What torque would have to be applied to a rod with a length of #1 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #7 Hz# over #5 s#?

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Answer 1 · 2017-04-27T06:34:23

The torque for the rod rotating about the center is #=1.47Nm#
The torque for the rod rotating about one end is #=5.86Nm#

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

The rate of change of angular velocity is

#(domega)/dt=(7)/5*2pi#

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

So the torque is #tau=1/6*(14/5pi) Nm=7/15piNm=1.47Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=2/3*(14/5pi)=28/15pi=5.86Nm#