An object with a mass of # 2 kg# is traveling in a circular path of a radius of #4 m#. If the object's angular velocity changes from # 2 Hz# to # 6 Hz# in # 2 s#, what torque was applied to the object?

1 Answer
Feb 6, 2017

Answer:

The torque was #=402.1 Nm#

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#

#=2*4^2= 32 kgm^2#

The rate of change of angular velocity is

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

#=(4pi) rads^(-2)#

So the torque is #tau=32*(4pi) Nm=128piNm=402.1Nm#