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 #5 Hz# over #8 s#?

Question

1080 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-26T05:55:09

The torque, for the rod rotating about the center is #=35.3Nm#
The torque, for the rod rotating about one end is #=141.4Nm#

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*3*6^2= 9 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=9*(5/4pi) Nm=(45/4pi)Nm=35.3Nm#

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

#I=1/3*mL^2#

#=1/3*3*6^2=36#

So,

The torque is #tau=36*(5/4pi)=45pi=141.4Nm#