum

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

The moment of inertiai is #=I#

For the object, #I=(mr^2)#

So, #I=2*(2)^2=8kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(6-3)/1*2pi#

#=(6pi) rads^(-2)#

So the torque is #tau=8*(6pi) Nm=150.8Nm#