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

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Answer 1 · 2018-01-24T18:23:21

The torque for the rod rotating about the center is #=307.9Nm#
The torque for the rod rotating about one end is #=1231.5Nm#

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

The mass of the rod is #m=9kg#

The length of the rod is #L=7m#

The moment of inertia of a rod, rotating about the center is

#I=1/12*mL^2#

#=1/12*9*7^2= 36.75 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(12)/9*2pi#

#=(8/3pi) rads^(-2)#

So the torque is #tau=36.75*(8/3pi) Nm=307.9Nm#

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

#I=1/3*mL^2#

#=1/3*9*7^2=147kgm^2#

So,

The torque is #tau=147*(8/3pi)=1231.5Nm#