Consider the circle of radius $1$ with its centre at the point $(0,1)$. From this initial position, the circle is rolled along the positive $x$-axis without slipping. Find the locus of the point $P$ on the circumference of the circle which is on the origin at the initial position of the circle.
My work:
I found out the equation of the circle to be
$x^2+(y-1)^2=1$.
I have a weak feeling that the locus traced by the point $P$ might be helical.
But, I cannot do anything. Please help out, I don't have much idea about this type of problems.
[Math] Rolling of a circle along the positive $x$-axis without slipping and finding the locus of a point lying on the circumference of the circle.
coordinate systemsgeometry
Related Solutions
Let the center be at $(\theta,1)$. To reach this position, the circle must have rolled by the angle $\theta$, clockwise. Hence the absolute position of $P$ is the position of the center plus the position of $P$ relative to the center,
$$(x,y)=(\theta,1)+\left(-\sin\theta,-\cos\theta\right)=\left(\theta-\sin\theta,1-\cos\theta\right).$$
You can eliminate $\theta$ using
$$\cos\theta=1-y$$ and
$$x=\pm\arccos(1-y)\mp\sqrt{1-(1-y)^2}+2k\pi.$$
Best Answer
It is a bit unfortunate to use a unit circle-I will do it with a circle of radius $r$ which makes it more clear where things come from. We can set $r=1$ at the end.
When the circle has rotated through an angle $t$, the point of contact with the $x$ axis is $(rt,0)$. The center of the circle is at $(rt,r)$ The angle to P, measured counterclockwise from a vector from the center of the circle and pointing right, is $\frac {3\pi} 2-t$, so the offset from the center of the circle to P is $(r\cos (\frac {3\pi}2-t),r\sin (\frac {3\pi}2-t))$, so the location of P is $(rt+r\cos (\frac {3\pi}2-t),r+r\sin (\frac {3\pi}2-t))$. You can set $r$ to $1$ and use the angle-difference formulas to get rid of the $\frac {3\pi}2$