An object with a mass of # 3 kg# is traveling in a circular path of a radius of #15 m#. If the object's angular velocity changes from # 14 Hz# to # 23 Hz# in #9 s#, what torque was applied to the object?

1 Answer
Jan 21, 2017

Answer:

The torque was #4241.2Nm#

Explanation:

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*15^2= 675 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(23-14)/9*2pi#

#=((18pi)/9) =2pirads^(-2)#

So the torque is #tau=675*(2pi)Nm=1350piNm=4241.2Nm#