An object with a mass of # 2 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 1 Hz# to # 6 Hz# in # 1 s#, what torque was applied to the object?

1 Answer
May 21, 2017

Answer:

The torque was #=251.3Nm#

Explanation:

The torque is the rate of change of angular momentum

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

where #I# is the moment of inertia

For the object, #I=mr^2#

#I=2*2^2=8kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(6-1)/1*2pi#

#=(10pi) rads^(-2)#

So the torque is #tau=8*(10pi) Nm=80piNm=251.3Nm#