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 #15 Hz# over #6 s#?

1 Answer
Apr 14, 2017

Answer:

The torque, for the rod rotating about the center, is #=141.4Nm#
The torque, for the rod rotating about one end, is #=565.5Nm#

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=(15)/6*2pi#

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

So the torque is #tau=9*(5pi) Nm=45piNm=141.4Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=36*(5pi)=180pi=565.5Nm#