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

1 Answer
Mar 29, 2017

Answer:

The torque, for the rod rotating about the center is #=4.71Nm#
The torque, for the rod rotating about one end is #=18.85Nm#

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*3^2=0.75 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=0.75*(2pi) Nm=4.71Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=3*(2pi)=6pi=18.85Nm#