What torque would have to be applied to a rod with a length of #4 m# and a mass of #4 kg# to change its horizontal spin by a frequency #7 Hz# over #1 s#?

Question

1107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-17T06:54:29

The torque is #=234.6Nm#

The torque is the rate of change of angular momentum

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

The moment of inertia of a rod is #I=1/12*mL^2#

#=1/12*4*4^2= 16/3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(7)/1*2pi#

#=(14pi) rads^(-2)#

So the torque is #tau=16/3*(14pi) Nm=224/3piNm=234.6Nm#