An object with a mass of # 3 kg# is traveling in a circular path of a radius of #15 m#. If the object's angular velocity changes from # 4 Hz# to # 34 Hz# in #9 s#, what torque was applied to the object?

1 Answer
Feb 23, 2018

Answer:

The torque is #=7068.6Nm#

Explanation:

The torque is the rate of change of angular momentum

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

where #I# is the moment of inertia

For the object, #I=mr^2#

The mass is #m=3 kg#

The radius of the path is #r=15m#

So the moment of inertia is

,#I=3*(15)^2=675kgm^2#

The angular acceleration is

#alpha=(2pif_1-2pif_0)/t=(34-4)pi/9=10.47rads^-2#

So,

The torque is

#tau=Ialpha=675*10.47=7068.6Nm#