What torque would have to be applied to a rod with a length of #7 m# and a mass of #9 kg# to change its horizontal spin by a frequency #9 Hz# over #16 s#?

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Answer 1 · 2017-01-02T07:05:43

The torque is #=129.9Nm#

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*9*7^2= 147/4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(9)/16*2pi#

#=((9pi)/8) rads^(-2)#

So the torque is #tau=147/4*(9pi)/8 Nm=1323/32piNm=129.9Nm#