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Assume we have a skateboarder in an halfpipe with radius R.

enter image description here

The equation of motion is: $\ddot{x}=-g\frac{x}{R}\sqrt{1-(\frac{x}{R})^2}$

Now, I want to find v(x) for the skateboarder, if we drag him up to a height $h_0$ and let him skate from there. We also now that $U(\phi)=mgR(1-\cos(\phi))$

I had two idea. First one was using conversation of energie, like: $U=E_{kin}+E_{pot}$ and then solve for $v$. (That's what the solution does, so it would be okay)

The other idea was: $v(x)=\dot{x}=\int_0^{x_0} \ddot{x}$.

Is there a reason, that this approach, integration the eq. of motion, wouldn't work?

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    What's $r$ in the equation of motion? – 2017-01-29
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    Is it $v(t)$ of $v(x)$? I know velocity is a function of time. – 2017-01-29
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    @JosephQuarcoo And $x = x(t)$ hence $v = v(x(t))$. – 2017-01-29
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    velocity can be a function of everything. In my case, it's x. You let the skateboarder start at specific height and then you want a function which gives you the speed at a current position. So $v(x)$. You could also do $v(\phi)$ But anyway, everythings clear again :) Edit: Alan Turing was faster – 2017-01-29
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    Ok thanks for the clarification. – 2017-01-29

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The conservation of Energy is the most easy feature you may use.

For what concerns the integral, what you'd get would be:

$$\int -g \frac{x}{R}\sqrt{1 - \frac{x^2}{r^2}}\ \text{d}x = -\frac{g}{R}\left(-\frac{1}{3} r^2 \left(1-\frac{x^2}{r^2}\right)^{3/2}\right)$$

That is

$$\frac{gr^2}{3R} \left(1-\frac{x^2}{r^2}\right)^{3/2}$$

What do you think?

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    What's $r$ in your solution.? – 2017-01-29
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    @MyGlasses I actually just solved his integral, without caring about variables lol. I think it shale $R$ indeed, but the OP wrote $r$... – 2017-01-29
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    that's atypo. sorry about that. okay good, thanks. Just was confused since I got different results. Everythings clear again, thanks :) – 2017-01-29
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    @xotix You're welcome! – 2017-01-29