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Thread: simple centrifugal force problem plz help

  1. #1 simple centrifugal force problem plz help 
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    i have a 2m long tube spinning around its center at w=1

    theres a 1 kg projectile at 1 m distance from the axe inside the tube that is released at a certain moment

    what speed does the projectile have when it leaves the tube?

    plz show me how you do the calculations


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  3. #2 Re: simple centrifugal force problem plz help 
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    Quote Originally Posted by luxtpm
    i have a 2m long tube spinning around its center at w=1

    theres a 1 kg projectile at 1 m distance from the axe inside the tube that is released at a certain moment

    what speed does the projectile have when it leaves the tube?

    plz show me how you do the calculations
    I do this sort of trivial calculation in my head. For the harder ones I frown a bit and rest my head on my chin, but these don't rise to that level..


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  4. #3  
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    and the solution is?
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    speed = iirc
    Wise men speak because they have something to say; Fools, because they have to say something.
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    iirc?
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  7. #6  
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    nobody has ever told me directly, but I assume that it's "if I remember correctly".

    Back ton the original post.

    Here's a hint.

    W tells you how many times to end point of the pole rotates (ie covers a set distance) every second...

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    Quote Originally Posted by Arcane_Mathematician
    speed = iirc
    Think you mgiht have made a wee mistake there... :P

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    The OPís question is not trivial to me. I donít know the answer off hand and Iím not afraid to say so.

    This is what I do know off hand using cylindrical coordinates:

    When the mass leaves the tube its velocity must have a tangential and a radial component. (Note: If it has no radial velocity, it can't get out of the tube.)

    The tangential component is easy. Itís the angular velocity times the radius of the tube.

    The radial component is not easy, to me at least.
    Since the mass starts out in the middle of the tube with no radial velocity, the radial velocity must be the integral of the radial acceleration over the period of time the mass is in the tube. (The magnitude of the mass seems to be irrelevant. I assume whatever is rotating the tube is powerful enough to keep the angular velocity constant as the mass slides outward.) The radial acceleration would be the opposite of the centripetal acceleration, the angular velocity squared times the radial distance to the mass. But the radial acceleration changes with time because the radial distance changes with time.

    Am I all mixed up? Is there a simple way to calculate this? Please share.
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  10. #9  
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    The questionn just asks for speed, not velocity.

    I thought the answer was approximately just over 6m/s?

    Been a while since I've done this sort of stuff though so I might ahve done it all wrong.

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    Quote Originally Posted by mikelizzi
    The OPís question is not trivial to me. I donít know the answer off hand and Iím not afraid to say so.

    This is what I do know off hand using cylindrical coordinates:

    When the mass leaves the tube its velocity must have a tangential and a radial component. (Note: If it has no radial velocity, it can't get out of the tube.)

    The tangential component is easy. Itís the angular velocity times the radius of the tube.

    The radial component is not easy, to me at least.
    Since the mass starts out in the middle of the tube with no radial velocity, the radial velocity must be the integral of the radial acceleration over the period of time the mass is in the tube. (The magnitude of the mass seems to be irrelevant. I assume whatever is rotating the tube is powerful enough to keep the angular velocity constant as the mass slides outward.) The radial acceleration would be the opposite of the centripetal acceleration, the angular velocity squared times the radial distance to the mass. But the radial acceleration changes with time because the radial distance changes with time.

    Am I all mixed up? Is there a simple way to calculate this? Please share.
    I don't think the mass is located in the cntre of the tube. The tube is 2m long and rotates around its centre. The mass is located 1m from the centre of the tube, ie at the end of the tube.

    At least thats what I think is being described.

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    Quote Originally Posted by mikelizzi
    Since the mass starts out in the middle of the tube with no radial velocity,
    Does it? The tube is 2 meters in length and rotates about its center. The projectile is 1 meter from the "axe" whatever that is. If it's the axis of rotation, that puts the projectile at the perimeter of the circle to start with, so all it would have is a tangential component.
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    sorry i missed it:

    the tube is 4 m long and the projectile starts at 1 m radius so leaves at 2 m radius

    the projectile is porpelled by centrifugal force
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    I had supposed that the projectile was simply attached to the end and then released since he noted that the pole was 2 meters long, spinning around its center and noted that the projectile was 1 meter from the axis of rotation. So, doesn't this mean that the projectile is attached to the end? If it were 1 meter from the end, the projectile would need an initial velocity. If it were inside of the tube, near the end, we would also need to know an initial radial velocity. If it were traveling down the tube, one does have to account for several things which aren't exactly trivial, one of which being the Coriolis Effect.

