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 #6 Hz# over #5 s#?

1 Answer
Jun 11, 2017

Answer:

The torque for the rod rotating about the center is #=67.86Nm#
The torque for the rod rotating about one end is #=271.43Nm#

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

The rate of change of angular velocity is

#(domega)/dt=(6)/5*2pi#

#=(12/5pi) rads^(-2)#

So the torque is #tau=9*(12/5pi) Nm=108/5piNm=67.86Nm#

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

#I=1/3*mL^2#

#=1/3*3*6^2=36kgm^2#

So,

The torque is #tau=36*(12/5pi)=432/5pi=271.43Nm#