What torque would have to be applied to a rod with a length of #6 m# and a mass of #8 kg# to change its horizontal spin by a frequency #4 Hz# over #2 s#?

1 Answer
Feb 9, 2017

Answer:

The torque is #=301.6Nm#

Explanation:

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*8*6^2= 24 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=24*(4pi) Nm=96piNm=301.6Nm#