    However, I think this is just about leaving the end tangentially.
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by mikelizzi
    Since the mass starts out in the middle of the tube with no radial velocity,
    Does it? The tube is 2 meters in length and rotates about its center. The projectile is 1 meter from the "axe" whatever that is. If it's the axis of rotation, that puts the projectile at the perimeter of the circle to start with, so all it would have is a tangential component.

    Hmm. The initial problem states the tube is 2 meters long rotating about its center and the projectile starts 1 meter from the center. That would put the projectile at the end of the tube as you say.

    I assumed the tube was rotating about one end. Thatís why I thought the projectile started in the middle.

    If the projectile starts at the end and is just let go, that would be a trivial problem indeed. But then there would be no point in the launching device being a tube. Maybe I was trying to find a serious problem there when none existed.
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    Let F(x) be the force acting on the projectile at position x. Then F(x)=mxw^2. The work on the projectile is given by the integral of the force with respect to x. So if the projectile's final kinetic energy is E, then E=[(1/2)mx^2w^2]2_1 (hope you understand my notation)=(3/4)mw^2=mv^2/2, which yields v=sqrt((3/2)w^2).

    Anybody, feel free to check for mistakes.

    I agree, the question isn't trivial.
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  17. #16  
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    Quote Originally Posted by sox
    Quote Originally Posted by Arcane_Mathematician
    speed = iirc
    Think you mgiht have made a wee mistake there... :P
    Agree :-D The dimensions don't work out :-D
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    Fc=m*w*w*r

    Fc=1*1*1r

    F=m*a F=1a


    a=r

    dv/dt=r


    ...
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  19. #18 Re: simple centrifugal force problem plz help 
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by luxtpm
    i have a 2m long tube spinning around its center at w=1

    theres a 1 kg projectile at 1 m distance from the axe inside the tube that is released at a certain moment

    what speed does the projectile have when it leaves the tube?

    plz show me how you do the calculations
    I do this sort of trivial calculation in my head. For the harder ones I frown a bit and rest my head on my chin, but these don't rise to that level..

    RUBISH°°°°

    in the begining i thought you could do this calculation with your head,after all there are people who can calculate 2^22 with their minds, now i doubt it

    the final velocity has a normal and a tangential component with which the solution is even difficult for a rare human calculator

    so you thought wrongly:

    i have top physics knowledge like v=wr and i can do top calculation like 1*2=2 and i wont help this guy with this top mind°°°°

    haha you not only think yourself a genius but got to think all the rest is stupid, you just proved it

    man and if a person knowing so much physics as you cant solve a problem that is not in no physics text book but is extreamly easy which i solved with just a little help studying basic phyiscs you just proved youre a parrot with predilection for the words rubish and go to pseudoscience

    i hope you have learnt some and know now how to solve a problem which you didnt have memorized

    do another try with your privileged mind:

    w=1 m=1 r=from 1 to 2
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    i just solved the problem and the solution has infinite decimals

    i hope you can calculate up to 3 decimals of the final solution with your privileged mind to prove me wrong

    edit:

    in the case the mods thinks im being offensive....

    well i feel offended by a guy who says he knows the solution but wont tell and laughs at me
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    Quote Originally Posted by thyristor
    Let F(x) be the force acting on the projectile at position x. Then F(x)=mxw^2. The work on the projectile is given by the integral of the force with respect to x. So if the projectile's final kinetic energy is E, then E=[(1/2)mx^2w^2]2_1 (hope you understand my notation)=(3/4)mw^2=mv^2/2, which yields v=sqrt((3/2)w^2).

    Anybody, feel free to check for mistakes.

    I agree, the question isn't trivial.
    Excellent!

