What torque would have to be applied to a rod with a length of #4 m# and a mass of #3 kg# to change its horizontal spin by a frequency of #12 Hz# over #4 s#?

1 Answer
Mar 24, 2017

Answer:

The torque, for the rod rotating about the center, is #=75.4Nm#
The torque, for the rod rotating about one end, is #=30.1.6Nm#

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*4^2= 4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(12)/4*2pi#

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

So the torque is #tau=4*(6pi) Nm=24piNm=75.4Nm#

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

#I=1/3*mL^2#

#=1/3*3*4^2=16#

So,

The torque is #tau=16*(6pi)=96pi=301.6Nm#