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 #7 s#?

Question

1241 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-02T07:01:03

The torque is #=29.9 Nm#

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)/7*2pi#

#=((16pi)/7) rads^(-2)#

So the torque is #tau=25/6*(16pi)/7 Nm=400/42piNm=29.9Nm#