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 # 2 Hz# to # 12 Hz# in # 2 s#, what torque was applied to the object?

1 Answer
Jul 4, 2018

Answer:

The torque is #=1005.3Nm#

Explanation:

The torque is the rate of change of angular momentum

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

For the object, the moment of inertia is

#I=mr^2#

The mass of the object is #m=2 kg#

The radius of the circular path is #r=4m#

The moment of inertia is

#I=2*4^2=32kg m^2#

The angular acceleration is

#alpha=2pi(f_2-f_1)/t=2pi(12-2)/2=10pirads^-2#

The torque is

#tau=I omega=32*10pi=1005.3Nm#