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 #6 Hz# over #9 s#?

Question

1212 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T09:32:41

The torque is #=178.7Nm#

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= 42.67 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(6)/9*2pi#

#=((4pi)/3) rads^(-2)#

So the torque is #tau=42.67*(4pi)/3 Nm=178.7Nm#