What torque would have to be applied to a rod with a length of #1 m# and a mass of #1 kg# to change its horizontal spin by a frequency of #6 Hz# over #2 s#?

Question

1480 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-22T13:08:04

The torque for the rod rotating about the center is #=0.52Nm#
The torque for the rod rotating about one end is #=2.09Nm#

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*1*1^2= 1/12 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(6)/6*2pi#

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

So the torque is #tau=1/12*(2pi) Nm=1/6piNm=0.52Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=1/3*(2pi)=2/3pi=2.09Nm