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

Question

1337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-25T11:36:29

The torque is #=294.5 Nm#

The torque is the rate of change of angular momentum

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

The moment of inertia (spinning around the middle of the rod) is #I=1/12ml^2#

#=1/2*5*5^2= 125/2 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=125/2*3/2pi Nm=375/4piNm=294.5Nm#