How Is It Going To Move?

There are only three possible paths on which a point on the rod can move:

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• $\underline{\text{Straight line:}}$

Consider the two end points $A$ and $B$ of the rod.

At any given instant we can observe that the point $A$ moves vertically downward on the $\text{y - axis}$ and the point $B$ moves horizontally on the $\text{x - axis}$ . Hence, the locus of path of $A$ and $B$ are the straight lines $x=0$ and $y=0$ respectively.

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• $\underline{\text{Circle:}}$

Consider the coordinates of center of mass be $(x,y)$ .

From the frame of the rod (which is uniform), it is clear that its center of mass will lie exactly in the middle. Hence, the point $(x,y)$ is at a distance of $\dfrac{\ell}{2}$ from either end of the rod. Let us take the angle of inclination of rod from the horizontal surface be $\theta$ . Take components $\dfrac{\ell}{2} \cos \theta$ and $\dfrac{\ell}{2} \sin \theta$ of the length vector $\dfrac{\ell}{2}$ on the $\text{x - axis}$ and $\text{y - axis}$ respectively. From this, we conclude that,

$\begin{aligned} x & = & \dfrac{\ell}{2} \cos \theta \\ y & = & \dfrac{\ell}{2} \sin \theta \end{aligned}$

Eliminating $\theta$ gives,

$\begin{aligned} x^{2} + y^{2} & = & \dfrac{{\ell}^{2}}{4} \\ \\ \implies 4x^{2} + 4y^{2} & = & {\ell}^{2} \end{aligned}$

Which is the equation of a circle.

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• $\underline{\text{Ellipse:}}$

Consider any other variable point $(x,y)$ on the rod situated at a distance $a$ from the end $A$ of the rod and $b$ from the end $B$ of the rod. Again take the angle of inclination as $\theta$ and take component $a \cos \theta$ of length vector $a$ along the $\text{x - axis}$ and component $b \sin \theta$ of length vector $b$ along the $\text{y - axis}$ . From here, we conclude that,

$\begin{aligned} x & = & a \cos \theta \\ y & = & b \sin \theta \end{aligned}$

Eliminating $\theta$ gives,

$\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$

Which is the equation of an ellipse.

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Important Note: We have not considered the situation when the rod losses contact with the vertical surface and starts slipping on the horizontal surface as after this moment, all points on the rod will have the same path, i.e., a straight line $y=0$ .

Let a point $P$ on the rod $AB$ be $a$ from $A$ and $b$ from $B$ . That is $a+b= \ell$ . Then the coordinates of $P (x,y)$ (from the origin $O$ ), when the rod is inclined at $\angle ABO = \theta$ are given by:

$\begin{cases} \dfrac xa = \cos \theta \\ \dfrac yb = \sin \theta \end{cases} \implies \dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = \cos^2 \theta + \sin^2 \theta = 1$

$\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ is the equation of a ellipse. When $a=b$ , the equation is a circle. We note that $x=a\cos \theta$ and $y=b\sin \theta$ . When $a = 0$ , we get $P(0, b\sin \theta)$ , which is a vertical straight line, and when $b=0$ , we get $P(a\cos \theta, 0)$ , a horizontal straight line.

Therefore, the answer is $\boxed{\text{Straight line, Circle, Ellipse}}$ .