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In the book of Dynamics by Horace Lamb, at page 103, is it given

that for a motion on a smooth curve, the equation of motion is given by $$mv \frac{dv }{ ds} = -mg \sin \theta \qquad \frac{ mv^2 }{r} = -mg \cos\theta + R, $$ where $\theta$ is the angle between the surface normal and vertical direction, and $R$ is the "pressure" exerted by the curve.

[...]

Then, we have $$\cos \theta = \frac{dx}{ds} \qquad \sin\theta = \frac{dy }{ds}, $$ where $s$ is length of the path taken over the surface, and $x,y$ are usual cartesian coordinates as $y$ is taken as upward.

However, I cannot understand how does the author derives the latter equations between $\theta$ and the derivatives of $x,y$ wrt $s$.

Our
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2 Answers2

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Now apply the rule of $\sin \theta$ and $\cos \theta$

Rakibul Islam Prince
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Hint Write it as $dx=ds\cos\theta$ and $dy=ds\sin\theta$. It is just a decomposition of $ds$ into components along $x$ and $y$ axes

Andrei
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  • but the path is not a straight line; even though we can approximate it to be like that. – Our Nov 17 '18 at 18:51
  • The full path is not straight, but a small element $ds$ can be considered as such. – Andrei Nov 17 '18 at 18:53
  • Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ? – Our Nov 17 '18 at 18:55
  • Yes. In the limit when $ds\to 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $\theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis. – Andrei Nov 17 '18 at 18:57
  • Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK – Andrei Nov 17 '18 at 18:59