What torque would have to be applied to a rod with a length of #6 m# and a mass of #2 kg# to change its horizontal spin by a frequency #2 Hz# over #5 s#?

1 Answer
Jan 11, 2017

Answer:

The torque is #=15.1Nm#

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*2*6^2= 6 kgm^2#

The rate of change of angular velocity is

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

#=((4pi)/5) rads^(-2)#

So the torque is #tau=6*(4pi)/5 Nm=24/5piNm=15.1Nm#