Problem 3: Trajectory of Point A
1. Kinematics & Direction Analysis
The disc of radius $R$ rolls inside a fixed ring of radius $2R$. The center of the disc, $C$, stays at a distance $2R – R = R$ from the origin.
Let’s rigorously define the signs based on the diagram:
- Translational Motion ($C$): The center starts at $(-R, 0)$. The velocity vector $\vec{u}$ points in the $+y$ direction (up). This implies the center is orbiting Clockwise (CW) around the origin.
- Rotational Motion (Disc): For the disc to roll without slipping on the inside of the fixed ring, the contact point must have zero instantaneous velocity. Since the center moves “up” ($+y$) at the left-hand position, the spin must propel the leftmost point “down” ($-y$) relative to the center. This requires a Counter-Clockwise (CCW) rotation.
Spin Angular Velocity (CCW): $\quad \omega = \frac{u}{R}$
2. Coordinate Calculation
We analyze the position at time $t$ by superposition of vectors: $\vec{r}_A = \vec{r}_C + \vec{r}_{A/C}$.
Position of Center $C$:
$C$ orbits Clockwise starting from angle $\pi$ (the negative x-axis).
Angle at time $t$: $\theta_{orb} = \pi – \Omega t$.
Position of A relative to C:
The vector $\vec{CA}$ starts pointing left (angle $\pi$) and rotates Counter-Clockwise (CCW) with rate $\omega$.
Angle at time $t$: $\theta_{spin} = \pi + \omega t$.
3. Total Trajectory
Adding the components:
$$ x_A = x_C + x_{A/C} = -R\cos\left(\frac{u}{R}t\right) – R\cos\left(\frac{u}{R}t\right) = -2R\cos\left(\frac{u}{R}t\right) $$ $$ y_A = y_C + y_{A/C} = R\sin\left(\frac{u}{R}t\right) – R\sin\left(\frac{u}{R}t\right) = 0 $$$$ x(t) = -2R \cos\left(\frac{u}{R}t\right) $$ $$ y(t) = 0 $$
The point A oscillates strictly along the x-axis (a diameter of the large ring), confirming the Tusi Couple mechanism.
