What torque would have to be applied to a rod with a length of #6 m# and a mass of #3 kg# to change its horizontal spin by a frequency #4 Hz# over #6 s#?

Question

1178 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-23T11:09:48

The torque is #=37.7Nm#

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*3*6^2= 9 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(4)/6*2pi#

#=4/3pi rads^(-2)#

So the torque is #tau=9*4/3pi Nm=12piNm=37.7Nm#