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 #18 Hz# over #6 s#?

1 Answer
May 27, 2017

Answer:

The torque for the rod rotating about the center is #=78.54Nm#
The torque for the rod rotating about one end is #=314.16Nm#

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= 4.17 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(18)/6*2pi#

#=(6pi) rads^(-2)#

So the torque is #tau=4.17*(6pi) Nm=78.54Nm#

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*(6pi)=100pi=314.16Nm#