What torque would have to be applied to a rod with a length of #8 m# and a mass of #8 kg# to change its horizontal spin by a frequency #16 Hz# over #4 s#?

Question

1083 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-19T15:30:52

The torque is #=1072.3Nm#

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*8*8^2=128/3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(16)/4*2pi#

#=(8pi) rads^(-2)#

So the torque is #tau=128/3*(8pi) Nm=1072.3Nm#