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TIME TO SLIDE DOWN FRICTION LESS CURVE

SOLVE FOR $\int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy$

THEORY BELOW

Let there be a block of unit mass sliding down a circular curve of radius $R$ such that it started with $0$ initial speed from top most point as shown.

I attempted to find the time which it should take in sliding down the friction less curve, a given.

Now, from the principle of conservation of energy, the velocity $V$ of the block at a distance $y$ below the starting point should be that $V^{2} = 2*g*y$, where $g$ is gravity

Now, considering only the vertical motion, we have $\frac{dy}{dt} = V \sin \theta = V \sin A $ and that $\sin A = \sqrt{1 - (y/R)^2}$

or Time to slide = $T = \int_0^{T} dt = \int_0^{R} \frac{1}{V \sin A} dy = \int_0^{R} \frac{1}{\sqrt{2*g*y} \sqrt{1 - (y/R)^2}} dy = \frac{R}{\sqrt(2*g)} * \int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy$

i.e. it essentially breaks down to solving $\int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy$, which seems unintegrable!

KINDLY HELP/GUIDE!

Anshuman Dwivedi
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    have you tried converting to polar coordinates to integrate? – Angelica Dec 23 '21 at 17:58
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    $\int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy=,(y=Rx),\frac{1}{\sqrt R}\int_0^{1} \frac{1}{\sqrt{x(1 - x^2)}} dx=,(x=\sqrt t),,\frac{1}{2\sqrt R}\int_0^{1} \frac{t^{-\frac{3}{4}}}{\sqrt{(1 - t)}} dt=\frac{1}{2\sqrt R}B\Big(\frac{1}{4};\frac{1}{2}\Big)$ – Svyatoslav Dec 23 '21 at 18:59
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    Thanks, I will study this! – Anshuman Dwivedi Dec 24 '21 at 03:20

1 Answers1

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You will not have any solution in terms of elementary functions. Instead you will obtain an elliptic integral.

Render $y =R\cos\theta$, then the integral becomes

$R^{-1/2}\int_0^{\pi/2}\dfrac{d\theta}{\sqrt{\cos\theta}}$

Next put in $\cos\theta=1-2\sin^2(\theta/2)$ and compare with the complete elliptic integral of the first kind described in Wikipedia.

Oscar Lanzi
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