What torque would have to be applied to a rod with a length of #4 m# and a mass of #9 kg# to change its horizontal spin by a frequency #7 Hz# over #18 s#?

1 Answer
Sep 1, 2017

Answer:

The torque for the rod rotating about the center is #=29.3Nm#
The torque for the rod rotating about one end is #=117.3Nm#

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*9*4^2= 12 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(7)/18*2pi#

#=(7/9pi) rads^(-2)#

So the torque is #tau=12*(7/9pi) Nm=28/3piNm=29.3Nm#

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

#I=1/3*mL^2#

#=1/3*9*4^2=48kgm^2#

So,

The torque is #tau=48*(7/9pi)=112/3pi=117.3Nm#