Find the locus of point $P$ which lies on a circle.

Question

QUESTION: Consider the circle with radius $1$ and cente 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 ATTEMPT: Pardon me, for this question has been asked before. But I was unable to understand the explanation.. I applied the concept that no matter where the circle is, the distance of $P$ from the (then) centre of the circle is always equal to the radius, $1$. For this I first found out the locus of the centre, which is clearly $y=1$. But, this does not help. I don't understand how to approach this problem. I think there must be some smarter way to solve this. Can anyone please help me out?

Thank you.

Answer 1

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,

$$(x,y)=(\theta,1)+\left(-\sin\theta,-\cos\theta\right)=\left(\theta-\sin\theta,1-\cos\theta\right).$$

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You can eliminate $\theta$ using

$$\cos\theta=1-y$$ and

$$x=\pm\arccos(1-y)\mp\sqrt{1-(1-y)^2}+2k\pi.$$

Answer 2

Your locus is this beautiful shape called a cycloid

It's actually better for you to explore the explanation given there.

Btw, the best way to describe it is with a parametric equation.

Hint: for a rotation of an angle $\theta$ of that circle, consider the point where the circle touches the x axis. If you drew a radius from that point, what angle would it form with the x axis?

Answer 3

When the circle turns clockwise a full circle, it moves to the right by the length of its perimeter, if it does not slip.

If we use $\theta$ for the turning angle and $p = 2 \pi r$ for the perimeter for a circle of radius $r$, remembering that $2 \pi$ radians is a full turn, $$\theta = 2 \pi \frac{x}{p} = 2 \pi \frac{x}{2 \pi r} = \frac{x}{r}$$

The location of the point, when $(x, r)$ is the center of the circle, and $\theta$ is measured clockwise, and the point being at origin when $x = 0$ and $\theta = 0$, is $$\begin{cases} x_p = x - r \sin\theta \\ y_p = r - r \cos\theta \\ \end{cases}$$

To find the locus of the point as a function of position $x$ or rotated angle $\theta$, all you need to do is to substitute the other into the pair of equations.

In this particular case, $r = 1$ so $\theta = x$: $$\begin{cases} x_p = \theta - \sin\theta \\ y_p = 1 - \cos\theta \\ \end{cases} \quad \iff \quad \begin{cases} x_p = x - \sin x \\ y_p = 1 - \cos x \end{cases}$$