# Attached rigid bodies

1. May 15, 2006

### RazorTM

Well, this is my first post--hopefully I'm posting in the correct forum...

I'm developing a space simulator computer application in which a rigid body's motion is calculated using its linear momentum and rotational momentum. I'm simulating multiple bodies attached together (for example, a single space craft with missiles attached or a freighter with cargo attached), and I want to be able to release the attached objects and update the mass, center of mass, moment of inertia for all axes, linear momentum, etc. of the remaining group of objects. The only thing I'm having trouble with is understanding what would happen to the rotational momentum when one of the objects is released from a rotating group (imagine the freighter dropping its cargo in an emergency). My guess is that the group will lose rotational momentum proportionally to how much mass was dropped. I can use the parallel axis theorem to calculate the change in inertia, but is there a similar theorem or rule which I can use to determine the change in rotational momentum?

2. May 15, 2006

### LURCH

Just a suggestion, but if you treat the two objects as separate from one another, each having its own center of mass, you could then measure the distance from the center of rotation for the two conjoined objects to the center of mass for the object being released (the "cargo" mentioned in your example). This would give you the velocity at which the cargo was traveling when it was released, and that velocity times the mass of the cargo should tell you how much momentum to subtract from the rotational momentum of the remaining vessel.

Hope that made sense.

3. May 16, 2006

### RazorTM

So the linear momentum of the released object would be equal to both the amount of rotational momentum and the amount of linear momentum lost by the group that released the object--that makes sense. One more question: would the change in the location of the center of mass affect the rotational momentum or simply change the point around which the group of objects rotates?

Thanks for the help!

4. May 17, 2006

### pervect

Staff Emeritus
The structure can be thought of as being made up of a large number of small pieces, each with a velocity v(r) and a mass dm.

The important thing is that the instantaneous velocity of each piece will be the same before and after the separation as there are no impulsive forces. Because dt=0, dP = F dt must be zero.

Therfore you know the velocity of each piece of the system at the moment of separation- it was the same velocity that the piece had before the separation.

You can then re-compute the total linear momentum that each piece has (the sum of the linear momentum of the two pieces will be the total linear momentum that the whole assembly had).

Next up, re-compute the center of mass of each piece.

Finally re-compute the total angular momentum that each piece has around it's new center of mass.

As a cross-check, the total angular momentum around the original center of mass should be unchanged.

IIRC There's a pretty simple formula for the angular momentum of a body around a point other than its center of mass that can be used for this cross-check

I *think* the formula is hat this is the angular momentum of a body around a point different from its center of mass is the sum of the angular momentum of the body around it's center of mass plus the angular momentum of the center of mass around the new point.

5. May 17, 2006

### arildno

With the aid of Reynolds' transport theorem, in the absence of external torques, the rate of change of the angular momentum* of the geometric system designated as the "main body" will balance the angular momentum flux out of the body.
This is the same as saying that the angular momentum is constant for the material system consisting of the stuff included in the main body plus the stuff that are leaving it.

*with respect to an intelligently chosen reference point.

6. Jun 8, 2006

### RazorTM

Well, after a great deal of searching, I finally found a web page which I think properly explains the answer to my second question:
http://www.engin.brown.edu/courses/en4/notes/RigidDynamics/dynmod.htm#rb523 -- Scroll down to figure 5.2.11
(From http://www.engin.brown.edu/courses/en4/notes.htm)

It says that the angular momentum of a body around a point different from its center of mass is equal to the angular momentum of the body around its center of mass plus the cross product of the vector representing the radius of the rotation (the distance between the point of rotation and the center of mass) and the linear momentum of the center of mass, which is very close to what pervect said.

This basically tells me that as long as the space ship and cargo aren't rotating with respect to each other and no external forces are being applied when the ship drops the cargo, the ship *and* the cargo after being separated will both have the same angular velocity as the whole system had before the separation. Also, the velocity of the ship (or the cargo) after release will be the velocity of the center of mass plus the tangental velocity of the ship (or the cargo) around that center of mass.
Is my understanding correct?

Last edited: Jun 8, 2006
7. Jun 8, 2006

### reilly

Not so easy. Any system in which linear and rotational momentum are coupled is often very difficult. First I'd suggest you study the motion of an orbiting gyroscope, which you can find probably with a Google, or dope it out from Goldstein, Whittaker or ..... Or, look at the motion of a gyroscope on a plane, with before and after configurations, symmetrical or not, in which the inertia tensor changes. No doubt you will need the Euler equations for rigid body motion, which are developed in body centered coordinates. (Note that this often leads to solutions involving elliptic integrals, or worse.) I suspect you will need to make lots of approximations to make much progress -- for example, going from a cylindrically symmetrical ship to a non-symmetric ship after release of something will require highly heroic analytic efforts, symmetric to symmetric will not be so bad. Or, life will be simpler if the released mass is much smaller than the ship mass.

Once something is released, the inertia tensor of the total system will be varying in time.In virtually all circumstances, the ship and released object will be in relative rotational motion. Ouch.

