What torque would have to be applied to a rod with a length of #8 m# and a mass of #8 kg# to change its horizontal spin by a frequency #15 Hz# over #9 s#?

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Answer 1 · 2017-03-02T20:10:41

The torque (for the rod rotating about the center) is #=446.8Nm#
The torque (for the rod rotating about one end) is #=1782.2Nm#

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*8^2= 128/3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(15)/9*2pi#

#=(10/3pi) rads^(-2)#

So the torque is #tau=128/3*(10/3pi) Nm=1280/9piNm=446.8Nm#

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

#I=1/3*mL^2#

#=1/3*8*8^2=512/3kgm^2#

So,

The torque is #tau=512/3*(10/3pi)=5120/9pi=1787.2Nm#