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Dynamics Polar Coordinates question

  1. Oct 6, 2012 #1
    Hi everyone. Im a little desperated cause my exam is on monday and still much stuff to do.
    I dont get when Im supposed to use/consider radial and tranversal forces. Most excercises say "it rotates on the horizontal or vertical" I guess this is the info that tells me if there is radial/tranversal forces but i dont get it. Some time they consider the forces and sometimes they dont. I attach an example

    Ps. Please keep it simple cause the profesor isnt that good and basically im learning by my self
    Thanks!!
     

    Attached Files:

  2. jcsd
  3. Oct 6, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Marchese_alex! Welcome to PF! :smile:
    Usually, you have an F = ma equation for each,

    but one of them involves an unknown force …

    in that case, you have to use the other one!​

    in this case, the only radial force is the spring (because there's no friction), and you know what that is, so you can use the radial equation

    you can't use the tangential equation because you don't know what the tangential force is (until you work it out from the radial equation!) :wink:
     
  4. Oct 6, 2012 #3
    What ive gather from internet(because my profesor never teach us this) is like you say... i will have two forces radial (Fr=ma) in the direction of "r" (according to book) and angular force, that will be tangential to the path or just perpendicular to Fr (Fbeta=ma) correct me if wrong. If i see that the object angel changes I say that theres an angular force. For example, if the rod goes from horizontal (flat on floor) to vertical, i can say that a angular force is present. Correct?

    And since theres a change in angle it will be going in a circular motion, and if theres a circular motion I have a centripetal force. Is that is the same thing as radial force or is the samething as radial aceleration(Ar)--> Fr=mAr
     
  5. Oct 6, 2012 #4

    tiny-tim

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    correct, but confusing :confused:

    there are two F = ma equations, one radial and one tangential

    (essentially, there's only one F = ma vector equation, of which you can take the components in any two independent directions, which you usually take to be perpendicular)​

    there may be two forces, or more than two forces, it's the number of equations that matters
    if the angel o:) accelerates then yes, there must be a force in the tangential direction

    (but if the angle changes uniformly, no)
    correct :smile:

    (even if the angle changes uniformly)

    except you really ought to call it centripetal acceleration, and not mention centripetal force unless you're using a rotating frame of reference
    if there is angular motion, then there is always a centripetal acceleration

    this is separate from the radial acceleration, which is d2r/dt2

    you must subtract the centripetal acceleration (because it's always negative :wink:) from the radial acceleration to get the total "a" to put in the radial F = ma :smile:
     
  6. Oct 6, 2012 #5

    tiny-tim

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    to illustrate the last point, try this problem

    a smooth rod rotates uniformly and horizontally with angular velocity ω

    a bead is free to slide (without friction) on the rod … obviously, it will move away from the centre!

    but the only force on it (there's no spring or anything else) is the normal force from the rod, which is tangential

    so why does it have a radial motion, and what is the equation? :smile:
     
  7. Oct 6, 2012 #6
    if the angel o:) accelerates then yes, there must be a force in the tangential direction

    (but if the angle changes uniformly, no)


    if there is angular motion, then there is always a centripetal acceleration

    this is separate from the radial acceleration, which is d2r/dt2

    you must subtract the centripetal acceleration (because it's always negative :wink:) from the radial acceleration to get the total "a" to put in the radial F = ma :smile:
    [/QUOTE]


    When you say substract centripetal acceleration you mean ~~> see part III picture

    Is my diagram correct? Is radial force(Fr) always in that direction?

    http://i174.photobucket.com/albums/w105/marchese_alexander/cd953eb0b2ce1c91d11cce442007042f.jpg
     
  8. Oct 6, 2012 #7

    tiny-tim

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    yes, the radial component of acceleration is always ar = r'' - r(θ')2 :smile:

    r'' is what i call the radial acceleration (i don't know whether other people call it that, or whether they call the whole of ar the radial acceleration)

    and r(θ')2 is the centripetal acceleration
    there isn't necessarily a radial force

    for example, if a weight on a string attached to the ceiling moves in a horizontal circle, the only forces are vertical (the weight) and along the string (the tension) …

    but the radial direction is horizontal :wink:
     
  9. Oct 6, 2012 #8
    What exactly make this excercises different that they say Fr exist?
     

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  10. Oct 6, 2012 #9

    tiny-tim

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    no, they don't call it the radial force, they call it the radial component of the force :wink:

    the (net) force always has a radial component,

    but there isn't always an individual force along the radial direction :smile:
     
  11. Oct 6, 2012 #10
    and since there angular motion I will have centripetal acceleration. Right?
     
  12. Oct 6, 2012 #11
    Ok i get the centri acceleration. Now i just dont get
    1.the radial and tranversal component part of a force... Is it radial for "x" component and tranversal for y

    2.what exactly is F and Fr representing in the free body diagram
     
  13. Oct 6, 2012 #12

    tiny-tim

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    Right! :smile:
    In the diagram, "F" seems to be a mistake for "Fθ".

    The problem asks for the "radial and transverse components of the force exerted on A", which are Fr and Fθ.

    Radial is the r component, and transverse is the θ component.

    (tangential isn't often used, it's the component of the force along the line of motion, which won't be transverse unless r is constant)
     
  14. Oct 7, 2012 #13
    Got it!
     
  15. Oct 7, 2012 #14
    How about this excercise.

    The only part I dont get is why I cant say that the first derivative of the angle is 0 and they put that is w(angular velocity). I know that the first derivative is angular velocity and the second derivative is angular acceleration. Also I know if I say is 0 then I would not be able to find what the excercise is asking. Im just trying to understand the "why".


    Heres my data http://i174.photobucket.com/albums/w105/marchese_alexander/5ce37c38be157e984f3f113fe1a3a65f.jpg
     

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  16. Oct 8, 2012 #15

    tiny-tim

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    θ' is the angular velocity, ω (same as x' is the ordinary velocity, v)

    i don't understand why you think ω = 0 :confused:

    if it's moving, then its speed isn't 0, and its angular speed isn't 0
     
  17. Oct 8, 2012 #16
    Ok! Thanks
     
  18. Oct 8, 2012 #17
  19. Oct 8, 2012 #18
    Bump... Need to know please
     
  20. Oct 8, 2012 #19

    tiny-tim

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    14.94 is in the x-y plane (ie horizontal) :wink:
     
  21. Oct 8, 2012 #20
    What do you mean? Still dumb lol

    The #95 #97 (can see in pictures) they dont say anything about x,y plane and they put it the same way
     
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