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 # 2 Hz# to # 9 Hz# in # 2 s#, what torque was applied to the object?

Question

1459 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-16T10:34:32

The torque is #=703.7Nm#

The torque is the rate of change of angular momentum

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

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

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

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

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

The rate of change of angular velocity is

#(domega)/dt=(f_2-f_1)/t*2pi=(9-2)/2*2pi#

#=(7pi)rads^(-2)#

So,

The torque is #tau= 32*7pi=703.7Nm#