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

1 Answer
Mar 6, 2017

Answer:

The torque was #=50.27Nm#

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*2^2= 8 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=8*(2pi) Nm=16piNm=50.27Nm#