What torque would have to be applied to a rod with a length of #12 m# and a mass of #7 kg# to change its horizontal spin by a frequency #4 Hz# over #8 s#?

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Answer 1 · 2017-07-16T09:04:02

The torque for the rod rotating about the center is #=263.9Nm#
The torque for the rod rotating about one end is #=1055.6Nm#

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*7*12^2= 84 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(4)/8*2pi#

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

So the torque is #tau=84*(pi) Nm=84piNm=263.9Nm#

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

#I=1/3*mL^2#

#=1/3*7*12^2=336kgm^2#

So,

The torque is #tau=336*(pi)=336pi=1055.6Nm#