What torque would have to be applied to a rod with a length of #5 m# and a mass of #5 kg# to change its horizontal spin by a frequency of #2 Hz# over #2 s#?

1 Answer
Feb 21, 2017

Answer:

The torque (rod rotating about the center) #=65.45Nm#
The torque (rod rotating about one end) #=261.8Nm#

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*5*5^2= 125/12 kgm^2#

The rate of change of angular velocity is

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

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

So the torque is #tau=125/12*(2pi) Nm=125/6piNm=65.45Nm#

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

#I=1/3*mL^2#

#=1/3*5*5^2=125/3#

So,

The torque is #tau=125/3*(2pi)=250/3pi=261.8Nm#