Disc going along a parabola

[QuantumDuality]

Homework Statement

A disc of radius R rolls without slipping along the parabola y= ax². Obtain the constrain equation

Homework Equations

Because there's no slipping, then:
$R d \theta = ds (1)$
Where $\theta$ is the angle between the line from the center of the disc to a fixed point and the line from the center of the disc to the contract point with the parabola
Also:
$ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + 4a^2 x^2} dx (2)$

The Attempt at a Solution

So I just have to equate (1) and (2) to get the constraint equation?, Because I have seen a generalization of the constraint equation for an arbitrary curve that use

$ds = R(d\theta + d\phi)$

Where $\theta$ is an angle between a radius to a fixed point and the radius parallel to the y axis, while $\phi$ is an angle between the radius parallel to the y-axis and the radius to the contact point
Here's the generalization:

https://campus.mst.edu/physics/courses/409/Problem-Solutions/HW#4/HW4_prob3_ hoop on curve.pdf

 

[Dr.D]

Your problem statement says, "... rolls along ... the parabola..." Is the disk in the plane of the parabola, perhaps normal to the plane of the parabola, or something else again?

 

[haruspex]

Because there's no slipping, then:
$R d \theta = ds (1)$
Where $\theta$ is the angle between the line from the center of the disc to a fixed point and the line from the center of the disc to the contract point with the parabola

Took me a while to realize you meant a point fixed on the perimeter of the disc.

Where θ is an angle between a radius to a fixed point and the radius parallel to the y axis, while ϕ is an angle between the radius parallel to the y-axis and the radius to the contact point

I think if you draw a diagram you will find that your θ equals their θ+φ+constant.