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 #1 Hz# over #2 s#?

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Answer 1 · 2017-03-29T06:36:40

The torque, for the rod rotating about the center is #=134.04Nm#
The torque, for the rod rotating about one end is #=536.17Nm#

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=42.67 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(1)/2*2pi#

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

So the torque is #tau=42.67*(pi) Nm=134.04Nm#

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

#I=1/3*mL^2#

#=1/3*8*8^2=170.67#

So,

The torque is #tau=170.67*(pi)=504/5pi=536.17Nm#