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Angular momentum newbie question

  1. May 6, 2007 #1
    i have a question which you will surely be reluntant to answer, since it is basicly what any book that explains angular momentum tries to examine, through several equations such as torque, and i could just check one of those books and i'd have my answer but, believe me, i have read them.
    so, i wanted to know about the conservation of angular momentum. the law says that m(rxv) is constant unless axr is not zero. this means that if a particle is moving with radius R and changes to a smaller radius r the velocity will increase by the same factor thus requiring energy input (a normal force). i wanted to know the explanation to this. i know torque is the variation of those quantities and that without torque they should remain the same but i just don't see the relation between radius and velocity. why does velocity increase? i won't probably get much of an answer but if you have read so far, thanks :P.
     
    Last edited: May 6, 2007
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  3. May 6, 2007 #2
    I think if the radius r is shortened to half its value, without any torque being applied, the linear (tangential) velocity of the particle does not change, although its angular velocity w increases. If the original velocity of the particles was rw, its new velocity is (1/2)r(2w), which is the same. If this were not so, Newton's second law would be violated, which is not possible.
     
  4. May 6, 2007 #3
    Pick up a circular object. Spin it around. Notice that the outside moves faster than a point near the inside. Therefore, for a spinning object, velocity is greater if you are further away from the axis that it's spinning around.

    To put it another way, picture this: the distance all the way around the outer rim of a CD is maybe 15 inches. The distance around the inner rim of the shiny part is maybe 2 inches. Imagine the CD spins all the way around, one time, in one second. If you were sitting on the outer rim of the CD, you would travel 15 inches in a circle. That's 15 inches per second. If you were sitting on the inner rim, you would only travel in a 2 inch circle. That's 2 inches per second.

    EDIT: looking back, you were asking a much more sophisticated question than what I thought. Sorry, guess this post was useless.
     
    Last edited: May 6, 2007
  5. May 6, 2007 #4
    that is what i used to think, and it would explain why angular velocity increases when the radius is smaller but it seems it is proportional to r^2 (mrv) so velocity increases too (angular velocity increases with 1/mr^2, being r^2 the moment of inertia). this is why it needs energy. if you have Serway's book (physics for scientists and engeneers) check out example 11.8 page 347).
     
  6. May 6, 2007 #5
    Second try: Newton's second law is not violated. You have to apply a force to pull the object in, and that force accelerates the object. That's why it has a greater velocity (not just angular velocity) after you pull it in.

    But I think Daniel already knows that. The question is why do you have to increase the tangential velocity when you pull it in, that is, what's up with this whole "angular momentum" conservation thing? Let's see if I can help now.

    When the object is spinning around a circle, it constantly "wants" to keep going in a straight line, and the force that holds it on a circular path keeps pulling it in, always constantly bending its trajectory into a circle. The tighter the circle, the more its preferred straight path has to be bent, so the harder we have to pull to keep it going in a circle.

    For this reason, if you want to make a tighter circle you have to pull harder. When you pull it in, you are doing work (applying a force over a distance), so you must be speeding the object up, increasing its linear momentum as well as angular momentum.

    Helpful?
     
  7. May 6, 2007 #6
    Let me add a little more. The force you apply when you pull the object in adds a radial component to its velocity (obviously, since it moves inward). The object's total linear velocity is a combination of its radial and tangential velocity. It ends up with a linear velocity that is greater than before, and pointed not quite tangential, but also a little bit inward.

    Now, as you pull it in some more, you are accelerating it linearly, because its velocity has an inward-pointing component which is getting larger (in other words, the circle it's pointing along is getting tighter). So its linear velocity (which is not all tangential) keeps going up. The way the inward velocity is converted to tangential velocity is obvious: since the whole situation is spinning, what is "inward" one moment is closer to "tangential" the next.

    That last sentence is maybe the best.
     
  8. May 6, 2007 #7
    very clever. that sure is giving me a lot to think. it will give a little trouble when i read the book again and try to figure out how this fits but it sure explains it. it explains why a normal force would only increase or decrease the radius according to the law of conservation of angular momentum. tangencial forces change only the tangencial velocity so this wouldn't apply.
    thank you, you were of a great help. it's funny to see how a small set of simple equations hides such complexity. i guess that is why in some sites this concept is introduced in a somewhat empirical way. i have much to think about but at least now i can proceed with my studies. thanks.
     
  9. May 7, 2007 #8
  10. May 7, 2007 #9
    According to Newton's second law F(net) = dp/dt, only a net force in the direction of motion of an object, can change the magnitude of its velocity. A force perpendicular to the direction of motion of an object, does not change the magnitude of its velocity. If this were not true, a rotating object at the end of a string would increase its speed constantly, because it is subject to a constant centripetal force.
     
