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 #5 Hz# over #6 s#?

1 Answer
Mar 23, 2017

Answer:

The torque, for the rod rotating about the center is #=21.82Nm#
The torque, for the rod rotating about one end is #=87.28Nm#

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=(5)/6*2pi#

#=(5/3pi) rads^(-2)#

So the torque is #tau=25/6*(5/3pi) Nm=125/18piNm=21.82Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=50/3(5/3pi)=250/9pi=87.28Nm#