Total Angular Momentum of 2 connected falling bodies.

In summary, In this conversation, a summary of the contents is provided. The first two bodies have masses and moments of inertia, and are connected by a rotational joint. There's gravity. The total angular momentum of the system is calculated to be L=I\cdot\omega. However, when trying to find the total angular momentum, the attempted solution doesn't seem to work. The final solution is to account for the rotation of the masses around the COM.
  • #1
xtinch
4
0
I have two 2 rigid bodies with masses m1 and m2 and Moments of Inertia I1 and I2, they are connected by a free rotational joint at some point, their coms lie at c1 and c2. There's gravity.
In the beginning both have some angular velocity [itex]\omega_i[/itex]
Questions:
- Total angular momentum of the system
[STRIKE]- COM trajectory[/STRIKE]

Attempt:
- Calculating L of the two bodies is no Problem with [itex]L=I\cdot\omega[/itex], Also if they're unconnected L is simply [itex]L_1+L_2[/itex]. However my attempts at finding the total angular Momentum don't seem to work out.
[STRIKE]- I thought that the COM trajectory should simply be gravity dependent, i.e. [itex]\frac{1}{2}gt^2[/itex], but somehow it wobbles in my physics simulator. (I'm using PhysX 2.8.3)[/STRIKE] (The COM trajectory is what it should be :) )
- Also I thought to calculate the total Angular Momentum I would have to shift the two Momenta to the COM via the parallel axis thm, and then build the sum.

So I'm stuck - any help would be very much appreciated.

xtin
 
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  • #2
Hi xtinch!
Welcome to PF !

xtinch said:
I have two 2 rigid bodies with masses m1 and m2 and Moments of Inertia I1 and I2, they are connected by a free rotational joint at some point, their coms lie at c1 and c2. There's gravity.

I can't aacually understand how they are connected but whatever axis that may be, you need to find moment of inertia's along a common axis,

Use [itex]I = I_{COM} + md^2[/itex]

Here d is the distance b/w COM of body and axis along which you need to find I
 
  • #3
I see, [itex]I=I_{COM}+md^2[/itex] is the parallel axis thm i mentioned, also I use [itex]R_{ig}*I*R_{ig}^T[/itex] to rotate the MoI.

So my final attempt was to rotate the MoI so they coincide with the global coordinate frame, then translate them to the global COM. (Let [itex]T(v)[/itex] be the Matrix that shifts the MoI by the vector v)
[itex]L_{total} = R^T_1*I_1*R_1+T(c_{total}-c_1)+R^T_2*I_2*R_2+T(c_{total}-c_2)[/itex]
I also tried [itex] R^T_1*(I_1+T(c_{total}-c_1))*R_1[/itex] to no avail. =\
 
  • #4
xtinch said:
I see, [itex]I=I_{COM}+md^2[/itex] is the parallel axis thm i mentioned, also I use [itex]R_{ig}*I*R_{ig}^T[/itex] to rotate the MoI.

So my final attempt was to rotate the MoI so they coincide with the global coordinate frame, then translate them to the global COM. (Let T(v) be the Matrix that shifts the MoI by the vector v)
[itex]L_total = R^T_1*I_1*R_1+T(c_{total}-c_1)+R^T_2*I_2*R_2+T(c_{total}-c_2)[/itex]
I also tried [itex] R^T_1*(I_1+T(c_1-c_{total}))*R_1[/itex] to no avail. =\

Do you have any pic related to the ques?
I can't really imagine the case ! :shy:
 
  • #5
1zch9nq.jpg


Here's an image, m3 is simply a mass that puts the system into rotation. I'm only interested in the Momentum after the collision. Below is a plot of the L calculated with my above suggestion - it's clearly not constant.

So m1+m2 fall together as it's basically a box with an attached, freely rotateable rod (there are no collisions between the rod and the box), hit m3 which is fixed in space, start to rotate and then AngVel should be constant.

(Note: ignore that the plot is labeled AngVel it's the momentum ;) )
 
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  • #6
Ok, I found the solution. I forgot to account for the rotation of the masses around the com.

The solution is simply:

[itex]L_{total} = I_1\cdot\omega_1+m_1*(c_1-c_{total})\times(v_{com}-v_1)+I_2\cdot\omega_2+m_2*(c_2-c_{total})\times(v_{com}-v_2)[/itex]

Thanks for your help and I hope this will help others who search for this simple but annoying thing in vain ;)
 

What is the Total Angular Momentum of 2 connected falling bodies?

The Total Angular Momentum of 2 connected falling bodies refers to the combined rotational momentum of two objects that are connected and falling towards a common center of gravity.

How is the Total Angular Momentum of 2 connected falling bodies calculated?

The Total Angular Momentum of 2 connected falling bodies can be calculated by multiplying the moment of inertia of the system by its angular velocity. The moment of inertia is the measure of an object's resistance to change in rotation and is dependent on the mass and distribution of the objects.

What factors affect the Total Angular Momentum of 2 connected falling bodies?

The main factors that affect the Total Angular Momentum of 2 connected falling bodies are the mass and distribution of the objects, as well as the distance between them. The more massive and spread out the objects are, the greater their moment of inertia and subsequently, their total angular momentum.

Does the Total Angular Momentum of 2 connected falling bodies change during the fall?

According to the law of conservation of angular momentum, the Total Angular Momentum of 2 connected falling bodies will remain constant unless an external torque is applied. This means that as the objects fall, their angular velocity may increase, but their total angular momentum will remain the same.

How is the Total Angular Momentum of 2 connected falling bodies useful in studying motion?

The concept of Total Angular Momentum of 2 connected falling bodies is useful in understanding the rotational motion of objects and systems. It can also be applied in various fields such as astrophysics, engineering, and mechanics to analyze and predict the behavior of rotating systems.

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