What torque would have to be applied to a rod with a length of #3 m# and a mass of #1 kg# to change its horizontal spin by a frequency of #6 Hz# over #2 s#?

1 Answer
Feb 10, 2017

Answer:

The torque is #=14.1Nm#

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*1*3^2= 3/4 kgm^2#

The rate of change of angular velocity is

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

#=(6pi) rads^(-2)#

So the torque is #tau=3/4*(6pi) Nm=9/2piNm=14.1Nm#