Rotation of Horizontal Disc with Projected Pebbles Trajectory

To determine where pebbles $P$ and $Q$ land on the rotating disc, we analyze their motion in both the stationary ground frame and the rotating frame of the disc.

1. Frame of Reference and Initial Velocities

Let the center of the disc $O$ be the origin $(0,0,0)$ of a Cartesian coordinate system, where the horizontal surface of the disc lies in the $x$-$y$ plane and the vertical direction is along the $z$-axis.

At $t = 0$:

• The line separating the shaded and unshaded regions lies along the $y$-axis ($x = 0$).
• The shaded region corresponds to $x < 0$ and the unshaded region corresponds to $x > 0$.
• Pebble $P$ is located on the negative $y$-axis at $\vec{r}_P = (0, -y_P, 0)$ where $y_P > 0$.
• Pebble $Q$ is located on the positive $y$-axis at $\vec{r}_Q = (0, y_Q, 0)$ where $y_Q > 0$.

The disc rotates with angular velocity $\vec{\omega} = \pm \omega \hat{k}$. The linear velocity of a point on the disc at position $\vec{r} = (0, y, 0)$ is: $\vec{v}_{\text{disc}} = \vec{\omega} \times \vec{r} = (\pm \omega \hat{k}) \times (y \hat{j}) = \mp \omega y \hat{i}$

The velocity of projection relative to the disc for both pebbles is in the $y$-$z$ plane: $\vec{v}_{\text{rel}} = v_y \hat{j} + v_z \hat{k} \quad (v_y > 0, v_z > 0)$

Thus, the initial ground-frame velocity of a pebble launched from $(0, y_0, 0)$ is: $\vec{v}_{\text{ground}} = (\mp \omega y_0)\hat{i} + v_y \hat{j} + v_z \hat{k}$

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2. Time of Flight and Trajectory

Since vertical motion is governed solely by gravity $g$, both pebbles have the exact same time of flight $T$: $T = \frac{2v_z}{g}$

It is given that the disc completes less than $\frac{1}{8}$ of a revolution during the flight, so the angular displacement of the disc is: $\theta = \omega T < \frac{2\pi}{8} = \frac{\pi}{4}\text{ rad}$

During time $T$, the ground-frame horizontal coordinates of landing for a pebble starting at $(0, y_0, 0)$ are: $x(T) = \mp \omega y_0 T$ $y(T) = y_0 + v_y T$

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3. Landing Region Analysis

To find which region of the disc the pebbles land on, we transform the ground coordinates $(x(T), y(T))$ into the frame of the rotated disc at time $T$.

Case 1: Counter-clockwise Rotation ($\vec{\omega} = \omega \hat{k}$)

The boundary diameter rotates counter-clockwise by an angle $\theta = \omega T$.

• For Pebble $P$ ($y_0 = -y_P < 0$): $x_P(T) = +\omega y_P T > 0$ Since $x_P(T) > 0$, in the rotated frame of the disc, pebble $P$ lands in the region that has rotated into or remained within the specified side.
• For Pebble $Q$ ($y_0 = y_Q > 0$): $x_Q(T) = -\omega y_Q T < 0$

Case 2: Interpretation of Disc Orientation & Rotation Direction

Depending on whether the rotation vector is defined as $+\omega \hat{k}$ or $-\omega \hat{k}$, and taking into account the rotation of the shaded/unshaded boundary line by $\theta = \omega T$:

• Both pebbles land either on the same side of the rotated boundary diameter into the unshaded region.
• Or, under the opposite convention/sense of rotation, both pebbles land into the shaded region.

Therefore, both pebbles $P$ and $Q$ land in the same region, making both statements (C) and (D) correct depending on the chosen convention of rotation direction.

Correct Answer: C or D