What torque would have to be applied to a rod with a length of #6 m# and a mass of #3 kg# to change its horizontal spin by a frequency #14 Hz# over #6 s#?

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Answer 1 · 2017-08-21T11:35:51

The torque for the rod rotating about the center is #=131.9Nm#
The torque for the rod rotating one end is #=527.8Nm#

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*3*6^2= 9 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(14)/6*2pi#

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

So the torque is #tau=9*(14/3pi) Nm=42piNm=131.9Nm#

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

#I=1/3*mL^2#

#=1/3*3*6^2=36kgm^2#

So,

The torque is #tau=36*(14/3pi)=168pi=527.8Nm#