What torque would have to be applied to a rod with a length of #6 m# and a mass of #2 kg# to change its horizontal spin by a frequency #2 Hz# over #8 s#?

Question

1049 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-01T17:07:01

The torque, for the rod rotating about the center is #=9.42Nm#
The torque, for the rod rotating about one end is #=37.7Nm#

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*6^2= 6 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=6*(pi/2) Nm=9.42Nm#

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

#I=1/3*mL^2#

#=1/3*2*6^2=24kgm^2#

So,

The torque is #tau=24*(pi/2)=37.7Nm#