What torque would have to be applied to a rod with a length of #3 m# and a mass of #8 kg# to change its horizontal spin by a frequency #8 Hz# over #9 s#?

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Answer 1 · 2017-02-11T17:46:49

The torque is #=33.5Nm#

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*3^2=6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(8)/9*2pi#

#=((16pi)/9) rads^(-2)#

So the torque is #tau=6*(16pi)/9 Nm=33.5Nm#