What torque would have to be applied to a rod with a length of #7 m# and a mass of #12 kg# to change its horizontal spin by a frequency #14 Hz# over #9 s#?

Question

883 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-07T05:57:10

The torque is #=478.9Nm#

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*12*7^2= 49 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(14)/9*2pi#

#=((28pi)/9) rads^(-2)#

So the torque is #tau=49*(28pi)/9 Nm=478.9Nm#