What torque would have to be applied to a rod with a length of #3 m# and a mass of #8 kg# to change its horizontal spin by a frequency #8 Hz# over #5 s#?

Question

1026 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T05:55:15

The torque for the rod rotating about the center is #=60.3Nm#
The torque for the rod rotating about one end is #=241.3Nm#

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*8*3^2= 6 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=6*(16/5pi) Nm=96/5piNm=60.3Nm#

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

#I=1/3*mL^2#

#=1/3*8*3^2=24kgm^2#

So,

The torque is #tau=24*(16/5pi)=3844/5pi=241.3Nm#