What torque would have to be applied to a rod with a length of #4 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #4 Hz# over #3 s#?

1 Answer
Jun 3, 2017

Answer:

The torque for the rod rotating about the center is #=22.34Nm#
The torque for the rod rotating about one end is #=89.36Nm#

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*2*4^2= 8/3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(4)/3*2pi#

#=(8/3pi) rads^(-2)#

So the torque is #tau=8/3*(8/3pi) Nm=64/9piNm=22.34Nm#

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

#I=1/3*mL^2#

#=1/3*2*4^2=32/3kgm^2#

So,

The torque is #tau=32/3*(8/3pi)=256/9pi=89.36Nm#