    If I understand correctly, you start the projectile in the rotating tube, at rest, at 1 meter radius and have the projectile exit the tube at 2 meters radius.

    Youíve calculated the radial component of the exit velocity. So I just add the tangential component to it to get the net velocity?
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    Quote Originally Posted by luxtpm
    Fc=m*w*w*r

    Fc=1*1*1r

    F=m*a F=1a


    a=r

    dv/dt=r


    ...
    I fail to see how this is a solution. To find v, you must integrate r dt. But how do you know r as a functino of t?
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  23. #22  
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    Quote Originally Posted by mikelizzi
    Quote Originally Posted by thyristor
    Let F(x) be the force acting on the projectile at position x. Then F(x)=mxw^2. The work on the projectile is given by the integral of the force with respect to x. So if the projectile's final kinetic energy is E, then E=[(1/2)mx^2w^2]2_1 (hope you understand my notation)=(3/4)mw^2=mv^2/2, which yields v=sqrt((3/2)w^2).

    Anybody, feel free to check for mistakes.

    I agree, the question isn't trivial.
    Excellent!

    If I understand correctly, you start the projectile in the rotating tube, at rest, at 1 meter radius and have the projectile exit the tube at 2 meters radius.

    Youíve calculated the radial component of the exit velocity. So I just add the tangential component to it to get the net velocity?
    If by "add" you mean vector addition, then yes
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    Quote Originally Posted by thyristor
    Quote Originally Posted by mikelizzi
    Quote Originally Posted by thyristor
    Let F(x) be the force acting on the projectile at position x. Then F(x)=mxw^2. The work on the projectile is given by the integral of the force with respect to x. So if the projectile's final kinetic energy is E, then E=[(1/2)mx^2w^2]2_1 (hope you understand my notation)=(3/4)mw^2=mv^2/2, which yields v=sqrt((3/2)w^2).

    Anybody, feel free to check for mistakes.

    I agree, the question isn't trivial.
    Excellent!

    If I understand correctly, you start the projectile in the rotating tube, at rest, at 1 meter radius and have the projectile exit the tube at 2 meters radius.

    Youíve calculated the radial component of the exit velocity. So I just add the tangential component to it to get the net velocity?
    If by "add" you mean vector addition, then yes

    Many Thanks !
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    acceleration = r

    we can write that: dv/dt= r

    to solve this, we use a trick: dv/dt = dv/dr dr/dt
    = v dv/dr (because dr/dt = v)

    = d(v2/2)/dr Ö

    so now the original equation can be rewritten = d(v2/2)/dr = r Ö
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    Something that has not been mentioned is that the problem, as stated, is insolvable. There is information omitted.

    Either the OP forgot to mention that the tube was being spun by a power source that is able to maintain a constant angular velocity for the tube throughout, or he needed to provide the mass plus inner and outer radii of the tube.

    Because, if the tube is not powered, this becomes a conservation of angular momentum problem, and you need the moment of inertia of the tube to determine the final angular velocity of the tube/projectile.
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    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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    oh the tube is powered at a constant w=1
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    Quote Originally Posted by sox
    Quote Originally Posted by Arcane_Mathematician
    speed = iirc
    Think you mgiht have made a wee mistake there... :P
    I could swear the standard for rotational speed was radians per second.
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    Quote Originally Posted by thyristor
    Quote Originally Posted by sox
    Quote Originally Posted by Arcane_Mathematician
    speed = iirc
    Think you mgiht have made a wee mistake there... :P
    Agree :-D The dimensions don't work out :-D
    and, what dimesions aren't working? speed is measured in meters per second, and a radius, which in meters, multiplied by angular velocity, which is radians per second, gives meters per second.

    Wheres the error?
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  30. #29  
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    The original question was just asking for good old fashioned speed.

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  31. #30  
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    The Lagrangian for this problem can be taken to be
    where is the radial distance of the mass from the center of rotation of the tube.

