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 #3 Hz# over #5 s#?

Question

1090 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-11T07:01:02

The torque (for rod rotating about the center ) is #=33.9Nm#
The torque (for rod rotating about onr end ) is #=135.7Nm#

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=(3)/5*2pi#

#=(6/5pi) rads^(-2)#

So the torque is #tau=9*(6/5pi) Nm=54/5piNm=33.9Nm#

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*(6/5pi)=135.7Nm#