  11. May 7, 2007 #10

    russ_watters

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    I tried to merge/split a couple of threads here - I hope I didn't mangle them too much.
     
  12. May 7, 2007 #11
    interesting stuff. i just figured out the same principle applies to a negative work, i.e. when the particle moves outside (by the same principle a perpendicular component is gained eliminated by centripetal force).

    now i'm thinking about the 'top'. the top goes in circules in precessional motion by the action of gravity. the common explanation is that gravity applies a delta-L addiction of angular momentum. i figured out that by rotating the top the motion of the particles seen in the same position looks a bit more like an elipse, and that means that the top gains motion around the origin (like a torque had been acting - and it is). i probably didn't make myself clear, but i'm just saying one could see how it happens without labeling it "conservation of angular momentum". not relevant anyway.
    i have 2 questions about this, if anyone would be so kind to answer or at least share a thought.one is: angular momentum changes direction - is that all that happens? i mean, does the top gain L, or it's L is just shifted? it would be kind of like gravity that only changes the satellite's direction. could it be said that gravity does no work?
    i would also like to know if there is a simple way to discribe this proccess of shifting direction, like the one Xezlec posted?
     
  13. May 7, 2007 #12
    I was trying (however poorly) to get across the fact that there is some component of the radial force that is in the direction of motion of the object, because obviously the object is moving inward somewhat. If it weren't moving inward a little bit at some point, its radius would never decrease. The whole point that we are discussing is the case where we pull an object in to a tighter circle, decreasing its radius. I'm just struggling to put this clearly into words.
     
  14. May 8, 2007 #13
    I suggest that you read the post #8 and have a look to the linked images. A force F has been exerted on the object during a time [tex]\Delta t[/tex]. In the drawing the impulsion is exaggerated, but with small impulsions Fdt, the change in radius and speed is continuous.

    If you prefer a more mathematical demonstration, you can take an object which has a circular trajectory of radius R and a speed V. The centripetal force to maintain this trajectory is F=mV^2/R. If you now exert a bigger force, the instantaneous radius of the trajectory will be smaller. The object will begin to turn around a point situated between the previous center of rotation and the actual position of the object. This new trajectory approaches the object to the "old" rotation center and changes the direction of the moment, adding a component directed to the center of the trajectory.
     
    Last edited: May 8, 2007
  15. May 8, 2007 #14
    Your example of a point object rotating about the end of a string going through a tube is very good. But I have a question. You know the radial acceleration of the object is a = v^2/r; therefore, the force holding the other end of the string must be ma = mv^2/r = mw^2 r. If the force pulling the string is increased to twice its original value, the length of r diminished to half its value, and we have m2a = m x^2 r/2, where x is the new value of the angular velocity. My question, what is the value of x? Thank you.
     
  16. May 8, 2007 #15

    Hootenanny

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    Assuming your x is the new value of w, we would have x = 2w.
     
  17. May 8, 2007 #16

    Doc Al

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    If the distance r is decreased to r/2, the speed doubles (and the angular speed quadruples). The required force increases accordingly--it doesn't just double.
     
  18. May 8, 2007 #17

    Hootenanny

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    I took Pete's question to mean that "if we halve r and double the centripetal force, what would be the new angular velocity?"

    Edit: In any case, this may be an attempt by Pete to discuss his new 'theory' we've all been waiting for...
     
    Last edited: May 8, 2007
  19. May 8, 2007 #18

    Doc Al

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    Cool. (I was assuming the question had to do with conservation of angular momentum: where the object starts with a given speed and distance and you pull it in.)
     
  20. May 8, 2007 #19
    No, that is the catch!

    When you increase the force, the radius will diminish but, as the radius diminish, the linear speed increases due to the conservation of angular momentum. If you want to compute the final radius and speed to have the double of the force you must write the equations:
    [tex]{V_2^2\over R_2}=2{V_1^2\over R_1}[/tex] for the force.
    [tex]V_2R_2=V_1R_1[/tex] for the conservation of angular momentum.
    The solution is
    [tex]V_2=\root 3 \of 2 V_1[/tex]
    and
    [tex]R_2={1\over\root 3 \of 2}R_1[/tex]
    The new angular velocity is:
    [tex]\omega_2=\root 3 \of 4 \,\,\,\omega_1[/tex]
     
    Last edited: May 8, 2007
  21. May 8, 2007 #20
    Thank you. The original tangential velocity of the object was v1 = rw, and the new tangential velocity is then v2 = (r/2)2w, but isn't v1 = v2?
     
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