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 #12 Hz# over #4 s#?

Question

1026 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-01T06:30:37

The torque for the rod rotating about the center is #=169.6Nm#
The torque for the rod rotating about one end is #=678.6Nm#

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=(12)/4*2pi#

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

So the torque is #tau=9*(6pi) Nm=54piNm=169.6Nm#

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*(6pi)=216pi=678.6Nm#