An object with a mass of # 2 kg# is traveling in a circular path of a radius of #4 m#. If the object's angular velocity changes from # 1 Hz# to # 5 Hz# in # 2 s#, what torque was applied to the object?

1 Answer
Feb 14, 2018

Answer:

The torque is #=402.1Nm#

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

The mass of the object is #m=2kg#

The radius of the path is #r=4m#

For the object, #I=(mr^2)#

So, #I=2*(4)^2=32kgm^2#

The initial angular velocity is

#omega_0=2pif_0=2*pi*1=2pirads^-1#

The final angular velocity is

#omega_1=2pif_1=2*pi*5=10pirads^-1#

The time is #t=2s#

The angular acceleration is

#alpha=(omega_1-omega_0)/t=(10pi-2pi)/2=4pirads^-2#

So,

The torque is #tau= 32*4pi=402.1Nm#