What torque would have to be applied to a rod with a length of #7 m# and a mass of #9 kg# to change its horizontal spin by a frequency #12 Hz# over #16 s#?

1 Answer
Feb 21, 2017

Answer:

The torque (about the center) is #=173.2Nm#
The torque (about one end) is #=692.7Nm#

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 abour the center is

#I=1/12*mL^2#

#=1/12*9*7^2= 147/4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(12)/16*2pi#

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

So the torque is #tau=147/4*(3pi)/2 Nm=441/8piNm=173.2Nm#

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

#I=1/3*mL^2#

#=1/3*9*7^2=147#

So,

The torque is #tau=147*(3/2pi)=441/2pi=692.7Nm#