Equation of various cycloids: when it is regular?

Question

This is an exercise from a book on Differential Geometry.

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Let a disc of radius 1 roll on the x-axis in the x–y plane. Consider a second disc of radius $r>0$ with the same centre and rigidly connected with the first.

(a) Describe the path on which a point on the rim of the second disc moves by a parametrised curve. This curve is called cycloid.

(b) Sketch the curve for $r<1$, $r=1$, and $r>1$

(c) In which cases is the parametrised curve regular?

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My answer: Unit circle is placed on $x$-axis with touching point $(0,0)$, so the center of both circles is $(0,1)$.

Then the "down-point" $P$ of second circle of radius $r$ is $(0,1-r)$.

To describe the locus of $P$ as unit circle rolls on $x$-axis, it is equivalent to do (rotate anticlockwise around $(0,1)$ by $\theta$ and then translate along $x$-axis by $r\theta$.)

The position $P'$ occupied by $P$ is then $(r\theta-r\sin\theta, 1-r\cos\theta)$. [my Q 1: Is this correct?]

For checking regularity of the curve, we need to check that the velocity vector $ (x'(\theta), y'(\theta))$ is not $(0,0)$ for any $\theta$.

But (if answer to Q.1 is yes then), we can see that for $\theta=2\pi n$, the velocity vector is $0$, whether it is $r<1$ or $r=1$ or $r>1$. So curve is not regular at some points for all $r$. Then (my Q.2) what is the meaning of the question c) above?

Answer 1

The equation of the center of the circle is $C = (\theta,1)$. It is controlled by the rotation of the unit circle.

The position of P is $(x,y) = C + (-r\sin{\theta},-r\cos{\theta}) = (\theta -r\sin{\theta},1-r\cos{\theta})$

Octave code:

function plotcycloids(r,theta1,theta2)
  theta = [theta1:0.01:theta2];
  x = theta -r*sin(theta);
  y = 1 - r*cos(theta);
  plot(x/pi,y); grid minor on;
end

figure 1;
plotcycloids(0.5,0,4*pi);
title("r = 0.5","fontsize",20);
print "-S1000,1000" r_half.jpg
    
figure 2;
plotcycloids(1,0,4*pi);
title("r = 1","fontsize",20);
print "-S1000,1000" r_1.jpg

figure 3;
plotcycloids(3,0,4*pi);
title("r = 2","fontsize",20);
print "-S1000,1000" r_2.jpg

$$\frac{\partial (x,y)}{\partial \theta} = (1 -r\cos{\theta},r\sin{\theta})$$ If $r < 1$ then $\displaystyle \frac{\partial x}{\partial \theta} \gt 0$.

If $r = 1$ then $\theta = 0 + 2 \pi k, k \in \mathbb{J}$ gives a zero velocity.

If $r > 1$ then the curve folds back on itself when $\displaystyle \frac{\partial x}{\partial \theta} \lt 0$. The $x$ coordinate moves in the negative direction.

So $r < 1$ is required for a regular curve.