What torque would have to be applied to a rod with a length of #6 m# and a mass of #2 kg# to change its horizontal spin by a frequency #7 Hz# over #8 s#?

1 Answer
Oct 30, 2017

Answer:

The torque for the rod rotating about the center is #=33Nm#
The torque for the rod rotating about one end is #=132Nm#

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#

The mass of the rod is #m=2kg#

The length of the rod is #L=6m#

#=1/12*2*6^2= 6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(7)/8*2pi#

#=(7/4pi) rads^(-2)#

So the torque is #tau=6*(7/4pi) Nm=21/2piNm=33Nm#

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

#I=1/3*mL^2#

#=1/3*2*6^2=24kgm^2#

So,

The torque is #tau=24*(7/4pi)=42pi=132Nm#