What torque would have to be applied to a rod with a length of #5 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #3 Hz# over #6 s#?

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Answer 1 · 2017-02-15T15:56:05

The torque is #=13.1Nm#

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*2*5^2= 25/6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(3)/6*2pi#

#=pi rads^(-2)#

So the torque is #tau=25/6*pi Nm=13.1Nm#