An object with a mass of # 2 kg# is traveling in a circular path of a radius of #7 m#. If the object's angular velocity changes from # 7 Hz# to # 12 Hz# in # 3 s#, what torque was applied to the object?

Question

1309 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-24T20:22:59

The torque was #=1026.3Nm#

The torque is the rate of change of angular momentum

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

The moment of the object is

#I=mr^2#

#=2*7^2= 98 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(12-7)/3*2pi#

#=(10/3pi) rads^(-2)#

So the torque is #tau=98*(10/3pi) Nm=980/3piNm=1026.3Nm#