Computing angular speed wrt CM I get a contradiction

In summary, the conversation discusses the computation of angular speed in a rigid body system using the split of kinetic energy and angular momentum. A simple case of two particles is first considered, where the correct answer is obtained. However, when a disk is introduced, the results are contradictory. The conversation then moves on to discussing the concept of angular momentum around a specific point and how it can be split into the center of mass point and rotational angular momentum. The conclusion is that the results for the disk example are incorrect due to the fact that two independent point masses do not form a rigid body.
  • #1
arestes
80
3
Hello!
I have been brushing up my Rigid Body Dynamics.
I tried computing the angular speed with respect the Center of Mass (CM) using the usual split of kinetic energy and also the split of Angular momentum using the CM.
First, a simple case: Two particles of mass M each separated by a distance 2R. One going to the positive x direction with speed v and the other at rest.
preguntame_1.png

The Kinetic Energy, computed separately is:
[tex] KE=\frac{1}{2} M v^2 [/tex]
whereas the same KE computed using the split:
[tex]KE=\frac{1}{2} (2M) (v/2)^2 + \frac{1}{2} I_{CM} \omega_{CM}^2[/tex]
Where, obviously, the speed of the CM is [itex] \frac{v}{2} [/itex]. Also, it's easy to see that [tex]I_{CM} =2MR^2[/tex]
therefore, equation both expressions for KE:
[tex]\omega_{CM} = \pm \frac{v}{2R}[/tex]

Then, to check, I compute the angular momentum of the system (with respect to the point where the particle in rest is), using the same trick of splitting it using the CM (which clearly is exactly in the middle, at a distance R of each particle):
Separately: [tex] L_O=-Mv(2R) [/tex]
As a system: [tex] L_O=-MvR+ I_{CM}\omega_{CM}=-MvR+(2MR^2)\omega_{CM} [/tex]
Equating both expressions I get the right answer expected (this time with the correct sign):
[tex]\omega_{CM} = - \frac{v}{2R}[/tex]

So far so good. Now comes the problem. I repeat the ordeal this time with a disk moving with speed v to the right but rotating with angular speed [itex] \omega_0 [/itex]. The CM is again exactly in the middle of the particle at rest and the center of the disk.
preguntamelo_2.png

The Kinetic Energy of the system, computed separately (KE of the disk + 0 of the particle at rest):
[tex] KE=\frac{1}{2}Mv^2+\frac{1}{2} I_{0} \omega_0^2=\frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2 )\omega_0^2 =\frac{1}{2}Mv^2+\frac{1}{4}MR^2\omega_0^2 [/tex]
where [itex]I_{0} [/itex] is just the moment of inertia of the disk wrt to its own cm, not of the system. Now I compute the same KE using the split and the CM:
[tex]KE=\frac{1}{2} (2M) (v/2)^2 + \frac{1}{2} I_{CM} \omega_{CM}^2=\frac{1}{2} (2M) (v/2)^2 + \frac{1}{2} (MR^2+(\frac{1}{2}MR^2+MR^2)) \omega_{CM}^2=\frac{1}{2} (2M) (v/2)^2 + \frac{1}{2} (\frac{5}{2}MR^2) \omega_{CM}^2=\frac{1}{4} M v^2 + \frac{5}{4} (MR^2) \omega_{CM}^2[/tex]
using the parallel axis theorem (terms in parentheses) for the moment of inertia of the disk.
Equating both expressions I solve for [itex]\omega_{CM}[/itex]
[tex]\omega_{CM}=(\frac{1}{5}((v/R)^2+\omega_0^2))^{1/2}[/tex]

Something already seems odd. A square root of a sum... I then check what the angular momentum split tells me. First the angular momentum computed separately. The disk + 0 of the particle at rest:
Separately: [tex] L_O=-Mv(2R) +(I_{CM} \omega_0)=-2MvR +(\frac{1}{2} M R^2 \omega_0)[/tex]
As a system: [tex] L_O=-(2M)(v/2)R+ I_{CM}\omega_{CM}=-MvR+((\frac{1}{2}MR^2+MR^2)+MR^2)\omega_{CM} =-MvR+ I_{CM}\omega_{CM}=-MvR+(\frac{5}{2}MR^2)\omega_{CM}[/tex]
Equating both expressions I get
[tex]\omega_{CM} =\frac{1}{5}(-(v/R)+\frac{1}{2}\omega_0)[/tex]

And clearly it's a contradiction. What is the correct result for [itex]\omega_{CM}[/itex]?
Any help to resolve this?
 

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  • #2
arestes said:
Then, to check, I compute the angular momentum of the system (with respect to the point where the particle in rest is), using the same trick of splitting it using the CM (which clearly is exactly in the middle, at a distance R of each particle):
Separately: [tex] L_O=-Mv(2R) [/tex]
That is not the angular momentum in the CM system, where you have 2 particles of mass M at distance R moving with v/2 for a total of L=MvR.
It is the angular momentum around the lower object, for example, but that is a different quantity.
 
  • #3
mfb said:
That is not the angular momentum in the CM system, where you have 2 particles of mass M at distance R moving with v/2 for a total of L=MvR.
It is the angular momentum around the lower object, for example, but that is a different quantity.
Yes, It is in fact the angular momentum around the lower mass (at the origin).
Then I compute the same angular momentum but viewed as a system, therefore (as a system) there's a moment of inertia and I understand that the angular momentum around any point (in particular, again around O) can be split into L (of the center of mass point) _(around the chosen point) + rotational L.

I still get the right answer in the case of the point particles but something is wrong when I put forth the second example.
 
  • #4
arestes said:
Yes, It is in fact the angular momentum around the lower mass (at the origin).
Then I compute the same angular momentum but viewed as a system, therefore (as a system) there's a moment of inertia and I understand that the angular momentum around any point (in particular, again around O) can be split into L (of the center of mass point) _(around the chosen point) + rotational L.

I still get the right answer in the case of the point particles but something is wrong when I put forth the second example.

I didn't read all your post, but two independent point masses don't form a rigid body. In a rigid body, by definition, the distance between any two points must remain fixed.
 

Related to Computing angular speed wrt CM I get a contradiction

1. What does "computing angular speed wrt CM" mean?

Computing angular speed wrt CM refers to calculating the angular speed or rotation rate of an object about its center of mass. This measurement is commonly used in physics and engineering to describe the rotational motion of objects.

2. How is angular speed wrt CM different from angular velocity?

Angular speed wrt CM and angular velocity are essentially the same concept. However, angular speed is a scalar quantity that only describes the magnitude of the rotation, while angular velocity is a vector quantity that also includes the direction of the rotation.

3. What is a contradiction in this context?

A contradiction in this context refers to a discrepancy or inconsistency in the results obtained when calculating the angular speed wrt CM of an object. This could be due to errors in the measurements or incorrect assumptions in the calculations.

4. How can a contradiction be resolved when computing angular speed wrt CM?

If a contradiction occurs when computing angular speed wrt CM, it is important to carefully check all the measurements and calculations to identify any potential errors. It may also be helpful to consult with colleagues or refer to reliable sources for guidance.

5. What are some common sources of contradictions when computing angular speed wrt CM?

Some common sources of contradictions when computing angular speed wrt CM include incorrect measurements of the object's mass or radius, using incorrect formulas or equations, and neglecting to account for external forces acting on the object.

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