    Lagrange's equation of motion is then subject to


    This yields and

    Let Then the desired final speed of the mass is



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  32. #31  
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    Quote Originally Posted by Arcane_Mathematician
    Quote Originally Posted by thyristor
    Quote Originally Posted by sox
    Quote Originally Posted by Arcane_Mathematician
    speed = iirc
    Think you mgiht have made a wee mistake there... :P
    Agree :-D The dimensions don't work out :-D
    and, what dimesions aren't working? speed is measured in meters per second, and a radius, which in meters, multiplied by angular velocity, which is radians per second, gives meters per second.

    Wheres the error?
    I'm sorry. I was confused by the "m/s" after "rw". I thought you meant mass by "m" and what "s" stood for I didn't really know. Usually (at least this is what I've been taught), you shouldn't put units in italic font
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    Quote Originally Posted by DrRocket
    The Lagrangian for this problem can be taken to be
    where is the radial distance of the mass from the center of rotation of the tube.

    Lagrange's equation of motion is then subject to


    This yields and

    Let Then the desired final speed of the mass is



    I see our answers differ. Is there anything wrong to my solution?
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  34. #33  
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    Quote Originally Posted by thyristor
    Let F(x) be the force acting on the projectile at position x. Then F(x)=mxw^2. .
    your mistake is here. I'm not sure where this came from , but in any case the force on the mass is normal to the radius so your work calculation does not work.

    This problem is a poster child example of the simplification offered by the Lagrangian formulation for such problems of constrained motion -- a fairly trivial calculation using the variational approach, vs a difficult mess if one tries to find all of the forces involved.

    With a little concentration you can do the Lagrangian calculation in your head, or failing that a couple of lines on a piece of paper.
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    Dr.rocket i think he was talking about the centripetal force which is represented by the formula:



    Dr.Rocket. how does Lagrangian mechanics actually differ from classical mechanics?? i haven't come across Lagrangian mechanics quite yet.
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    Quote Originally Posted by Heinsbergrelatz
    Dr.rocket i think he was talking about the centripetal force which is represented by the formula:



    Dr.Rocket. how does Lagrangian mechanics actually differ from classical mechanics?? i haven't come across Lagrangian mechanics quite yet.
    Lagrangian mechanics is classical mechanics.

    The Lagrangian , Hamiltonian and ordinary Newtonian formulations are equivalent, but sometimes one formulation is easier to work with.

    Centripetal force is a convenient fiction for circular motion, but there is really no such thing, and in any case what we have in this problem is not circular motion. You have a ball in a tube, and the force exerted by the tube on the ball is perpendicular to the axis of the tube.
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    Lagrangian mechanics is classical mechanics.

    The Lagrangian , Hamiltonian and ordinary Newtonian formulations are equivalent, but sometimes one formulation is easier to work with.

    Centripetal force is a convenient fiction for circular motion, but there is really no such thing, and in any case what we have in this problem is not circular motion. You have a ball in a tube, and the force exerted by the tube on the ball is perpendicular to the axis of the tube.
    So they would be equal in terms of precision in calculations??

    Wow, what im learning now in Physics, we are taught that centripetal force exists, seems like the physics I learn(high school physics) seems to contradict alot of physics we are going to learn in the future.
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    Quote Originally Posted by Heinsbergrelatz
    ]

    So they would be equal in terms of precision in calculations??
    Precision in calculation is a question usually posed in regard to numerical methods/approximations.

    The various formulations of classical mechanics describe exactly the same physics.

    If you have circular motion at constant speed, then you have a centrepetal acceleration, and that acceleration is real. To produce that acceleration requires a corresponding force directedv radially inward. But it is not some new kind of force, and in this case we do not have circular motion and there is no radial force.
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Let F(x) be the force acting on the projectile at position x. Then F(x)=mxw^2. .
    your mistake is here. I'm not sure where this came from , but in any case the force on the mass is normal to the radius so your work calculation does not work.
    I am prepared to be believe you, but I still don't understand why my approach doesn't work; that is, why can you not calculate the work on the projectile in a rotating reference frame exerted by the centrifugal force?
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    Quote Originally Posted by DrRocket
    The Lagrangian for this problem can be taken to be
    where is the radial distance of the mass from the center of rotation of the tube.

