What torque would have to be applied to a rod with a length of #5 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #2 Hz# over #6 s#?

1 Answer
Feb 23, 2017

Answer:

The torque (for the rod rotating about the center) is #=8.73Nm#
The torque (for the rod rotating about one end) is #=34.91Nm#

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*5^2= 25/6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(2)/6*2pi#

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

So the torque is #tau=25/6*(2/3pi) Nm=25/9piNm=8.73Nm#

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

#I=1/3*mL^2#

#=1/3*2*5^2=50/3kgm^2#

So,

The torque is #tau=50/3*(2/3pi)=100/9pi=34.91Nm#