# One object moves along the cycloid at a constant rate, how about its acceleration? [closed]

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 $$\theta$$?

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First we use the following parametric equation of cycloid $$$$\begin{split} x & \e R\plr{\theta\m\sin\theta}\\ y & \e R\plr{1\m\cos\theta}\\ \end{split} \tag{01}\label{01}$$$$ The velocity vector is $$$$\mathbf v\e c\plr{\cos\phi,\sin\phi} \tag{02}\label{02}$$$$ where $$$$\tan\phi\e\dfrac{\mathrm dy}{\mathrm dx}\e\dfrac{\sin\theta}{1\m\cos\theta}\e \cot\plr{\cdots} \tag{03}\label{03}$$$$ so $$$$\mathbf v\e c\blr{\sin\plr\cdots,\cos\plr\cdots} \tag{04}\label{04}$$$$ For the acceleration vector $$\:\mathbf a\:$$ we have $$$$\mathbf a\e\dfrac{\mathrm d\mathbf v}{\mathrm dt}\e\cdots\:\dot{\!\!\theta}\left(\cdots,\cdots\right) \tag{05}\label{05}$$$$ But $$$$\begin{split} c\e & \dfrac{\mathrm ds}{\mathrm dt}\e\sqrt{1\p\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2}\dfrac{\mathrm dx}{\mathrm dt}\\ & \e\cdots\cdots\cdots\e \cdots\cdots\cdots\:\dot{\!\!\theta}\\ \end{split} \tag{06}\label{06}$$$$ that is $$$$\dot{\!\!\theta}\e \dfrac{c}{\cdots\cdots} \tag{07}\label{07}$$$$ Inserting this expression in equation \eqref{05} we have $$$$\mathbf a\e\left(\mathrm a_{\,x},\mathrm a_{\,y}\right) \tag{08}\label{08}$$$$ so $$$$a_{\,y}\e\cdots\e\texttt{constant ???} \tag{09}\label{09}$$$$

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.

• @ 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. Nov 13 '21 at 12:18
• @ 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 $\frac{c^2}{R}$ Nov 13 '21 at 12:25
• Thanks for your answer, I have got the answer just like@Frobenius has gived by converting the parametric equation to the cartesian equation
– Joy
Nov 13 '21 at 15:26
• @Frobenius you are right, but could you write the steps?
– Joy
Nov 13 '21 at 15:27
• @JohnHunter your idea is not wrong, but it seems like the derivative has some troubles
– Joy
Nov 13 '21 at 15:29