What torque would have to be applied to a rod with a length of #1 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #5 Hz# over #4 s#?

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Answer 1 · 2017-01-18T14:19:35

The torque is #=1.31Nm#

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*1^2= 1/6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(5)/4*2pi#

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

So the torque is #tau=1/6*(5pi)/2 Nm=5/12piNm=1.31Nm#