What torque would have to be applied to a rod with a length of #1 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #3 Hz# over #4 s#?

Question

1155 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-09T14:25:46

The torque (rod rotating about the center) is #=0.785Nm#
The torque (rod rotating about one end) is #=3.14Nm#

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*2*1^2= 1/6 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=1/6*(3/2pi) Nm=1/4piNm=0.785Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=2/3*(3/2pi)=pi=3.14Nm#