Modeling a tossed turkey leg (angular momentum)

In summary: 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 really understand what angular momentum is.You would be adding the impulse (force x time) to obtain the coordinates of the masses.
  • #1
besjbo
15
0

Homework Statement



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.



Homework Equations



Ltrans = r x p
Lrot = I[itex]\omega[/itex] = MR2[itex]\omega[/itex]


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.
 
Physics news on Phys.org
  • #2
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
rcgldr said:
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).

How do I account for spring force?

And in terms of calculating and updating values, what should I be computing first?
 
  • #4
besjbo said:
How do I account for spring force?
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
rcgldr said:
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.

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:
  • #6
besjbo said:
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?
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
haruspex said:
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.

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
besjbo said:
Sorry for my display of daftness, but what would I be adding to obtain the two sets of coordinates?
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.
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.
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
besjbo said:
We were told this is computationally much easier than having the two masses connected by a rigid "stick."
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.

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

besjbo said:
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.
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:
  • #10
rcgldr said:
This makes the situation much more complicated. I'm wondering if this was the intent of the teacher when creating this problem.

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
besjbo said:
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.
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:

https://www.physicsforums.com/showthread.php?t=635188l&page=2

The rotating masses connected by a oscillating spring situation is even more complicated.
 
Last edited:
  • #12
rcgldr said:
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:

https://www.physicsforums.com/showthread.php?t=635188l&page=2

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

Wow, that does seem beyond what I think we would be expected to be able to do, and definitely beyond my capabilities.
 
  • #13
rcgldr said:
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.
Integrations will not be necessary. The software will model the differential equations.
 
  • #14
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
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
haruspex said:
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} ##?

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.
 

Attachments

  • turkey leg.png
    turkey leg.png
    8 KB · Views: 436
  • #17
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
haruspex said:
Integrations will not be necessary. The software will model the differential equations.
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
rcgldr said:
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.

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

Related to Modeling a tossed turkey leg (angular momentum)

1) What is angular momentum and how does it apply to a tossed turkey leg?

Angular momentum is a measure of the rotational motion of an object. In the case of a tossed turkey leg, its angular momentum would be determined by its mass, velocity, and distance from the axis of rotation.

2) How do you model the trajectory of a tossed turkey leg?

To model the trajectory of a tossed turkey leg, we would use mathematical equations and principles such as projectile motion and conservation of angular momentum. We would also consider external factors such as air resistance and the shape of the turkey leg itself.

3) What factors affect the angular momentum of a tossed turkey leg?

The angular momentum of a tossed turkey leg is affected by its mass, velocity, and distance from the axis of rotation. Other factors such as the shape of the turkey leg and the air resistance it experiences can also impact its angular momentum.

4) Can you predict where a tossed turkey leg will land using angular momentum?

Yes, by using mathematical models and equations, we can predict the trajectory and landing location of a tossed turkey leg based on its initial angular momentum. However, external factors such as wind or unexpected collisions may affect the actual outcome.

5) How can understanding angular momentum help us in other areas of science?

Angular momentum is a fundamental principle in physics and can be applied to various fields such as astronomy, mechanics, and fluid dynamics. Understanding angular momentum can help us explain and predict the behavior of objects in motion, from the rotation of planets to the flight of a tossed turkey leg.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
967
  • Introductory Physics Homework Help
Replies
3
Views
275
  • Introductory Physics Homework Help
Replies
4
Views
832
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
846
  • Introductory Physics Homework Help
Replies
9
Views
2K
Replies
13
Views
992
  • Introductory Physics Homework Help
Replies
12
Views
978
  • Introductory Physics Homework Help
Replies
2
Views
972
  • Introductory Physics Homework Help
Replies
14
Views
2K
Back
Top