What torque would have to be applied to a rod with a length of #4 m# and a mass of #5 kg# to change its horizontal spin by a frequency of #3 Hz# over #7 s#?

Question

1305 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-30T06:30:04

The torque for the rod rotating about the center is #=17.95Nm#
The torque for the rod rotating about one end is #=71.81Nm#

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*5*4^2= 6.67 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(3)/7*2pi#

#=(6/7pi) rads^(-2)#

So the torque is #tau=6.67*(6/7pi) Nm=17.95Nm#

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

#I=1/3*mL^2#

#=1/3*5*4^2=26.67kgm^2#

So,

The torque is #tau=26.67*(6/7pi)=71.81Nm#