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

1 Answer
Oct 11, 2017

Answer:

The torque was #=60.3Nm#

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

The mass of the object is #m=3kg#

The radius is #r=2m#

For the object, #I=mr^2#

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

And the rate of change of angular velocity is

#(d omega)/dt=(Deltaomega)/t=(10pi-2pi)/5=(8/5pi)rads^-2#

So,

The torque is

#tau=12*8/5pi=60.3Nm#