cycloid generated by a circle

rmpatel5 · #1 $Old$ 11-22-2008, 08:10 PM

See attachment for figure.

Moo · #2 $Old$ 11-23-2008, 02:25 AM

Hello,

Consider the line from A to the y-axis. Its length is $\pi a$

It is also equal to $x+s+AP$
But $x=a(t-sin(t))$

So we have $\pi a=a(t-\sin(t))+s+AP \Leftrightarrow \boxed{s=\pi a-a(t-\sin(t))-AP}$

Consider the triangle ABP. It's a right angle triangle.
Length of BP is a.
Angle ABP is pi-t.
So since sin(ABP)=AP/BP, we get :
$\sin(\pi-t)=\frac{AP}{a}$

by simple trigonometry, $\sin(\pi-t)=\sin(t)$
Hence $\boxed{AP=a \sin(t)}$

Finally :
$\begin{aligned} s&=\pi a-a(t-\sin(t))-AP \\ &=\pi a-at+a \sin(t)-a \sin(t) \\ &=\boxed{a(\pi-t)} \end{aligned}$