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 #2 Hz# over #8 s#?

1 Answer
May 12, 2017

Answer:

The torque for the rod rotating about the center is #=67.02N#
The torque for the rod rotating about one end is #=268.08N#

Explanation:

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=(2)/8*2pi#

#=(1/2pi) rads^(-2)#

So the torque is #tau=42.67*(1/2pi) Nm=67.02Nm#

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=36*(1/2pi)=268.08Nm#