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

Question

1499 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-05T07:21:45

The torque is #=226.2Nm#

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=(17-2)/5*2pi#

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

So the torque is #tau=12*(6pi) Nm=72piNm=226.2Nm#