# Homework Help: Modeling a tossed turkey leg (angular momentum)

1. Mar 17, 2013

### besjbo

1. The problem statement, all variables and given/known data

Imagine that you toss the drumstick of a turkey to your friend across a table. Approximate the leg as a baton consisting of two equal masses connected by a stiff spring. Let the masses, m and M, each equal 0.1 kg (we’ll modify this assumption later) and connect them with a spring of negligible mass and spring constant, k = 105 N/m. Let the initial distance between the centers of the two masses be the rest length of the spring, d0 = 0.15m.

You give the leg an impulse so that, as it leaves your hand, mass m has v1=<2.77, 1.25, 0> m/s and M starts with v2=<1.25, 4.0, 0> m/s.

Take the origin of the coordinate system be the initial location of mass m. Let the second mass be initially located at <d0, 0, 0>. Write a code to evolve the flight of the leg under the action of gravity, but ignore air resistance.

2. Relevant equations

Ltrans = r x p
Lrot = I$\omega$ = MR2$\omega$

3. The attempt at a solution

This is where I'm struggling. I don't quite know how to approach this problem, at least not practically. Reading the book, while somewhat helpful, has not really seemed to make me capable of understanding how to go about solving this particular problem.

BTW, the code in question is in VPython, which is largely a logical language, the difficulties of which I'll deal with myself. My questions are more about the logic of this problem.

I will admit I cannot say I (fully) understand angular momentum, much less its interaction with spring force in this situation. I'm not looking for anything to be handed to me, but I'm confused enough by all of the concepts I'm trying to grasp that I feel I need guidance in baby steps. Any help would be greatly appreciated.

2. Mar 17, 2013

### rcgldr

The change in linear momentum equals the impulse (force x time) in the direction of the impulse, regardless of the point of application of the impulse. The change in angular momenum equals the tangental component of impulse x radius (distance from center of mass which should be same as distance from center of rotation since the object is not attached to anything while it travels). You should assume the impluse is instantaneous, so that the spring does not expand or shrink during the application of the impulse.

3. Mar 17, 2013

### besjbo

How do I account for spring force?

And in terms of calculating and updating values, what should I be computing first?

4. Mar 17, 2013

### rcgldr

Initially you can ignore the spring if you only calculate the path of the center of mass of the system, and it's angular momentum, a one time calculation since angular momentum is conserved after the impulse.

With the undamped spring present, the two masses will oscillate inwards and outwards about the center of mass as the spring expands and shrinks. You'll have to consider the spring as compressable as well as expandable from it's "rest" state. I'm not sure why this was added to the problem, since it makes the situation much more complex if the goal is to calculate the path of both mass m and M.

5. Mar 17, 2013

### besjbo

We were told this is computationally much easier than having the two masses connected by a rigid "stick." Not entirely how or why, either.

Also, if I'm only calculating the motion of the center of mass, does that not mean that I will not be obtaining the expected rotation of the two masses about the center of mass?

Last edited: Mar 17, 2013
6. Mar 17, 2013

### haruspex

As rcgldr indicates, you can treat the motion of the common mass centre separately from the rotation/oscillation and simply add them together at the end to find the coordinates of the two masses.

7. Mar 17, 2013

### besjbo

Sorry for my display of daftness, but what would I be adding to obtain the two sets of coordinates?

Also, I still don't have a practical idea of how to start this problem, and I still don't know what "[considering] the spring as compressable as well as expandable from it's "rest" state" entails, computationally.

8. Mar 17, 2013

### haruspex

Model the movement of the common mass centre. This is a simple 2D trajectory problem.
Model the motion of the two masses relative to their common mass centre. For this you can treat the common mass centre as stationary, so you only have to consider the two masses in a kind of orbit around each other.
To find the coordinates of a given mass at a given time, add its relative position to the position of the common mass centre.
That just means the force in the spring is always proportional to the difference between its current length and its relaxed length. Otherwise you have nonlinearity where the spring becomes fully compressed.
The hard part is going to be the rotation/oscillation. See if you can write some equations around that. Bear in mind that the common mass centre stays put for this, and there's a fixed relationship between the two distances from it.

9. Mar 17, 2013

### rcgldr

I don't think so, a rigid massless stick would have been easier, since the velocities of the masses with respect to the center of mass would not be osciallating as the spring expands and compresses.

