What torque would have to be applied to a rod with a length of #8 m# and a mass of #3 kg# to change its horizontal spin by a frequency #18 Hz# over #9 s#?

1 Answer
Mar 2, 2017

Answer:

The torque (for the rod rotating about the center) is #=201.1Nm#
The torque (for the rod rotating about one end) is #=804.2Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

The moment of inertiaof a rod, rotating about the center is

#I=1/12*mL^2#

#=1/12*3*8^2= 16 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(18)/9*2pi#

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

So the torque is #tau=16*(4pi) Nm=64piNm=201.1Nm#

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

#I=1/3*mL^2#

#=1/3*3*8^2=64kgm^2#

So,

The torque is #tau=64*(4pi)=804.2Nm#