What torque would have to be applied to a rod with a length of #6 m# and a mass of #8 kg# to change its horizontal spin by a frequency #8 Hz# over #5 s#?

Question

950 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T15:24:26

The torque for the rod rotating about the center is #=241.3Nm#
The torque for the rod rotating about one end is #=965.1Nm#

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt=Ialpha#

The mass of the rod is #m=8kg#

The length of the rod is #L=6m#

The moment of inertia of a rod, rotating about the center is

#I=1/12*mL^2#

#=1/12*8*6^2= 24 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(8)/5*2pi#

#alpha=(16/5pi) rads^(-2)#

So the torque is #tau=24*(16/5pi) Nm=241.3Nm#

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

#I=1/3*mL^2#

#=1/3*8*6^2=96kgm^2#

So,

The torque is #tau=96*(16/5pi)=504/5pi=965.1Nm#