Rotating and kinetic energy kinematics

  • #1

Main Question or Discussion Point

Hello everyone,
I've run across an interesting Newtonian physics problem that I'd like some input on. The problem begins with a rotating object. Let's assume it is a slender rod rotating about one end with a given mass (m), length (L) and rotational speed (ω). This results in the rod having the energy:

KERot=1/6*mL2ω2

The question is what happens when this object is no longer held in rotational motion but becomes completely free of external forces.

Originally I had thought that the object must travel linearly maintaining the same speed of the CG as when rotating, such that:

v=rCGω
rCG=L/2
v=Lω/2

However, this would mean that energy is not conserved because:

KELin=1/2*m*(Lω/2)2=1/8*mL2ω2

This leads me to suspect that when the object is released, the whole object (including the CG) takes on the requisite speed for energy conservation in pure linear motion. In the case above this would be:

v=sqrt(1/3)*Lω

But notice that this speed is greater than the speed of the CG when rotating. It just seems wrong that the CG would instantly obtain a greater speed when releasing the object from the external force which kept it rotating.

Another intresting aspect is that this increase in speed is not a given. If the object were a flywheel with a radius F which then bursts into 3 120° sectors (a common failure mode), the sector CG would have been traveling at the following speed before burst:

v=4Fsin3(60°)ω/(2∏-3sin(60°))=0.705*Fω

But after bursting the speed would have to be (conserving energy in purely linear motion):

v=sqrt(1/3)Fω=0.577*Fω

This time the sector CG is traveling slower than before release. I guess I am having a hard time accepting that CG would change speeds. I also am cognizant of the fact that when released, these objects may well not travel in purely linear motion, but might also take on some new rotational speed about the CG. Could this be the component I am missing? This could make sense in the scenario where the CG wants to speed up (the rod). Maybe the CG maintains the same speed and the leftover energy turns into rotation about the CG. Visually this makes sense if I envsision a batter letting go of a bat. But this wouldn't help with the scenario where the CG wants to slow down (burst flywheel)... What characteristic about an object would make it spin after release? Maybe the radius of gyration being greater than that of the CG? This is true of the slender rod:

rG=sqrt(1/3)L>L/2

But not true of the burst flywheel:

rG=sqrt(1/6)F<0.705F

I welcome your thoughts!
 

Answers and Replies

  • #2
Doc Al
Mentor
44,892
1,144
However, this would mean that energy is not conserved because:

KELin=1/2*m*(Lω/2)2=1/8*mL2ω2
You left out the rotational KE about the center of mass.

As soon as you remove all external forces, the velocity of the center of mass will not change. But there is still rotation about the center of mass.
 

Related Threads on Rotating and kinetic energy kinematics

Replies
13
Views
775
Replies
1
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
367
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
18
Views
1K
Replies
4
Views
6K
  • Last Post
Replies
8
Views
1K
Replies
2
Views
2K
Top