What torque would have to be applied to a rod with a length of #4 m# and a mass of #3 kg# to change its horizontal spin by a frequency of #4 Hz# over #3 s#?

Question

1045 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-06T09:18:31

The torque for the rod rotating about the center is #=33.5Nm#
The torque for the rod rotating about one end is #=134.0Nm#

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*4^2= 4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(4)/3*2pi#

#=(8/3pi) rads^(-2)#

So the torque is #tau=4*(8/3pi) Nm=32/3piNm=33.5Nm#

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

#I=1/3*mL^2#

#=1/3*3*4^2=16#

So,

The torque is #tau=16*(8/3pi)=128/3pi=134.0Nm#