What torque would have to be applied to a rod with a length of #4 m# and a mass of #3 kg# to change its horizontal spin by a frequency of #9 Hz# over #4 s#?

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Answer 1 · 2017-06-19T06:12:22

The torque for the rod rotating about the center is #=56.5Nm#
The torque for the rod rotating about the center is #=226.2Nm#

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*4^2= 4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(9)/4*2pi#

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

So the torque is #tau=4*(9/2pi) Nm=18piNm=56.5Nm#

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

#I=1/3*mL^2#

#=1/3*3*4^2=16kg m^2#

So,

The torque is #tau=16*(9/2pi)=72pi=226.2Nm#