What torque would have to be applied to a rod with a length of #8 m# and a mass of #8 kg# to change its horizontal spin by a frequency #6 Hz# over #9 s#?

1 Answer
Feb 18, 2017

Answer:

The torque is #=178.7Nm#

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 is #I=1/12*mL^2#

#=1/12*8*8^2= 42.67 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(6)/9*2pi#

#=((4pi)/3) rads^(-2)#

So the torque is #tau=42.67*(4pi)/3 Nm=178.7Nm#