You can calculate the angular momentum which is constant, but the stretch or compression of the spring will affect the angular velocity.

Take a simpler case, imagine the spring and it's 2 masses in outer space free from an external forces with no rotation. Start with the masses at some velocity directly away from the center of mass, no rotation involved, and calculate the path of the masses based on the spring cycling between expanding and compressing. Next consider what happens if the spring and masses are rotating. Angular momentum is conserved, and now the centripetal force exerted by the spring is opposed by the reactive force of the rotating masses (some call this reactive centrifugal force that the masses exert onto the ends of the spring). This results in a force that varies with distance from the center of mass, which requires you write equations in the form of derivatives based on acceleration versus distance from center of mass, then you have to figure out how to integrate the derivatives, to eventually translate this into position versus time (assuming you don't run into an integral that is problematic so solve). This makes the situation much more complicated. I'm wondering if this was the intent of the teacher when creating this problem.

Last edited: Mar 17, 2013
10. Mar 17, 2013

### besjbo

I don't think it was. Perhaps there's some limitation of VPython that I'm not aware of, but the professor specified that the spring is there because it would be much harder to model this situation had there been a rigid connection between the masses.

11. Mar 17, 2013

### rcgldr

I don't understand the issue of a ridgid connection. I updated my previous post to explain the issue with a spring, that you'll have to write a derivative for acceleration versus position, and then have to go through some difficult integrations (if possible) to convert this into position versus time for the two masses.

To give you an idea of what's involved, here's a thread that calculates how long it takes for two point masses to collide due to gravity and initial rest positions, in post 19 of this thread:

The rotating masses connected by a oscillating spring situation is even more complicated.

Last edited: Mar 17, 2013
12. Mar 17, 2013

### besjbo

Wow, that does seem beyond what I think we would be expected to be able to do, and definitely beyond my capabilities.

13. Mar 17, 2013

### haruspex

Integrations will not be necessary. The software will model the differential equations.

14. Mar 17, 2013

### besjbo

Would either of you be willing to grant me a concrete first step or two? I can't quite seem able to translate your descriptions into actual calculations, at least not with any confidence that I'm doing things the right way.

Things are still quite cloudy, despite my efforts to understand them. I'm willing to admit I may be in over my head, but I have no choice but to try to do this, despite my underdeveloped ability to apply the concepts in question. I don't want to seem like I want anyone to do my homework, but I've been struggling with this for some time.

15. Mar 17, 2013

### haruspex

Consider one mass in relation to the common mass centre (CMC). Let its distance from that centre be r = r(t), and the rest length of the distance be r0. Let its angular velocity about the CMC be ω = ω(t). (Motion is all in one plane.) What equations can you write relating r, ω, acceleration of the mass and the tension in the spring?
Are you familiar with these equations?
$F_r = m \ddot r -mr \dot {\theta}^2$
$F_{\theta} = mr \ddot \theta +2m \dot r \dot {\theta}$?

16. Mar 17, 2013

### besjbo

Those equations do not look familiar, but I'm admittedly a bit thrown off by the dots above some of your terms. I'm not sure what they represent. Nevertheless, I'm also not sure what the two forces represent. In notes I took when the problem was introduced, we were told the only two forces acting on the masses were gravity and spring.

Also, I attached an image that's part of our assignment which would probably help with some notation.

BTW, I do want to thank you for your patience and for spending your time helping me out. It's greatly appreciated.

#### Attached Files:

• ###### turkey leg.png
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17. Mar 17, 2013

### haruspex

The dots denote differentiation wrt to time, so a double dot shows an acceleration. In terms of angular velocity, ω, $\dot \theta = \omega$; $\ddot \theta = \dot\omega = \frac{d \omega}{dt}$. For the motion relative to the CMC you can ignore gravity, so the only force is that from the spring. That will be Fr in the equation. Fθ will be 0.
You also need an equation relating Fr to r and the spring constant.

18. Mar 18, 2013

### rcgldr

Its seems unlikely that it is intended for new students to implement a form of numerical intergration to solve this problem, at least a problem where the differential equations are complex due to modeling the dynamic path of masses connected by a spring. The original poster hasn't responded yet, so perhaps the assignment was changed.

19. Mar 19, 2013

### besjbo

No, nothing has changed, I've just put this assignment temporarily on the back-burner.