What torque would have to be applied to a rod with a length of #1 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #9 Hz# over #4 s#?

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Answer 1 · 2017-07-03T15:19:55

The torque for the rod rotating about the center is #=2.36Nm#
The torque for the rod rotating about one end is #=9.42Nm#

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

The rate of change of angular velocity is

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

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

So the torque is #tau=1/6*(9/2pi) Nm=3/4piNm=2.36Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=2/3*(9/2pi)=3pi=9.42Nm#