One object moves along the cycloid at a constant rate, how about its acceleration? We know that the parametric equation:
$$x=R(\theta+\sin(\theta))$$
$$y=-R(1+\cos(\theta))$$
and the constant velocity $c$.
How do I prove that the acceleration of the object in the $y$ direction is constant?
OK, this is not just a simple homework question, I want to learn something about mathematical physics from this question. I am suffered from $\frac{d^2y}{dt^2}$:why can't I get the second derivation of parametric equations $y=-R(1+\cos(\theta))$ , which gives the solution related to $\theta$?
 A: $\newcommand{\bl}[1]{\boldsymbol{#1}} 
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First we use the following parametric equation of cycloid
\begin{equation}
\begin{split}
x & \e R\plr{\theta\m\sin\theta}\\
y & \e R\plr{1\m\cos\theta}\\
\end{split} 
\tag{01}\label{01}
\end{equation}
The velocity vector is
\begin{equation}
\mathbf v\e c\plr{\cos\phi,\sin\phi}
\tag{02}\label{02}
\end{equation}
where
\begin{equation}
\tan\phi\e\dfrac{\mathrm dy}{\mathrm dx}\e\dfrac{\sin\theta}{1\m\cos\theta}\e \cot\plr{\cdots}
\tag{03}\label{03}
\end{equation}
so
\begin{equation}
\mathbf v\e c\blr{\sin\plr\cdots,\cos\plr\cdots}
\tag{04}\label{04}
\end{equation}
For  the acceleration vector $\:\mathbf a\:$ we have
\begin{equation}
\mathbf a\e\dfrac{\mathrm d\mathbf v}{\mathrm dt}\e\cdots\:\dot{\!\!\theta}\left(\cdots,\cdots\right)
\tag{05}\label{05}
\end{equation}
But
\begin{equation}
\begin{split}
c\e & \dfrac{\mathrm ds}{\mathrm dt}\e\sqrt{1\p\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2}\dfrac{\mathrm dx}{\mathrm dt}\\
& \e\cdots\cdots\cdots\e \cdots\cdots\cdots\:\dot{\!\!\theta}\\
\end{split} 
\tag{06}\label{06}
\end{equation}
that is
\begin{equation}
\dot{\!\!\theta}\e \dfrac{c}{\cdots\cdots}
\tag{07}\label{07}
\end{equation}
Inserting this expression in equation \eqref{05} we have
\begin{equation}
\mathbf a\e\left(\mathrm a_{\,x},\mathrm a_{\,y}\right)
\tag{08}\label{08}
\end{equation}
so
\begin{equation}
a_{\,y}\e\cdots\e\texttt{constant ???}
\tag{09}\label{09}
\end{equation}
A: Hints:
To find $\frac{d^2y}{dt^2}$ in the case where the speed along the cycloid is constant, using $\omega$ for $\frac{d\theta}{dt}$, the formulae in the question and using the chain rule $\frac{dy}{dt} = \frac{dy}{d\theta}\times \frac{d\theta}{dt}$
the speed is constant $v_x^2 +v_y^2 = 2\omega^2R^2(1+\cos\theta) = c^2$
so $$\omega^2 = \frac{c^2}{2R^2(1+\cos\theta)}\tag1$$
so $$2\omega\frac{d\omega}{dt} = \frac{c^2}{2R^2} \times \frac{sin\theta}{(1+\cos\theta)^2}\times \omega \tag2$$
and $\frac{d\omega}{dt}$ can be found in terms of $\theta$
best of luck with it.
