Problem 1: Coupled Motion of Ring and Ball
1. Analysis of the Center of Mass
Let the mass of the ring be $M$ and the mass of the ball be $m$. We are given that they have the same mass, so $M = m$.
Initially, the system is at rest. The ball is placed at a distance $0.5r$ from the center of the ring. Let’s define the initial positions on a coordinate system:
- Center of Ring ($C$): Origin $(0, 0)$
- Ball ($B$): $(0.5r, 0)$
The Center of Mass (CoM) of the system is located at:
$$X_{cm} = \frac{M(0) + m(0.5r)}{M+m} = \frac{m(0.5r)}{2m} = 0.25r$$Since the external forces on the horizontal plane are zero, and the initial momentum is zero (equal opposite velocities), the Center of Mass remains stationary throughout the motion.
2. Trajectories in the CoM Frame
Since the CoM is stationary at $x = 0.25r$, we can analyze the motion relative to it.
- Distance of Ring Center from CoM: $|0 – 0.25r| = 0.25r$
- Distance of Ball from CoM: $|0.5r – 0.25r| = 0.25r$
Both bodies are at an equal distance ($0.25r$) from the CoM. Since the collisions are perfectly elastic and the masses are equal, the bodies simply exchange velocity components along the line of impact. However, due to the circular geometry of the ring, the kinematics constrain the motion to specific paths.
The ball collides with the ring when the distance between the ball and the ring’s center is $r$. In the CoM frame, this corresponds to the bodies being on opposite sides of the CoM, separated by a total distance $r$. This implies each body is at a distance $0.5r$ from the CoM at the moment of impact.
Therefore, both the ball and the center of the ring are confined to move within a circle of radius $0.5r$ centered at the CoM. The specific initial conditions (velocities perpendicular to the line joining them) and the geometry of elastic reflection in a circle result in trajectories that form equilateral triangles.
Figure 1: The trajectories of the ball (solid red) and the ring center (dashed blue) relative to the stationary Center of Mass.
3. Separation Distances
The separation $d$ between the center of the ring and the ball is the distance between their instantaneous positions.
In the CoM frame, the positions are always diametrically opposite relative to the CoM (since $r_{ball, cm} = -r_{ring, cm}$).
Let $x$ be the distance of the ball from the CoM. The distance of the ring center from CoM is also $x$. Thus, the total separation is $2x$.
Minimum Separation:
The minimum distance from the CoM occurs when the velocity is perpendicular to the radius vector, which happens at the midpoint of the chord connecting collision points (the apothem of the triangle). For an equilateral triangle inscribed in a circle of radius $R’ = 0.5r$:
$$x_{min} = R’ \cos(60^\circ) = (0.5r) \cdot \frac{1}{2} = 0.25r$$Minimum separation between bodies $= 2 \cdot x_{min} = 2(0.25r) = \mathbf{0.5r}$.
Maximum Separation:
The maximum distance occurs at the vertices of the triangular path (the points of collision). At these points, the distance from CoM is the radius of the confinement circle, $R’ = 0.5r$.
Maximum separation between bodies $= 2 \cdot R’ = 2(0.5r) = \mathbf{r}$.
Note: This confirms the collision condition, as the ball hits the ring wall when their separation is exactly the radius of the ring $r$.