What torque would have to be applied to a rod with a length of #1 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #15 Hz# over #4 s#?

1 Answer
Jul 14, 2017

The torque for the rod rotating about the center is #=3.93Nm#
The torque for the rod rotating about one end is #=15.71Nm#

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*2*1^2= 1/6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(15)/4*2pi#

#=(15/2pi) rads^(-2)#

So the torque is #tau=1/6*(15/2pi) Nm=5/4piNm=3.93Nm#

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

#I=1/3*mL^2#

#=1/3*2*1^2=2/3kgm^2#

So,

The torque is #tau=2/3*(15/2pi)=5pi=15.71Nm#