What torque would have to be applied to a rod with a length of #3 m# and a mass of #4 kg# to change its horizontal spin by a frequency #5 Hz# over #1 s#?

Question

1081 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T06:31:54

The torque is #=94.2Nm#

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*4*3^2= 3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(5)/1*2pi#

#=(10pi) rads^(-2)#

So the torque is #tau=3*(10pi) Nm=30piNm=94.2Nm#