5.1.2 Example 2: Cycloid Motion

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  • Опубліковано 19 гру 2024

КОМЕНТАРІ • 26

  • @wreza97
    @wreza97 8 місяців тому +1

    00:01 Cycloid motion is observed when a charged particle experiences forces from both an electric field and a magnetic field.
    02:12 Lorence force law and cycloid motion
    04:26 Deriving equations of motion for cycloid motion.
    06:53 Derivatives of sine and cosine result in their original functions with a negative coefficient.
    09:09 Initial conditions and derivatives in Cycloid Motion
    11:21 Calculation error detected in the given equations
    14:05 Cycloid motion equations simplification
    16:42 Equation for cycloid motion with moving center

  • @saqarkhaleefah6159
    @saqarkhaleefah6159 5 років тому +8

    Shouldn't the cross product at 3:46 yield ByI - BxJ? The second term of the determinant should always get a - sign infront right?

  • @jg394
    @jg394  12 років тому +3

    See 5:30 where I introduce omega = QB/m
    I substitute in 6:20 to bring the equations together. Part of that is multiplying by omega since y double dot = x triple dot / omega.
    The first time derivative of (E/B)t is (E/B). The time derivative of that is 0. So the 2nd time derivative of (E/B)t is 0.

  • @mubashirhussain9176
    @mubashirhussain9176 2 роки тому +1

    How did you make the equation for x and y in 7:25 plz reply ma

  • @erikbeier9515
    @erikbeier9515 4 роки тому +2

    Cross product done incorrectly at 3:30

  • @nilimaibrahim8865
    @nilimaibrahim8865 10 років тому +5

    a correction needed... 2nd part of the last term will be (y-R)^2 (whole square) , not (y-R^2)

  • @radiuscubed
    @radiuscubed 4 роки тому

    Super cool that you can get the cycloid with electromagnetism. There seems to be quite a few ways of getting it. I got with Snell's Law, for the most part, and I know I've seen it done with calculus of variations too.

  • @claireworld_
    @claireworld_ 8 місяців тому

    Amazing explanation!!

  • @manojdeshpande2685
    @manojdeshpande2685 11 років тому

    At 12:43 derivative of sine is cosine and not (minus)cosine

    • @jg394
      @jg394  11 років тому

      Good catch. I think I fix it later... don't I?

    • @jg394
      @jg394  11 років тому

      jg394 Nope, I don't catch that. It ends up it doesn't matter.

    • @manojdeshpande2685
      @manojdeshpande2685 11 років тому

      yeah it doesn't matter
      i was just saying :)

  • @shenshen0912
    @shenshen0912 9 років тому +2

    I can understand the final x part , which has velocity E/B.
    But what about the Y part? Is R a constant?
    R = (Q*E)/(B*m) right?
    Then why y changes ?
    Well...i think the Q will rotate like the way you said, just can't explain the equation.

    • @shenshen0912
      @shenshen0912 9 років тому

      thanks for your lesson !!
      jg394

  • @thecules6891
    @thecules6891 2 роки тому

    E is upwards en B in into the page so by that and the RHR the magnetic force is to the right
    Even if velocity increases and Fmagnetic increases it should still point right so i do not understand why it returns to the yaxis qualitatively

  • @aahilmirza4684
    @aahilmirza4684 4 роки тому

    bro ,, you did it amazing . thanks a lot
    big fan of you.

  • @S식
    @S식 7 років тому

    So the motion is in direction perpendicular to E? I need added explanation about it :) thanks

  • @nelofarta1665
    @nelofarta1665 6 років тому +1

    Thanks a lot!

  • @persianmeme3530
    @persianmeme3530 2 роки тому

    Beautiful

  • @kaursingh637
    @kaursingh637 5 років тому

    SIR =BOTH B AND E GIVE RISE TO CYCLOID MOTION --PLEASE CONFIRM

  • @master_altaf_khan
    @master_altaf_khan 2 роки тому

    thanks

  • @kaursingh637
    @kaursingh637 5 років тому +1

    SIR - WHEN THERE IS ONLY B -THE MOTION IS CIRCLE ==WHEN WE HAVE BOTH B AND E =THE MOTION IS CYCLOID -- PLEASE CONFIRM -THANK U SIR

  • @manojdeshpande2685
    @manojdeshpande2685 11 років тому

    just saying :P