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 #7 Hz# over #3 s#?

Question

1059 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-12T15:20:17

The torque is #625.5Nm#

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= 512/12 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(7)/3*2pi#

#=((14pi)/3) rads^(-2)#

So the torque is #tau=512/12*(14pi)/3 Nm=625.5Nm#