An object with a mass of # 3 kg# is traveling in a circular path of a radius of #6 m#. If the object's angular velocity changes from # 6 Hz# to # 11 Hz# in #5 s#, what torque was applied to the object?

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Answer 1 · 2017-05-13T09:50:13

The torque was #=678.58Nm#

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=3*(6)^2=108kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(11-6)/5*2pi#

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

So the torque is #tau=108*(2pi) Nm=678.58Nm#