What torque would have to be applied to a rod with a length of #5 m# and a mass of #12 kg# to change its horizontal spin by a frequency #8 Hz# over #7 s#?

1 Answer
Oct 30, 2017

Answer:

The torque for the rod rotating about the center is #=179.5Nm#
The torque for the rod rotating about one end is #=718.1Nm#

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#

The mass of the rod is #m=12kg#

The length of the rod is #L=5m#

#=1/12*12*5^2= 25 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(8)/7*2pi#

#=(16/7pi) rads^(-2)#

So the torque is #tau=25*(16/7pi) Nm=400/7piNm=179.5Nm#

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

#I=1/3*mL^2#

#=1/3*12*5^2=100kgm^2#

So,

The torque is #tau=100*(16/7pi)=1600/7pi=718.1Nm#