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Separating angular and linear momentum

  1. Jan 15, 2010 #1
    Is it possible to separate the angular and linear momentum and treat them separately in a system such as an inverted pendulum?

    For example, a pole on a platform balancing on its end. If it is unbalanced so that it falls off the platform, it goes through motion like an inverted pendulum until it leaves the platform at which point it has angular momentum around its COM and linear momentum of its COM with vertical and horizontal components (correct me if I'm wrong here).

    Is the horizontal component of linear momentum of its COM the same just before it breaks contact with the platform as it is when it is falling?

    If so, does that mean that angular momentum could be ignored when calculating horizontal linear acceleration?

    Finally, can anyone point me to the kind of calculations I could do to discover the linear horizontal momentum change generated in this scenario?
     
  2. jcsd
  3. Jan 17, 2010 #2
    Maybe I didn't explain that very well, let me try again.

    Taking the scenario of a non-accelerating rod in space.
    2 equal and opposite forces are applied simultaneously to the rod. The first is applied at the rod's COM at an angle of 5* to the vertical. The second is applied at the rod's end at 5* to the vertical, so that both forces are directly opposite each other.
    I would expect the forces to combine and create a torque that would cause angular acceleration of the rod around its COM.

    Now if a third force is introduced at the end that prevents it from rotating, the rod will move like a pendulum, pivoting about its base.

    Once all the forces are removed, I would like to calculate the linear acceleration that the rod has experienced.

    I presume I can calculate the linear acceleration simply by considering force 3 and using F = ma?

    If I can get confirmation on that, I can go on to the second part of the problem. This all relates to a problem I am trying to solve regarding the mechanics of running, the effect of gravity and correcting the angular motion so the runner does not fall on his face.
     
  4. Jan 17, 2010 #3

    Doc Al

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    Sounds good to me. (The first two forces cancel out.) Using F = ma will give you the acceleration of the rod's center of mass.
     
  5. Jan 17, 2010 #4
    Doc Al,

    Thanks for the response. The idea of the first 2 forces is that they are acting at an angle to the rod so there is an effective lever arm between, probably wasn't clear from my description - that's what the 5* angle from the vertical for force application was for.


    Presuming that is right, the second part of the question is to relate the first scenario to an inverted pendulum. So, a simple rod resting vertically on the earth is tipped forwards so that it falls over.
    The forces would be gravity acting on the COM vs vertical GRF at the base, both equal and opposite but offset from each other so they produce rotation - like the first 2 forces in the previous example. The third force is horizontal GRF provided by friction that stops the base slipping as the rod rotates and falls.
    If ground contact instantaneously ceases part way through the fall, can I calculate the linear acceleration at that point from the sum of the horizontal impulses up to that point?

    I think that should be OK, and if so it leads on to the 3rd and possibly final question.
     
  6. Jan 17, 2010 #5

    Doc Al

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    Don't assume that the upward normal force exerted by the ground is equal and opposite to the gravitational force on the COM. (That's true while the rod is vertical, but not once it starts falling.)
     
  7. Jan 18, 2010 #6
    Doc Al,

    I did wonder about that as I typed it. Is it actually the case that the GRF acts along the rod with magnitude equal to the weight of the rod?
     
  8. Jan 18, 2010 #7

    HallsofIvy

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    Yes. You can either think of an "infinitesmal" force acting on an infinitesmal mass at each point on the rod (so that the integral along the length of the rod is equal to its weight) or you can think of the entire weight as acting at the center of gravity of the rod.
     
  9. Jan 18, 2010 #8

    Doc Al

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    No.

    I assume that GRF stands for 'Ground Reaction Force', which is your term for the contact force that the ground exerts on the rod? If I'm misinterpreting what you mean, let me know.
     
  10. Jan 18, 2010 #9
    Doc Al,

    Yes, GRF is ground reaction force, the force exerted by the ground at the contact point with an object.

    I presumed that the GRF would be equal in magnitude to the rod's weight but with its direction along the rod. When the rod is vertical, GRF and weight cancel and so there is no net external force.
    When the rod is leaning over, GRF would have a vertical component that is less than the rod's weight and a non-zero horizontal component (assuming sufficient friction).
    So that would create a change in momentum seen as the COM of the rod moving horizontally and downwards.

    Is that right?
     
  11. Jan 18, 2010 #10

    Doc Al

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    No. In general, that's not true.
    That's true.
    That's also true. But the magnitude of the GRF does not necessarily equal the weight of the rod, nor is the direction of the GRF necessarily parallel to the rod.
     
  12. Jan 18, 2010 #11
    Doc Al,

    OK, seems like I am on the wrong track then.

    So if take the example of a leaning rod, how do we calculate the GRF's magnitude and direction assuming that the ground does not yield and that friction is sufficient to stop the rod slipping?
     
