What torque would have to be applied to a rod with a length of #7 m# and a mass of #12 kg# to change its horizontal spin by a frequency #14 Hz# over #7 s#?

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Answer 1 · 2017-02-06T07:49:52

The torque is #=615.8Nm#

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*12*7^2= 49 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(14)/7*2pi#

#=(4pi) rads^(-2)#

So the torque is #tau=49*(4pi) Nm=615.8Nm#