An object with a mass of # 8 kg# is traveling in a circular path of a radius of #12 m#. If the object's angular velocity changes from # 15 Hz# to # 6 Hz# in # 6 s#, what torque was applied to the object?

1 Answer
Jan 25, 2017

Answer:

The answer is #=10857.3Nm#

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#

#=8*12^2= 1152 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(15-6)/6*2pi#

#=((18pi)/6)=3pi rads^(-2)#

So the torque is #tau=1152*(3pi) Nm=3456piNm=10857.3Nm#