    Lagrange's equation of motion is then subject to


    This yields and

    Let Then the desired final speed of the mass is




    DrRocket,

    I have no exposure to the Lagrangian Method so this question may be stupid.
    The solution you calculated appears to give the final velocity as a function of the final time. But the final time is an unknown in this problem. So is there more to be done to determine the final velocity?
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    Quote Originally Posted by mikelizzi
    Quote Originally Posted by DrRocket
    The Lagrangian for this problem can be taken to be
    where is the radial distance of the mass from the center of rotation of the tube.

    Lagrange's equation of motion is then subject to


    This yields and

    Let Then the desired final speed of the mass is




    DrRocket,

    I have no exposure to the Lagrangian Method so this question may be stupid.
    The solution you calculated appears to give the final velocity as a function of the final time. But the final time is an unknown in this problem. So is there more to be done to determine the final velocity?
    The final time is not undetermined.
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    Quote Originally Posted by thyristor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Let F(x) be the force acting on the projectile at position x. Then F(x)=mxw^2. .
    your mistake is here. I'm not sure where this came from , but in any case the force on the mass is normal to the radius so your work calculation does not work.
    I am prepared to be believe you, but I still don't understand why my approach doesn't work; that is, why can you not calculate the work on the projectile in a rotating reference frame exerted by the centrifugal force?
    You have to calculate a line integral and you know neither the line (the trajectory of the particle) nor the force in an inertial reference frame.

    There is probably some way to do this, ordinary Newtonian mechanics and the Lagrangian perspective are ultimately equivalent, but it will be a complicated mess.

    Also, the "centripetal" force is not a force in the usual sense -- ask yourself just what is applying this force. Rather it is the result of trying to apply Newtonian mechanics in a non-inertial reference frame, here a rotating frame. By definition, a non-inertial frame is one in which Newton's equations do not apply. To compensate you get pseudo-forces, like the "Coriolis force" and "centripetal force" and complicated expressions describing motion.

    Once you go down that path you have to take the whole dose of medicine that goes with it -- ugly. The way this is done is to actually solve things in an inertial frame and then work out the translation back to the rotating frame -- the pseudo forces show up in the translation.

    You have to play by ALL of the rules. That means using an inertial frame if you want to use straight Newtonian theory. The onus is on you to show that you are playing by the rules, not on someone else to show precisely where the error manifests itself when you don't.
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    DrRocket,

    Thank you for your response to my last post. I see now that your solution for was right in front of my face. Moving on, I have another question about your solution. It seems to me that the final velocity must be a function of the angular velocity of the tube. The faster the tube rotates the faster the exit velocity of the projectile. But I donít see an angular velocity term in either the setup or solution. I know the OP suggested using . Is your derivation for the OPís suggested case or have I missed something else?
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    i think to remember the answer was sqaure root of 7

    and its obtained quite simply by the way i pointed
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    Quote Originally Posted by mikelizzi
    DrRocket,

    Thank you for your response to my last post. I see now that your solution for was right in front of my face. Moving on, I have another question about your solution. It seems to me that the final velocity must be a function of the angular velocity of the tube. The faster the tube rotates the faster the exit velocity of the projectile. But I donít see an angular velocity term in either the setup or solution. I know the OP suggested using . Is your derivation for the OPís suggested case or have I missed something else?
    The effect of is in the Lagrangian. The Lagrangian that I quoted is specifically for which keeps things simple arithmetically.

    More generally the Lagrangian would be
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    If a euclidean system rotates with something rotating the anglular speed reverses direction. Clockwise becomes counterclockwise or vice versa. That also means two opposing directions for accelleration, plus and minus.
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    DrRocket, where did you get your E.O.M from?

    Did you put the Lagrangian into the Euler-Lagrange Equations?

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    Quote Originally Posted by sox
    DrRocket, where did you get your E.O.M from?

    Did you put the Lagrangian into the Euler-Lagrange Equations?
    E.O.M.? Equations of motion I presume.

    One takes the Lagrangian and using calculus of variations finds critical points -- that results in the Eulere-Lagrange equations which in the case of Lagrangian mechanics are the Lagrangian equations of motion.

    You should be able to find the development in any good book on classical mechanics -- the books by Marion or Goldstein are good sources.
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