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 # 23 Hz# in #9 s#, what torque was applied to the object?

1 Answer
Nov 24, 2017

Answer:

The torque is #=8953.5Nm#

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 of the path is #r=15m#

For the object, #I=mr^2#

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

And the rate of change of angular velocity is

#(d omega)/dt=(Deltaomega)/t=(2pif_1-2pif_2)/t#

#=(46pi-8pi)/9=(38/9pi)rads^-2#

So,

The torque is

#tau=675*38/9pi=8953.5Nm#