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

1 Answer
Apr 14, 2017

Answer:

The torque, for the rod rotating about the center, is #=0.35Nm#
The torque, for the rod rotating about one end, is #=1.40Nm#

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

The rate of change of angular velocity is

#(domega)/dt=(2)/3*2pi#

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

So the torque is #tau=1/12*(4/3pi) Nm=1/9piNm=0.35Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=1/3*(4/3pi)=4/9pi=1.40Nm#