What torque would have to be applied to a rod with a length of #4 m# and a mass of #3 kg# to change its horizontal spin by a frequency of #5 Hz# over #2 s#?

1 Answer
Dec 5, 2017

Answer:

The torque for the rod rotating about its center is #=62.83Nm#
The torque for the rod rotating about one end is #=251.33Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

The mass of the rod is #m=3kg#

The length of the rod is #L=4m#

The moment of inertia of a rod, rotating about the center is

#I=1/12*mL^2#

#=1/12*3*4^2= 4 kgm^2#

The frequency is #f=5 Hz#

The rate of change of angular velocity is

#(domega)/dt=(5)/2*2pi#

#=(5pi) rads^(-2)#

So the torque is #tau=4*(5pi) Nm=20piNm=62.83Nm#

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

#I=1/3*mL^2#

#=1/3*3*4^2=16kgm^2#

So,

The torque is #tau=16*(5pi)=80pi=251.33Nm#