We know that the parametric equation:



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 $\theta$?


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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}



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.

  • $\begingroup$ @ Frobenius, but the object oscillates up and down, surely that needs acceleration up and then acceleration down, so it doesn't seem as though it can be constant in the $y$ direction. $\endgroup$ Nov 13 '21 at 12:18
  • $\begingroup$ @ Frobenius even so the $y$ acceleration alternates up to down, however it isn't clear what $c$ means, presumably the velocity of the wheel, or perhaps that's the same anyway as the average speed of the object. Then the magnitude of the total acceleration is constant at $\frac{c^2}{R}$ $\endgroup$ Nov 13 '21 at 12:25
  • $\begingroup$ Thanks for your answer, I have got the answer just like@Frobenius has gived by converting the parametric equation to the cartesian equation $\endgroup$
    – Joy
    Nov 13 '21 at 15:26
  • $\begingroup$ @Frobenius you are right, but could you write the steps? $\endgroup$
    – Joy
    Nov 13 '21 at 15:27
  • $\begingroup$ @JohnHunter your idea is not wrong, but it seems like the derivative has some troubles $\endgroup$
    – Joy
    Nov 13 '21 at 15:29

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