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 #16 Hz# over #4 s#?

1 Answer
Jan 24, 2017

Answer:

The answer is #226.2Nm#

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

The rate of change of angular velocity is

#(domega)/dt=(16)/4*2pi#

#=(8pi) rads^(-2)#

So the torque is #tau=9*(8pi) Nm=72piNm=226.2Nm#