What torque would have to be applied to a rod with a length of #3 m# and a mass of #2 m# to change its horizontal spin by a frequency of #3 Hz# over #2 s#?

1 Answer
May 13, 2017

Answer:

The torque for the rod rotating about the center is #=14.14Nm#
The torque for the rod rotating about one end is #=56.55Nm#

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*2*3^2= 1.5 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(3)/2*2pi#

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

So the torque is #tau=1.5*(3pi) Nm=4.5piNm=14.14Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=6*(3pi)=18pi=56.55Nm#