What torque would have to be applied to a rod with a length of #5 # and a mass of #5 kg# to change its horizontal spin by a frequency of #2 Hz# over #4 s#?

1 Answer
Jun 27, 2017

Answer:

The torque for the rod rotating about the center is #=32.7Nm#
The torque for the rod rotating about one end is #=130.9Nm#

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)/4*2pi#

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

So the torque is #tau=125/12*(pi) Nm=32.7Nm#

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

#I=1/3*mL^2#

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

So,

The torque is #tau=125/3*(pi)=130.9Nm#