What torque would have to be applied to a rod with a length of #3 m# and a mass of #1 kg# to change its horizontal spin by a frequency of #5 Hz# over #2 s#?

1 Answer
Mar 30, 2018

Answer:

The torque for the rod rotating about the center is #=11.78Nm#
The torque for the rod rotating one end is #=47.12Nm#

Explanation:

The torque is the rate of change of angular momentum

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

The mass of the rod is #m=1kg#

The length of the rod is #L=3m#

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

#I=1/12*mL^2#

#=1/12*1*3^2= 0.75 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=0.75*(5pi) Nm=11.78Nm#

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

#I=1/3*mL^2#

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

So,

The torque is

#=3*(5pi)=47.12Nm#