What torque would have to be applied to a rod with a length of #1 m# and a mass of #7 kg# to change its horizontal spin by a frequency #5 Hz# over #4 s#?

Question

1130 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-31T10:21:44

The torque, for the rod rotating about the center, is #=4.58Nm#
The torque, for the rod rotating about one end, is #=18.33Nm#

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

The rate of change of angular velocity is

#(domega)/dt=(5)/4*2pi#

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

So the torque is #tau=7/12*(5/2pi) Nm=35/24pi=4.58Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=7/3*(5/2pi)=35/6pi=18.33Nm#