What torque would have to be applied to a rod with a length of #4 m# and a mass of #3 kg# to change its horizontal spin by a frequency of #2 Hz# over #1 s#?

1 Answer
Jan 3, 2017

Answer:

The torque is #=50.3Nm#

Explanation:

We asume that the rod is spinning around its centerm this is important for the moment of inertia.

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*3*4^2= 4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(2)/1*2pi#

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

So the torque is #tau=4*4pi Nm=16piNm=50.3Nm#