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 # 1 Hz# to # 8 Hz# in # 3 s#, what torque was applied to the object?

1 Answer
May 9, 2017

Answer:

The torque was #=469.1Nm#

Explanation:

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

The rate of change of angular velocity is

#(domega)/dt=(8-1)/3*2pi#

#=(14/3pi) rads^(-2)#

So the torque is #tau=32*(14/3pi) Nm=469.1Nm#