What torque would have to be applied to a rod with a length of #4 m# and a mass of #2 kg# to change its horizontal spin by a frequency #9 Hz# over #2 s#?

1 Answer
Apr 3, 2018

Answer:

The torque for the rod rotating about the center is #=75.49Nm#
The torque for the rod rotating about one end is #=301.69Nm#

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=2kg#

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*2*4^2= 2.67 kgm^2#

The rate of change of angular velocity is

#alpha=(domega)/dt=(9)/2*2pi#

#=(9pi) rads^(-2)#

So the torque is #tau=2.67*(9pi) Nm=75.49Nm#

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

#I=1/3*mL^2#

#=1/3*2*4^2=10.67kgm^2#

So,

The torque is #tau=10.67*(9pi)=301.69Nm#