Best of luck, and regards,
Reilly Atkinson

Last edited: Jun 8, 2006
8. Jun 11, 2006

### RazorTM

But once the ship releases the cargo, they become 2 separate systems, right?

Here's a situation I created as an example. Does my reasoning make sense?

The ship and cargo are rotating forward at 2rad/s and moving forward with a velocity of 5m/s. Their centers of mass are 3m apart as shown in the drawing below.

Before releasing the cargo

For the inertia of the ship around its own center of mass:
$$I_S = \frac{1}{12}*m_S(3r^2+l^2) = \frac{1}{12}*900(3*1^2 + 12^2) = 10875$$

Using the parallel axis theorem to find the inertia of the ship around the system's center of mass:
$$I_{S/M} = I_S + m_S * d^2 = 10875 + 900 * 2^2 = 11775$$

Doing the same thing for the cargo:
$$I_C = \frac{1}{12}*m_C(l^2 + h^2) = \frac{1}{12}*450(6^2 + 2^2) = 1500$$
$$I_{C/M} = I_C + m_C * d^2 = 1500 + 450 * 2^2 = 3300$$

The total inertia around the system's center of mass is
$$I_M = I_{S/M} + I_{C/M} = 15075$$
And the rotational momentum about the system's center of mass:
$$L_M = \omega_M * I_M = 2 * 15075 = 30150$$

Now, looking only at the ship, its rotational momentum about the center of mass should be equal to its rotational momentum plus the radius of the rotation times the mass times the tangental velocity, or
$$L_{S/M} = L_S + r_S * m_S v_S$$

Since the ship and cargo are locked together, meaning they do not rotate with respect to each other, the angular velocity of the ship about its own center of mass should be equal to the angular velocity of the system about the system's center of mass.
$$\omega_S = \omega_M$$

Also, the tangental velocity of the ship is:
$$v_S = \omega_M * r_S = 2$$

Now,
$$L_{S/M} = \omega_S * I_S + r_S * m_S v_S = 2 * 10875 + 1 * 900 * 2 = 23550$$

We can calculate the same thing for the cargo:
$$L_{C/M} = \omega_C * I_C + r_C * m_C v_C = 2 * 1500 + 2 * 450 * 4 = 6600$$

The rotational momentum of the ship relative to the system's center of mass plus the rotational of the cargo relative to the system's center of mass should equal the rotational momentum of the system, right?
$$L_M = L_{S/M} + L_{C/M} = 23550 + 6600 = 30150$$ Correct!

After the release

Let's say that the cargo is released when the ship's nose is pointing downward at pi/3 radians:

Based on the Before the release section above, the cargo and the ship should both continue rotating about their own centers of mass at 2rad/s after release, and
Edit: I was thinking about this for a while after posting, and I think it's wrong. I should say that the new momentum of the ship is the momentum of the ship relative to the system center of mass before release, or 23550 in this case. If you divide the momentum by the ship's inertia, you should get 23550/10875 = 2.166 rad/s, and, similarly, for the cargo, we get 6600/1500 = 4.4 rad/s, both being in the same direction of the original rotation.

We can determine their final velocities with some vector diagrams:

For the ship,

For the cargo,

Last edited: Jun 12, 2006
9. Jun 13, 2006

### reilly

WOW, RazorTM, nicely done -- I haven't gone through evrything, but I will.
Regards,
Reilly

10. Jun 15, 2006

### RazorTM

Thanks. The one thing that perplexes me is that the total momentum for all the pieces after separating doesn't equal the total momentum from the system before separation. I'm still rather confused on this whole concept.

11. Jun 15, 2006

### pervect

Staff Emeritus
I'm getting 11025!

900*12 + 900/4 = 11025

Your diagram labels SM as 1 meter, not two

11025 + 900 = 11925

check

check

With above, 11925+3300 = 15225

The L_S in this should not include any center-of-mass adjustment if you use this formula. You are double counting here.

Let us switch to a frame in which M is not moving, for simplicity.

Thus L due to the Ship about M in this frame in which M is stationary is:

11025*w^2 + 900*w^2

(m_S * v_s *r_S = m_S * w_s * r_S * r_S, r_S=1, M_s = 900)

The 11025*w^2 is the component of angular momentum due to the rotation of the ship. The 900*w^2 is the component of angular momentum due to the motion of the center of mass of the ship

Similarly, L due to the Cargo about M is just

1500*w^2 + 1800^w^2

Let's go back to the frame in your diagram.

After release, the ship COM should have a speed of r_s*W = 2 meters/second, having a downward component of

(2*1*sin(Pi/3)) = 1.73 m/s

and a rightward velocity of (5 + 2*1*cos(Pi/3)) = 6 m/s

and it should in additoin be rotating with an unchanged angular velocity of w = 2 rad/sec

The cargo COM should have an upward velocity of

2*2*sin(Pi/3) = 3.46 m/s

and a rightward velocity of (5-2*2*cos(Pi/3)) = 3 m/s

plus it should rotate around it's axis with an unchanged angular velocity of 2 rad/sec.

