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Circular motion - Force perpendicular to velocity does not change magnitude.

  1. Sep 11, 2010 #1
    Hi, my first post here.

    Right.
    Spent the past hour searching all over for a clear explanation. In Circular Motion, or in general, the force perpendicular to the velocity vector changes only the direction and not the magnitude. Why is that so?
    My reasoning against goes like this - please explain which of my statements is wrong.

    Assuming velocity to be along x-axis, force acting along y-axis :
    1/ Force produces acceleration.
    2/ Acceleration will create a velocity along y-axis (initially v in y =0)
    3/ Resultant of v along x(constant) and v along y(accelerated) will have different direction AND magnitude.

    I've broken it into 3 statements, please tell me which part is wrong.
    Much appreciated, thanks!
     
  2. jcsd
  3. Sep 11, 2010 #2

    rock.freak667

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    If the object is moving at constant speed, the change in velocity between any two points (the vector directions) points towards the center of rotation.

    If there is varying speed, then there will be two accelerations, one towards the center of rotation and one in the direction of the velocity vector.
     
  4. Sep 11, 2010 #3
    Can you be a little more clear? I didn't understand how to use that in this situation.
     
  5. Sep 11, 2010 #4

    Cleonis

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    Gold Member

    attachment.php?attachmentid=28159&stc=1&d=1284195223.png

    Luckily I had a diagram lying around that fits the bill.
    In the diagram uniform circular motion is represented as motion along a 12-sided polygon. At points A, B, C etc the circumnavigating object receives an impulse towards the center of attraction.

    The object travels from A to B and without a centripetal impulse it would proceed to point c in an equal interval of time. At point B the object receives precisely the impulse to make it travel the distance cC in an equal interval of time. The deflected velocity vector BC is the vector sum of Bc and cC.

    The shorter the time intervals, the closer the approximation to the actual continuous motion.

    I have a hunch that you are thinking in terms of vector composition of a tangential velocity vector with a radial velocity vector. But the polygon does not consist of tangential velocity vectors.
     

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    Last edited: Sep 11, 2010
  6. Sep 11, 2010 #5
    Erm, so the centripetal force IS influencing the velocity(in both M&D) isn't it? Isn't the velocity vector Cc the result of the force? And hence it will clearly influence the direction AND magnitude of vector Bc to BC. Or are BC and Bc the same in magnitude? If so - what geometric condition makes the resultant of 2 vectors equal to one of the vectors as in this case?
     
  7. Sep 11, 2010 #6

    Cleonis

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    Yes, the triangle BcC is an [STRIKE]equilateral[/STRIKE] isosceles triangle. In other words, Bc and BC are equal in length.

    Take a second circle, with identical diameter, that touches the first circle exactly at point B. Bc is equal in length to AB, therefore point c is a point on the second circle. It follows from symmetry that the triangle BcC is an [STRIKE]equilateral[/STRIKE] isosceles triangle.
     
    Last edited: Sep 11, 2010
  8. Sep 11, 2010 #7
    So Cc is also equal to Bc and BC right?

    Okay, let me get this straight. The centripetal force is actually creating a velocity vector (Cc) which is equal in magnitude to the original velocity vector Bc, and is at an angle of 60 degrees with Bc - which leads to the resultant also being equal in magnitude and at an angle of 60 degrees.
    Is that right?
     
  9. Sep 11, 2010 #8

    Cleonis

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    I made a language error in my previous message. I should have written 'isosceles triangle', but by mistake I wrote 'equilateral triangle'.
     
    Last edited: Sep 11, 2010
  10. Sep 11, 2010 #9
    Ah okay. So basically the magnitude of velocity is not changing because the direction and magnitude of the velocity created by the centripetal force is such that an isosceles triangle is formed. And this is true for every circular motion. Is that right?
     
  11. Sep 12, 2010 #10
    In circular motion the force applied is such that it is always perpendicular to the direction of motion. Its direction changes as velocity changes, so that it is always perpendicular to the velocity. At any moment the force only acts so as to cause a change in direction. Think about it like this. Take a circle of radius v(the tangential velocity). Consider this image in the folder: View attachment 28169 . Sorry, the writing isn't neat. It shows the force produces circular velocity change, but 0 velocity change in any given direction. Hope you understood. Here is a better one. View attachment circular motion.doc .
     
    Last edited: Sep 12, 2010
  12. Sep 12, 2010 #11
    I think you meant uniform circular motion :)

    Edit: The below is for the OP.
    The treatment of this topic (circular motion - not just uniform) using vectors in Kleppner's "An Introduction to Mechanics" is excellent. It's the first chapter. You should really read that chapter. It certainly helped clear a lot of things for me about circular motion.
     
    Last edited: Sep 12, 2010
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