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Today I had been reading a physics book. The book stated in the important points section that

The point of a point on circumference of rolling disc is a cycloid and the distance moved by this point in one full rotation is $8R$.

I tried to prove it but there was no progress. The book even doesn't provide the solution. Can anyone here provide a better proof over how did they get this value.

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  • $\begingroup$ This is done on the relevant Wikipedia page and that page is the very first link a web search returns for "cycloid" $\endgroup$ – StephenG Jan 14 '18 at 11:32
  • $\begingroup$ See my answer here : What is the position as a function of time for a mass falling down a cycloid curve?. Inserting $:\theta=2\,\pi\:$ in the first rhs of equation (b-04) you have $\:s=8R$. $\endgroup$ – Frobenius Jan 14 '18 at 13:40
  • $\begingroup$ Thanks for all your help. I have got the proof on Wikipedia as per the comment of StephenG $\endgroup$ – Rohan Shinde Jan 14 '18 at 13:43
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The parametric equations of a cycloid are given by $$x=r(t-\sin t)$$ $$y=r(1-\cos t)$$ The distance moved by the point is also known as the arc length and this can be computed by $$s=\int_{t_0}^{t} r\sqrt{\frac{d}{dt}(t-\sin t)^2+\frac{d}{dt}(1-\cos t)^2}\ dt$$Where $t_0=0$ and $t=2\pi$. We then evaluate what's inside the square root and get $$s=\int_{0}^{2\pi} r\sqrt{(1-\cos t)^2+(\sin t)^2}\ dt$$ $$s=\int_{0}^{2\pi} r\sqrt{4\sin^2(t/2)}\ dt$$ $$s=\int_{0}^{2\pi} 2r\sin (t/2)$$ Integrating we get $$s=[-4r\cos (t/2)]_0^{2\pi}$$ $$s=4r+4r$$ $$s=8r$$ Hence the distance moved by a singular point in one full rotation is $8r$.

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