An object with a mass of # 2 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 3 Hz# to # 8 Hz# in # 1 s#, what torque was applied to the object?

1 Answer
Mar 28, 2018

Answer:

The torque is #=251.3Nm#

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=2 kg#

The radius of the path is #r=2m#

So the moment of inertia is

,#I=2*(2)^2=8kgm^2#

The angular acceleration is

#alpha=(2pif_1-2pif_0)/t=(16-6)pi/1=10pirads^-2#

So,

The torque is

#tau=Ialpha=8*10pi=251.3m#