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 #7 Hz# over #8 s#?

1 Answer
Feb 21, 2018

Answer:

The torque for the rod rotating about the center is #=49.5Nm#
The torque for the rod rotating one end is #=197.9Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt=Ialpha#

The mass of the rod is #m=3kg#

The length of the rod is #L=6m#

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=(7)/8*2pi#

#=(7/4pi) rads^(-2)#

So the torque is #tau=9*(7/4pi) Nm=63/4piNm=49.5Nm#

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*(7/4pi)=63pi=197.9Nm#