What torque would have to be applied to a rod with a length of #6 m# and a mass of #3 kg# to change its horizontal spin by a frequency #6 Hz# over #8 s#?

1 Answer
Mar 15, 2017

Answer:

The torque (for the rod rotatingabout the center) is #=42.1Nm#
The torque (for the rod rotatingabout one end) is #=169.6Nm#

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*3*6^2= 9 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(6)/8*2pi#

#=(3/2pi) rads^(-2)#

So the torque is #tau=9*(3/2pi) Nm=27/2piNm=42.1Nm#

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

#I=1/3*mL^2#

#=1/3*3*6^2=36#

So,

The torque is #tau=36*(3/2pi)=54pi=169.6Nm#