What torque would have to be applied to a rod with a length of #6 m# and a mass of #3 kg# to change its horizontal spin by a frequency #18 Hz# over #9 s#?

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Answer 1 · 2017-05-01T06:07:29

The torque for the rod rotating about the center is #=113.1Nm#
The torque for the rod rotating about one end is #=452.4Nm#

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*3*6^2= 9 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(18)/9*2pi#

#=(4pi) rads^(-2)#

So the torque is #tau=9*(4pi) Nm=36piNm=113.1Nm#

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

#I=1/3*mL^2#

#=1/3*3*6^2=36kgm^2#

So,

The torque is #tau=36*(4pi)=144pi=452.4Nm#