What torque would have to be applied to a rod with a length of #1 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #4 Hz# over #3 s#?

1 Answer
May 17, 2017

Answer:

The torque for the rod rotating about the centre is #=1.4Nm#
The torque for the rod rotating about one end is #=5.6Nm#

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*2*1^2= 1/6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(4)/3*2pi#

#=(8/3pi) rads^(-2)#

So the torque is #tau=1/6*(8/3pi) Nm=4/9piNm=1.4Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=2/3*(8/3pi)=16/9pi=5.6Nm#