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

1 Answer
Jan 11, 2018

Answer:

The torque for the rod rotating about the center is #=329.9Nm#
The torque for the rod rotating about one end is #=1319.5Nm#

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=7kg#

The length of the rod is #L=12m#

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

#I=1/12*mL^2#

#=1/12*7*12^2= 84 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(5)/8*2pi#

#=(5/4pi) rads^(-2)#

So the torque is #tau=84*(5/4pi) Nm=329.9Nm#

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

#I=1/3*mL^2#

#=1/3*7*12^2=336kgm^2#

So,

The torque is #tau=336*(5/4pi)=1319.5Nm#