12. Jun 16, 2006

### RazorTM

Oops, just a few typos that I made--sorry!

Where does the w^2 come from?
m_S * w_s * r_S * r_S only has a single w_s in it, not two, and L_S = I_S * w_S, not I_S * w_S^2, right?

I think you're trying to say 1500 * w + 1800 * w, which is exactly what I had, just not in quite the same form... Let me expand on my main concern:

I know for a fact that the angular momentum of the system in my example is 2 * 15225 = 30450, and I know that the linear momentum of the system is (900+450) * 5 = 6750. Therefore, the total momentum of the system is 37200.

Once the ship and cargo separate, I can easily determine their linear momentums.
For the ship, the momentum is
$$\sqrt{39}*900$$ (velocity times mass)
And for the cargo,
$$\sqrt{21}*450$$

If the ship and cargo are both rotating at 2rad/s after the come apart, then their angular momentums are 2*11025 = 22050 and 2*1500 = 3000.

When I add all four of these numbers (both linear momentums and angular momentums), I should get the total momentum of the system before the ship and cargo separated, right?
$$\sqrt{39}*900 + \sqrt{21}*450 + 22050 + 3000 \approx 32732.6573$$
That's just over 4467 units of momentum less than the 37200 which the system had before it separated. Either the system lost momentum (even though no external forces were acting upon it) or the angular velocities of the ship and cargo should increase slightly.

I determined that if the ship and cargo both have an angular velocity of 2.35667.... rad/s after the separation, the momentum works out now, but is that correct? Or will the cargo and ship have slightly different angular velocities since they were different distances from the center of mass of the system, have different amounts of intertia about the axis, have different masses, etc.?

Last edited: Jun 17, 2006
13. Jun 17, 2006

### pervect

Staff Emeritus
Oops - right!

The ship and the cargo will not experience any torques, so they will have an unchanged angular velocity of 'w' when they are released.

The way I compute the results are this:

After the release, the 900kg ship at it's center of mass (COM) will have a rightward velocity of 6 m/s and a downward velocity of 1.73 m/s, plus it will be rotating around it's COM at 2 radians/second, the same as it was before the release.

This is due to the fact that the velocity of the COM of the ship after release is the same as the velocity of the COM of the ship before the release. The COM of the ship was initially accelerating because of the cable connecting it to the cargo. When the cable is released, the COM of the ship will stop accelerating, and move at a constant velocity.

In addition, there is no torque on the ship, so it's angular velocity will likewise remain unchanged before and after release. (The sole difference is that the cable never put any torque on the ship, because it was attached at the COM of the ship - the cable did, however, accelerate the ship until it was released, at which point the force to the cable disappears, along with the cable itself, and the ship stops accelerating).

The detailed trig computations of the intial velocity of the COM of the ship were in my previous post, perhaps a bit terse, but it's just the velocity 2 m/s (r*w) in the direction specified by your diagram, broken into its vertical and horizontal components.

Similarly, the 450 kg cargo COM will have a rightward velocity of 3 m/s, and an upward velocity of 3.46 m/s (computed in the same manner), and the cargo will also be rotating around it's COM at 2 radians/second.

The total linear momentum of this system is zero in the vertical direction after the release, and is (900*6) + (450*3) = 6750 (kg m / s) in the horizontal direction. This is the same as it was before the release: (1350*5 = 6750).

The total angular momentum of the system around point M is the same, too.

Before you worked out the moment of inertia I of the entire system:

I = 15225, and you get I*w = 2*15225 = 30450 (kg-m^2 / s) for the total angular momentum

After the separation, you have a ship with a spin angular momentum around it's COM of 11025*2 = 22050, whose total angular momentum around point M is it's spin angular momentum (22050) plus the angular momentum due to the motion of it's COM relative to M, which is just

900 kg * (2 m/s) * 1 m = 1800 (kg - m^2/s)

You have a cargo with a spin angular momentum around its COM of 1500*2 = 3000 plus an angular momentu due to the motion of it's COM of

450 kg * (4m / s) * 2m = 3600 (kg - m^2/s)

The total is 22050 + 1800 + 3000 + 3600 = 30450 (kg m^2 / s)

this is the same as before.

Note that you cannot add the linear momentum to the angular momentum - they don't even have the same units. Linear momentum is

mass * velocity = kg * m / s

Angular momentum is

mass * velocity * radius = kg * m/s * m = kg m^2 / s

Angular momentum and linear momentum are two separate quantites, both of which are conserved.

Last edited: Jun 17, 2006
14. Jun 18, 2006

### RazorTM

Wow, thank you all for your help! I think the most critical thing that was confusing me is this:

I will see what I can do about improving on my space simulator and show my results here later.

15. Jun 19, 2006

### RazorTM

One more quick question: You're saying that the 1800 kg*m^2/s angular momentum from the ship (and 3600 from the cargo) seems to "disappear" because the ship is no longer accelerating towards the COM of the system, right?

Just trying to make sure I completely understand the situation.