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 # 2 Hz# to # 6 Hz# in # 5 s#, what torque was applied to the object?

1 Answer
Jan 27, 2018

Answer:

#60.31 N.m#

Explanation:

Torque (#tau#) is defined as change in rate of angular momentum i.e #(dL)/dt# or, #(d(Iomega))/dt# or, #I (domega)/dt# (as #L=I omega#, where, #I# is the moment of inertia)

So, #I = mr^2#

Given , #m =3 Kg# and #r=2m#

So, #I= 12 Kg.m^2#

Now #(domega)/dt = ((nu 1-nu 2)/t)*2 pi# i.e #(6-2)/5*2pi (rad)/s^2#

So, #tau= (12*8/5 pi) N.m# i.e #60.31 N.m#