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 #18 Hz# over #9 s#?

Question

1198 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-05T14:25:48

The torque for the rod rotating about the center is #=536.2Nm#
The torque for the rod rotating about one end is #=2144.7Nm#

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*8*8^2= 42.7 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=42.7*(4pi) Nm=536.2Nm#

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

#I=1/3*mL^2#

#=1/3*8*8^2=170.7kgm^2#

So,

The torque is #tau=170.7*(4pi)=2144.7Nm#