An object with a mass of # 3 kg# is traveling in a circular path of a radius of #9 m#. If the object's angular velocity changes from # 2 Hz# to # 6 Hz# in # 8 s#, what torque was applied to the object?

1 Answer
Apr 24, 2017

Answer:

The torque was #=763.4Nm#

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*(9)^2)=243kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(6-2)/8*2pi#

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

So the torque is #tau=243*(pi) Nm=763.4Nm#