What torque would have to be applied to a rod with a length of #3 m# and a mass of #1 kg# to change its horizontal spin by a frequency of #3 Hz# over #2 s#?

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Answer 1 · 2017-02-04T17:52:23

The torque #=7.07Nm#

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 is #I=1/12*mL^2#

#=1/12*1*3^2= 3/4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(3)/2*2pi#

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

So the torque is #tau=3/4*(3pi) Nm=9/4piNm=7.07Nm#