What torque would have to be applied to a rod with a length of #7 m# and a mass of #12 kg# to change its horizontal spin by a frequency #15 Hz# over #9 s#?

1 Answer
Dec 26, 2017

Answer:

The torque for the rod rotating about the center is #=513.1Nm#
The torque for the rod rotating about one end is #=2052.5Nm#

Explanation:

The torque is the rate of change of angular momentum

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

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

The length of the rod is #L=7m#

The moment of inertia of a rod, rotating about the center is

#I=1/12*mL^2#

#=1/12*12*7^2= 49 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(15)/9*2pi#

#=(10/3pi) rads^(-2)#

So the torque is #tau=49*(10/3pi) Nm=513.1Nm#

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

#I=1/3*mL^2#

#=1/3*12*7^2=196kgm^2#

So,

The torque is #tau=196*(10/3pi)=2052.5Nm#