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 #7 Hz# over #5 s#?

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Answer 1 · 2017-02-22T10:16:52

The torque (rod rotating about the center ) is #=79.2Nm#
The torque (rod rotating about one end ) is #=316.7Nm#

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=(7)/5*2pi#

#=(14/5pi) rads^(-2)#

So the torque is #tau=9*(14/5pi) Nm=126/5piNm=79.2Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=36*(14/5pi)=504/5pi=316.7Nm#