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 # 5 Hz# in # 1 s#, what torque was applied to the object?

Question

1944 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-16T15:43:14

The torque was #=100.5Nm#

The torque is the rate of change of angular momentum

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

The moment of inertia of the object is #I=mr^2#

#=2*2^2= 8 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(5-3)/1*2pi#

#=(4pi) rads^(-2)#

So the torque is #tau=8*(4pi) Nm=32piNm=100.5Nm#