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Deduction of centripetal force

  1. Feb 9, 2010 #1
    I have questions about the deduction of law of centripetal force

    http://img42.imageshack.us/img42/5841/centripetalforce.gif [Broken]
    Why do we assume that the change in direction of velocity of the object is V(t+dt)
    And similarly the change in direction of radius?


    Also what is the triangle of velocity ? and how to draw it?
    http://img203.imageshack.us/img203/4669/triangleofvelocity2.gif [Broken]
    Do we use only the magnitude of these directions to express the sides triangle of velocity bec. The triangle sides can't be negative while the directions could be +ve or negative?

    Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 10, 2010 #2

    Doc Al

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    The velocity (for uniform circular motion) is always tangent to the circle. V(t+dt) is the velocity at time t+dt, not the change in velocity. The change in velocity is V(t+dt) - V(t).

    You already have a drawing of the triangle of velocity. It represents the vector addition: V(t+dt) = V(t) + ΔV. The "trick" is realizing that the angle between the two velocity vectors is θ.

    When drawing a vector, the length of the arrow represents the magnitude, which is always positive. The direction of the vector is represented by the direction of the arrow.

    Read this derivation: http://hyperphysics.phy-astr.gsu.edu/HBASE/cf.html#cf2"
     
    Last edited by a moderator: Apr 24, 2017
  4. Feb 10, 2010 #3
    Thanks this really helped me but actually i can't understand this point

    You already have a drawing of the triangle of velocity. It represents the vector addition: V(t+dt) = V(t) + ΔV. The "trick" is realizing that the angle between the two velocity vectors is θ.

    ----
    I have another question
    can we prove that the triangle of velocity in the second circle is similar to the triangular cab (ABC) in the first circle?

    Thanks again
     
  5. Feb 11, 2010 #4

    Doc Al

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    Two points:
    (1) The angle between the two velocity vectors is θ. To understand that, realize that the velocity vector is always perpendicular to the radius. Thus when the radius moves by an angle θ, so must the velocity vector.
    (2) The triangles are similar. In the first circle you have an isosceles triangle with angle θ formed by the radii; In the second circle you have an isosceles triangle with angle θ formed by the velocity vectors. Two isosceles triangles with the same angle are similar.

    Another point: For a small change in angle dθ, the length AB (in the first diagram) will equal rdθ.
     
  6. Feb 11, 2010 #5
    mmm actually i can't imagine can u draw an illustration?
    i'll put a proof later to make sure

    Thanks so much
     
    Last edited: Feb 11, 2010
  7. Feb 11, 2010 #6
    bec ΔV = V(t+dt) - V(t)
    therefore V(t+dt) = V(t) + ΔV ?Right?
     
  8. Feb 12, 2010 #7

    Doc Al

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    The velocity vector is always a fixed angle (90°) with respect to the radius. If the radius points in the +x direction, the velocity vector points in the +y direction. When the radius has moved to be 30° (for example) above the x-axis, where must the velocity vector point?
    Right.
     
  9. Feb 12, 2010 #8
    Thanks so much
    I want to ask if this works ?

    http://img215.imageshack.us/img215/3401/proof3f.jpg [Broken]
    http://img693.imageshack.us/img693/8039/proofe.jpg [Broken]
    http://img705.imageshack.us/img705/5425/proof2t.jpg [Broken]

    I'm very sorry 4 the handwriting but i'm in a hurry

    Thanks again
     
    Last edited by a moderator: May 4, 2017
  10. Feb 12, 2010 #9
    Also i made an animated picture to demonstrate ur words
    http://img30.imageshack.us/img30/1725/centripetalanimated.gif [Broken]
    Is my understanding right?

    I want to make sure if my two posts are right
     
    Last edited by a moderator: May 4, 2017
  11. Feb 12, 2010 #10

    Doc Al

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    Your animation looks fine to me.
     
  12. Feb 13, 2010 #11
    Good
    I think i got it now
    but i have another way and also i want to make sure if its right ?

    [url]http://img24.imageshack.us/i/prooffz.jpg/[/url]

    IS this right

    Thanks so much
     
  13. Feb 14, 2010 #12

    Doc Al

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    I'm afraid that I don't understand what you are doing in that image. Could you define your triangles and angles, and then describe in words what you are trying to prove?
     
  14. Feb 14, 2010 #13
    I'm very sorry its my fault I should have explained more
    Actually I don't really need to put this proof but I'm just eager to know if its right or wrong
    http://img246.imageshack.us/img246/745/centripetalforce2.jpg [Broken]
    This pink line is an extension to the velocity vector
    And I drew a green radius to help me in the proof
    I don't want to prove any thing new I just want to prove that the angle between the velocity vectors is equal to the angle between the two radii .
    I understand what u said but I want to make sure if this method also works .

    Since the velocity vector is at point b is parallel to ax
    Therefore m(<MXY)=m(MXA)=m(<b) = 90 degrees

    Therefore XA = XY (According to a theorem in geometry )

    And MX is a common side in the two trianles MXY and MXA
    Therefore triangle MXY is congruent to triangle MXA

    Therefore m(<2)=m(<3)

    Since m(<1) = 1/2 m(<AMY) (angle of tangency and centeral angle subtended by the same arc)

    Therefore m(<1) = m(<2)
    I hope its obvious now
    Thanks
     
    Last edited by a moderator: May 4, 2017
  15. Feb 14, 2010 #14

    Doc Al

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    Looks good to me. You can also say that m(<1) = m(<2) since they are both complementary to angle MAX.
     
  16. Feb 15, 2010 #15
    Yeah this way is better and much more easier
    this is a very interesting discussion
    nothing is well-illustrated in my txt book

    thanks so so much this really helped
     
  17. Feb 15, 2010 #16

    ideasrule

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    Homework Helper

    As an aside, there's a much easier derivation of this.

    [tex]\vec v \cdot \vec r =C[/tex]

    Derive both sides:

    [tex]\vec a \cdot \vec r + \vec v \cdot \vec v =0[/tex]

    Get rid of the vectors:

    [tex]-ar + v^2 = 0[/tex]
     
  18. Feb 15, 2010 #17

    Doc Al

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    Nice. Combine that with a similar treatment of [itex]\vec v \times \vec r = C[/itex] to show that the acceleration is radial.
     
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