What torque would have to be applied to a rod with a length of #8 m# and a mass of #8 kg# to change its horizontal spin by a frequency #1 Hz# over #5 s#?

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Answer 1 · 2017-01-29T16:09:50

The torque is #=53.62Nm#

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*8*8^2= 128/3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(1)/5*2pi#

#=((2pi)/5) rads^(-2)#

So the torque is #tau=128/3*(2pi)/5 Nm=256/15piNm=53.62Nm#