What torque would have to be applied to a rod with a length of #12 m# and a mass of #6 kg# to change its horizontal spin by a frequency #7 Hz# over #9 s#?

Question

813 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-11T06:34:58

The torque for the rod rotating about the center is #=351.9Nm#
The torque for the rod rotating about one end is #=1407.4Nm#

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, rotating about the center is

#I=1/12*mL^2#

#=1/12*6*12^2= 72 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(7)/9*2pi#

#=(14/9pi) rads^(-2)#

So the torque is #tau=72*(14/9pi) Nm=112piNm=351.9Nm#

The moment of inertia of a rod, rotating about one end is

#I=1/3*mL^2#

#=1/3*6*12^2=288kgm^2#

So,

The torque is #tau=288*(14/9pi)=1407.4Nm#