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 #5 Hz# over #8 s#?

Question

1186 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-12T06:15:51

The torquefor the rod rotating about the center is #=167.6Nm#
The torque for the rod rotating about one end is #=670.2Nm#

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= 42.67 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(5)/8*2pi#

#=(5/4pi) rads^(-2)#

So the torque is #tau=42.67*(5/4pi) Nm=167.6Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=170.67*(5/4pi)=670.2Nm#