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

1 Answer
Apr 26, 2017

The torque #=15.1Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

The moment of inertia is #=I#

For the object, #I=(mr^2)#

So, #I=3*(2)^2s=12kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(3-2)/5*2pi#

#=(2/5pi) rads^(-2)#

So the torque is #tau=12*(2/5pi) Nm=15.1Nm#