How can I understand the motion of a particle on a cycloid trajectory?

[fayled]

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.

 

[pbuk]

Does this help?

 

[BvU]

How can I see that the particle moves around this circle

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.

 

[fayled]

Does this help?

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.

 

[fayled]

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.

I see. Is there a way of getting the motion from the parametric equations then?

 

[BvU]

How can I see that the particle moves around this circle (and so appears to be a point on a rolling wheel)?

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
$\sin \left(\phi + \pi/2\right ) = \cos\phi$

 

[fayled]

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
$\sin \left(\phi + \pi/2\right ) = \cos\phi$

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.

 

[BvU]

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.