What torque would have to be applied to a rod with a length of #8 m# and a mass of #8 kg# to change its horizontal spin by a frequency #1 Hz# over #9 s#?

1 Answer
Feb 23, 2017

Answer:

The torque (for the rod rotating about the center) is #=29.8Nm#
The torque (for the rod rotating about one end) is #=119.1Nm#

Explanation:

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

#I=1/12*mL^2#

#=1/12*8*8^2= 128/3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(1)/9*2pi#

#=(2/9pi) rads^(-2)#

So the torque is #tau=128/3*(2/9pi) Nm=256/27piNm=29.8Nm#

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

#I=1/3*mL^2#

#=1/3*8*8^2=512/3kgm^2#

So,

The torque is #tau=512/3*(2/9pi)=1024/27pi=119.1Nm#