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 #8 Hz# over #8 s#?

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Answer 1 · 2017-02-05T07:33:21

The torque is #=26.2Nm#

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=(8)/8*2pi#

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

So the torque is #tau=25/6*(2pi) Nm=25/3piNm=26.2Nm#