  13. Jan 21, 2010 #12
    OK, as no answer is forthcoming, I guess what I asked has no simple answer.


    To wrap this question up, I will go on to the main issue.
    I am discussing running technique with a number of people and particularly the role of gravity in running. Gravity is neutral over an entire running cycle, but there are points when it could be helping with propulsion, namely when the runner is leaning forwards of the support point.
    If gravity does net work on the runner's COM, I presume that it will create an inverted pendulum type motion which will have elements of angular and linear acceleration. For the runner to avoid falling on their face from the angular motion, they need to counter it. To counter the angular motion, the runner can push back against the ground to produce a somewhat forwards acting GRF. This force will add linear acceleration and also counter the angular motion.
    The big question for me is, if angular motion is corrected instantaneously with horizontal GRF, will gravity still do linear work on the runner i.e. will the net propulsion at that point be the sum of horizontal GRF plus a component of linear acceleration from gravity? Or will the net propulsion only be from horizontal GRF? Or something else :smile:
     
  14. Jan 21, 2010 #13

    Doc Al

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    You can calculate the GRF on the falling rod by first figuring out the acceleration of the rod's center of mass. Treat the rod as being in pure rotation about the contact point and make use of conservation of energy. (Sorry about not answering sooner.)
     
  15. Jan 21, 2010 #14
    Thanks Doc Al, that makes sense and I can see why I cannot assume that the GRF will act along the rod.

    Onto my last point as I posted above. To abstract the runner into an easier physical model, we can consider a leaning rod (not vertical or horizontal) on a trolley.
    If the trolley is accelerated such that the lean of the rod does not increase or decrease, how can the linear acceleration of the system (trolley and rod) be calculated, or if it is easier in words, does the rod's COM experience linear acceleration by both gravity making it fall and the force applied to the trolley, or just by the force applied to the trolley? In essence, can it be said that gravity is doing work on the rod in this scenario?
     
  16. Jan 21, 2010 #15

    Doc Al

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    For a leaning rod on a trolley, it would be easy to calculate the acceleration required to maintain a given angle. (Viewed from an accelerated frame, there would be a non-inertial force on the rod equal to '-ma' acting at its center of mass.)

    As far as gravity doing work, realize that gravity acts vertically while the rod's displacement is horizontal. So no work is done by gravity.
     
  17. Jan 21, 2010 #16
    Thanks Doc Al.

    I appreciate that gravity acts vertically. My confusion and questions arise from the fact that when a rod falls over, it's COM is lowered and also moved horizontally so gravity is instrumental in a horizontal movement - when it acts on a leaning rod it creates a non-vertical GRF at the base through friction.

    If we assume that friction stops the base of an object from slipping, is it not true to say that the instantaneous horizontal force of a leaning rod is m.g.sin(x)?


    I get the impression though that gravity will not be doing any work in the accelerated trolley scenario as the rod would need to fall forwards for gravity to be adding anything, right?
     
  18. Jan 21, 2010 #17

    Doc Al

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    How did you deduce this? (By 'instantaneous horizontal force' you mean the horizontal component of the GRF, right?)

    Right.
     
  19. Jan 21, 2010 #18
    Yes, I mean the horizontal component of GRF, excuse my wandering language :)


    For the scenario of a simple leaning falling rod:

    Gravity acts on the COM and it's force can be expressed as m.g acting vertically downwards.

    There is a GRF acting at some angle (Θ) in the rod COM's quadrant assuming friction prevents the rod from slipping.

    I'm assuming from Newton's 3rd law, we can say that GRF = mg (1)

    The GRF component in the horizontal direction is GRF.sin(Θ) (2)

    substituting (1) and (2) gives

    the GRF in the horizontal direction = m.g.sin(Θ)
     
  20. Jan 21, 2010 #19

    Doc Al

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    Not true. All Newton's 3rd law tells you is that whatever force the ground exerts on the rod (what you call the GRF), the rod will exert an equal and opposite force on the ground. As I've pointed out several times, the GRF does not necessarily equal the weight of the rod.
     
  21. Jan 22, 2010 #20
    Doc Al,

    Yes, sorry you did mention that the magnitude of GRF is not equal to the rod's weight.

    To be precise maybe I should have said horizontal GRF is proportional to m.g.sin(Θ) rather than equal to it?


    I've seen hGRF = m.g.sin(Θ) mentioned before and looking at its derivation, they start with a ball rolling down a hill where Θ is the angle between the vertical and the contact point of the ball on the slope. They argue that the ball is 'rotated downhill' by gravity and that is proportional to the angle of the slope (which is also the deviation of the COM from the point of contact between the ball and the slope). They then make the leap to a horizontal surface and object leaning over to generate a horizontal component and back it up with a force diagram of a regular pendulum where the tangential force is m.g.sin(Θ), I'm not sure if an inverted pendulum works in the same way as a regular pendulum, but that is their assertion.
    So, for a regular pendulum, is the tangential force m.g.sin(Θ)? Can that be applied to an inverted pendulum?
    If it can, I guess we can say that the tangential force of the falling rod is m.g.sin(Θ) and then derive the horizontal force from there.

    Alternatively, we could look at the approach you outlined, "You can calculate the GRF on the falling rod by first figuring out the acceleration of the rod's center of mass. Treat the rod as being in pure rotation about the contact point and make use of conservation of energy", but I would need some help to start on that.

    Thanks for your help, it is really appreciated.
     
    Last edited: Jan 22, 2010
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