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Given a friction-less slide $y=x^2$, place a particle on the slide at $(1,1)$. The particle is acted upon by constant gravity $g= 9.8$ units/s/s. At what time does it reach bottom?

The following is work done so far: Potential energy lost = Kinetic energy gained. $$ mass \cdot g \cdot (1-y) = \frac{1}{2} \cdot mass \cdot \|\mathbf{v}\|^2 $$ $$ \sqrt{2g(1-y)}= \|\mathbf{v}\| $$ Here $m$ is slope of the curve at the particle's position $$ v_x= -\| \mathbf{v} \| \cos{ \left( \arctan{m} \right) } $$ $$ v_y= -\| \mathbf{v} \| \sin{ \left( \arctan{m} \right) } $$ Here using trig identities $$ \cos{ \left( \arctan{m} \right) } = \frac{1}{\sqrt{1+m^2}} $$

$$ \sin{ \left( \arctan{m} \right) } = \frac{m}{\sqrt{1+m^2}} $$

So $$ v_x = -\| \mathbf{v} \| \frac{1}{\sqrt{1+m^2}} $$ $$ x' = -\sqrt{2g(1-x^2)} \cdot \frac{1}{\sqrt{1+(2x)^2}} $$ and $$ v_y = -\| \mathbf{v} \| \frac{m}{\sqrt{1+m^2}} $$ $$ y' = -\sqrt{2g(1-y)} \cdot \frac{2\sqrt{y}}{\sqrt{1+(2\sqrt{y})^2}} $$

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  • $\begingroup$ Work done so far: Given conservation of energy under constant gravity, potential energy + kinetic energy = constant. $\endgroup$ Commented May 9, 2016 at 5:43
  • $\begingroup$ For a particle at (x,y=x^2), it has fallen by a height 1-y, hence mass*g*(1-y)= 1/2 * mass * v^2. So v= sqrt(2g(1-y)) $\endgroup$ Commented May 9, 2016 at 7:09

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So you found

$$ \frac{\mathrm ds}{\mathrm dt}=\sqrt{2g(1-x^2)}\;. $$

We also have

\begin{align} \frac{\mathrm ds}{\mathrm dx} &=\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2} \\ &=\sqrt{1+4x^2}\;. \end{align}

Thus

\begin{align} \frac{\mathrm dt}{\mathrm dx} &=\frac{\mathrm ds}{\mathrm dx}\cdot\frac{\mathrm dt}{\mathrm ds} \\ &=\sqrt{\frac{1+4x^2}{2g(1-x^2)}}\;, \end{align}

and

\begin{align} t&=\int_0^1\mathrm dx\frac{\mathrm dt}{\mathrm dx}\\ &=\int_0^1\mathrm dx\sqrt{\frac{1+4x^2}{2g(1-x^2)}} \\ &= \frac{E(-4)}{\sqrt{2g}} \\ &\approx \frac{2.63518}{\sqrt{2g}} \\ &\approx0.595\text s\;, \end{align}

(Wolfram|Alpha computation of the integral), where $E(m)$ is the complete elliptic integral of the second kind with parameter $m=k^2$.

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  • $\begingroup$ amazing answer, thanks a lot $\endgroup$ Commented May 9, 2016 at 8:18

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