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Cycloidal Path of Charged Particle

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    A charged particle (q), mass (m) is released from a frictionless inclined plane of angle θ under influence of earth's acceleration (g) and magnetic field (B) perpendicular to (g) and plane of motion of particle. The particle slides down distance "l" along incline and then follows a cycloidal path with vertical displacement between highest and lowest point (h). Prove that [itex]l = h\frac{cot^{2}θ}{4}[/itex]



    2. Relevant equations

    F= qv[itex]\times[/itex]B


    3. The attempt at a solution

    In vector notation I took g as -g j. B as -B k.
    Took instantaneous velocity as v= vxi + vy j.
    Then formed a differential equation to vx and vy in terms of time. The point where it leaves the incline is the point of inflection of the curve. I bashed whatever equations I had to finally get the answer, but after doing this I think there's got to be an easy way to look at this. Can someone help me out?
     
  2. jcsd
  3. Mar 13, 2014 #2

    BvU

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    Wouldn't life be a lot easier for you if the particle does NOT leave the incline ?
    [edit] Hmmm... weak moment. B has to be ##\bot## g AND the plane of motion. You don't happen to have a picture at hand ?

    Anyway, if the particle stays on the plane and also if it leaves, there has to be a g in one of the relevant equations.
    And also in 3: what is your differential eqn? Can you show some of the bashing?
     
    Last edited: Mar 13, 2014
  4. Mar 14, 2014 #3

    rude man

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    Looks like B is to be perpendicular to g and to the NORMAL of the inclined plane. This appears to be what the OP has assumed, and it would indeed seem to produce a spiral path as it works its way down the incline.

    For the first l distance there are 3 forces on the mass: gravity, magnetic and the plane pushing up on the mass. After that, when the mass becomes airborne, the last of these forces is absent. So it seems like you first have to determine l which is when the inclined components of magnetic and gravitational forces add up to zero. That should not be too bad.

    Then, determine x and y for the airborne path to get h. I would write F = ma equations in x and y and solve for the locus. That sounds more imposing ...

    EDIT: you can solve for just the y motion & then the max - min of this motion should be h I guess. Then compute l/h.

    Getting late, may look some more tomorrow but probably others will have joined in by then.
     
    Last edited: Mar 14, 2014
  5. Mar 14, 2014 #4
    @BvU....h and l both come in terms of m,q,g,B,θ etc. then u take ratio and only θ remains. I'll Show the bashing.

    @rude man.....it doesnt come to be spiral, i'll show that mathematically.

    The F=qvb=ma equation yield the following
    [itex]m\frac{d}{dt}(v_{x}i + v_{y}j)=-qBv_{y}i + (qBv_{x}-mg)j[/itex].

    Thus,
    [itex] \frac{dv_{x}}{dt}= -\frac{qBv_{y}}{m}[/itex] and [itex]\frac{dv_{y}}{dt}=\frac{qBv_{x}}{m} -g[/itex]

    [itex]\Rightarrow \frac{d^{2}v_{x}}{dt^{2}}=-(\frac{qB}{m})^{2}(v_{x}-\frac{mg}{qB})[/itex].
    Let t=0 when particle is at bottom most point and velocity be [itex]v_{0}i
    \Rightarrow v_{x}= (v_{0} -\frac{mg}{qB})cos(\frac{qBt}{m}) + \frac{mg}{qB}[/itex].

    Further, [itex]y= (v_{0} - \frac{mg}{qB})\frac{m}{qB}(1-cos(\frac{qBt}{m})))[/itex].
    Clearly, [itex]h=\frac{2m}{qB}(v_{0} - \frac{mg}{qB})[/itex]

    As can be seen [itex]x= aθ + bsin(kθ) and y=ccos(kθ) [/itex]. Thus it happens to be a cycloid.
     

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  6. Mar 14, 2014 #5

    BvU

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    OK, I'm inclined to do away with the particle staying on the incline :redface:
    Argument: B doesn't do any work, so potential energy at highest point = potential energy at releasing meaning h = l sinΘ, so not the cot2/4. I took ##\vec B \bot## the plane -- as I was told.
    So we continue with
    • the assumption ##\vec B## is NOT ##\bot## the plane but in the horizontal negative z direction
    • we look at free fall in a magnetic field with inital condition ##v_0^2 = 2 l g \,sin\theta/ m##. (So ##\vec v_0 \cdot \hat\imath = v_0 \cos\theta## and ##\vec v_0 \cdot \hat\jmath = -v_0 \sin\theta##)
    Do we agree so far ?
     
  7. Mar 14, 2014 #6

    rude man

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    I agree with BvU's v2 = 2l sinθ/m. This is the 'takeoff point'. I now see a system of two diff. eq. in x and y with new initial conditions on vx and vy. I would impose a new coordinate system with x=y=0 at this point to clean up the equations in x and y.

    I also got the OP's dvx/dt and dvy/dt equations, but then I don't follow the rest to where he/she gets h. If that h is correct then we have a problem getting the answer, seems like.

    I would pursue solving the system of two diff. eq. in x and y, solving for y(t) and getting h that way. (I would use Laplace, but whatever).

    @theQmechanic: I used the word 'spiral' loosely. The exact shape is of course determined by the equations. It's undoubtedly a cycloid as you and the question state.
     
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