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 #7 Hz# over #18 s#?

1 Answer
Jul 8, 2017

Answer:

The torque for the rod rotating about the center is #=89.8Nm#
The torque for the rod rotating about one end is #=359.2Nm#

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

The rate of change of angular velocity is

#(domega)/dt=(7)/18*2pi#

#=(7/9pi) rads^(-2)#

So the torque is #tau=147/4*(7/9pi) Nm=89.8Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=147*(7/9pi)=359.2Nm#