1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Understanding the cycloid

  1. Sep 8, 2014 #1
    Suppose we have a particle of charge Q, mass m at rest at the origin at t=0, with an electric field E=Ek and a magnetic field B=Bi with i,j and k the cartesian unit vectors. Then the solution for the trajectory of this particle is
    x=0,
    y=E(wt-sinwt)/wB,
    z=E(1-coswt)/wB,
    where w=QB/m.
    These parametric desribe the motion of the particle, but they can be recast into the form
    (y-Rwt)2+(z-R)2=R2
    where R=E/wB which also describes the particle motion. This form is that of a circle, radius R, with centre (0,Rwt,R) that travels in along the y-axis with speed Rw.

    Now my book states
    'The particle moves as though it were a spot on the rim of a wheel, rolling down the y axis at speed Rw. The curve generated in this way is called a cycloid.

    However I've always had trouble being able to properly see this from the equations, but haven't worried about it too much. However I think not understanding it is stopping me from solving a similar problem with a slightly more complicated motion.

    So my problem is this. The equation (y-Rwt)2+(z-R)2=R2 describes the particle trajectory. All I can see from this is that at time t, the particle must be on a circle of radius R, centre (0,Rwt,R), which tells me the circle is moving along the y-axis. How can I see that the particle moves around this circle (and so appears to be a point on a rolling wheel)? And in order to appear like a rolling wheel, the particle would need a certain speed around the circumference of the moving circle (I believe it would have to move around the circumference at speed Rw, whilst the circle itself moves with this speed along the y-axis) - how do I see this is satisfied? Thanks.
     
  2. jcsd
  3. Sep 8, 2014 #2
    Does this help?
     
  4. Sep 8, 2014 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can not, in the same way as, when you state that ##x^2+y^2=R^2##, that doesn't reveal anything about the position as a function of time. All you get is the locus of the positions: all positions are on the circle around the origin with radius R.

    Whereas ##x=R\cos\omega t,\ y = R\sin\omega t## expresses x and y as functions of time t.

    In other words, the solution you begin with expresses y and z as functions of time t. The recasting only gives a relationship between y and z.
     
  5. Sep 8, 2014 #4
    Nope, it just states (or show in the case of the animations) what the parametric equations/cartesian equation describe/describes - in short, I would like an explanation of why they describe what they do.
     
  6. Sep 8, 2014 #5
    I see. Is there a way of getting the motion from the parametric equations then?
     
  7. Sep 8, 2014 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, as we have seen, the relationship between y and z can't help you there. So we have to stick to the functions of time. You can kind of 'build it up', for example:$$y=R\cos\omega t$$
    $$z = R\sin\omega t$$describes a point moving counterclockwise around a circle around the origin with radius R. Agree? Next:$$y=R\cos\omega t$$
    $$z = -R\sin\omega t$$describes a point moving clockwise around the same circle. Then:$$y=R\cos\omega t$$
    $$z = R(1-\sin\omega t)$$lifts the circle so that it 'rolls on the y-axis'. And finally:$$y=\omega Rt + R\cos\omega t$$
    $$z = R(1-\sin\omega t)$$lets the circle roll in the positive y direction at the proper rate.

    If you want the functions in your post #1 that start at (0,0,0) at t=0, shift by ##\pi/2##:
    ##\cos \left(\phi + \pi/2\right ) = - \sin\phi## and
    ##\sin \left(\phi + \pi/2\right ) = \cos\phi##
     
  8. Sep 8, 2014 #7
    Great, this was exactly what I was looking for. The very last part is confusing me though - although adding pi/2 clearly gives the correct answer, looking at the scenario I would want to be subtracting pi/2 phase. If z is upwards, y is to the right, surely out θ=wt is measured from the y axis anticlockwise to be consistent with your first step of the anticlockwise circle. So then if I want things to start at (0,0,0), I need to shift everything back by pi/2, i.e subtract it.
     
  9. Sep 8, 2014 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    My thingy starts at R,R for t = 0. So I have to let it roll a quarter turn, ##\pi/2##, to let z=0 and then y is ##\pi R/2##. The transform needed is ##\omega t'= \omega t - \pi/2## and I realize that I also need to shift y: ##y'= y - \pi R/2##.

    A bit confusing, I admit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Understanding the cycloid
Loading...