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 #3 Hz# over #3 s#?

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Answer 1 · 2016-12-23T14:35:03

The torque is #=1.05Nm#

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

The is #I=1/12ml^2#

#=1/12*2*1^2= 1/6 kgm^2#

The rate of change of angular velocity is

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

#=2pi rads^(-2)#

So the torque is #tau=1/6*2pi Nm=pi/3Nm=1.05Nm#