What torque would have to be applied to a rod with a length of #4 m# and a mass of #5 kg# to change its horizontal spin by a frequency of #3 Hz# over #7 s#?

1 Answer
Aug 24, 2017

Answer:

The torque for the rod rotating about the center is #=17.95Nm#
The torque for the rod rotating about one end is #=71.81Nm#

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, rotating about the center is

#I=1/12*mL^2#

#=1/12*5*4^2= 20/3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(3)/7*2pi#

#=(6/7pi) rads^(-2)#

So the torque is #tau=20/3*(6/7pi) Nm=40/7piNm=17.95Nm#

The moment of inertia of a rod, rotating about one end is

#I=1/3*mL^2#

#=1/3*5*4^2=80/3kgm^2#

So,

The torque is #tau=80/3*(6/7pi)=160/7pi=71.81Nm#