Consider a disc rotating in the horizontal plane with a constant angular speed $\omega$ about its centre $O$. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles $P$ and $Q$ are simultaneously projected at an angle towards $R$. The velocity of projection is in the $y$-$z$ plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed $1/8$ rotation, (ii) their range is less than half the disc radius, and (iii) $\omega$ remains constant throughout. Then

Step 1: Define the coordinate system and initial parameters.

Let the disc lie in the horizontal $x$-$y$ plane, rotating with constant angular velocity $\vec{\omega} = \omega \hat{z}$ about its center $O$. The diameter $R$-$P$ lies along the $y$-axis, separating the shaded region ($x < 0$) from the unshaded region ($x > 0$).

Both pebbles are projected in the $y$-$z$ plane towards $R$ with the same relative velocity:

$$\vec{v}_{\text{rel}} = (0, u_y, u_z)$$

Step 2: Determine the initial velocity in the ground frame.

For a pebble projected from a point on the diameter at a distance $\rho$ from the center $O$, its initial position vector is $\vec{r} = (0, \rho, 0)$. The velocity of this point on the disc is:

$$\vec{v}_{\text{disc}} = \vec{\omega} \times \vec{r}$$ $$\vec{v}_{\text{disc}} = \omega \hat{z} \times (\rho \hat{y}) = -\omega \rho \hat{x}$$

The initial velocity of the pebble in the ground frame is:

$$\vec{v}_{\text{ground}} = \vec{v}_{\text{rel}} + \vec{v}_{\text{disc}}$$ $$\vec{v}_{\text{ground}} = (-\omega \rho, u_y, u_z)$$

Step 3: Find the landing position in the ground frame.

The time of flight $T$ is determined solely by the vertical motion and is identical for both pebbles:

$$T = \frac{2u_z}{g}$$

In the ground frame, the horizontal coordinates of the landing point after time $T$ are:

$$X = -\omega \rho T$$ $$Y = \rho + u_y T$$

Step 4: Transform the landing coordinates to the rotating disc frame.

During the flight, the disc rotates through an angle $\phi = \omega T$. To find the landing position $x'$ in the rotating frame of the disc, we rotate the ground coordinates $(X, Y)$ by $-\phi$:

$$x' = X \cos\phi + Y \sin\phi$$ $$x' = -\omega \rho T \cos(\omega T) + (\rho + u_y T) \sin(\omega T)$$

Step 5: Apply small-angle approximations.

Since the flight time is short ($T < \frac{1}{8}$ of a rotation), the angle $\phi = \omega T$ is small. Using the Taylor expansions $\sin\phi \approx \phi - \frac{\phi^3}{6}$ and $\cos\phi \approx 1 - \frac{\phi^2}{2}$:

$$x' \approx -\omega \rho T \left(1 - \frac{\omega^2 T^2}{2}\right) + (\rho + u_y T) \left(\omega T - \frac{\omega^3 T^3}{6}\right)$$ $$x' \approx -\omega \rho T + \frac{1}{2}\omega^3 \rho T^3 + \rho \omega T - \frac{1}{6}\omega^3 \rho T^3 + u_y \omega T^2 - \frac{1}{6} u_y \omega^3 T^3$$

Simplifying the expression by cancelling the leading terms:

$$x' \approx \omega u_y T^2 + \frac{1}{3}\omega^3 \rho T^3 - \frac{1}{6} u_y \omega^3 T^3$$

Since $\omega T \ll 1$, the dominant term is:

$$x' \approx \omega u_y T^2$$

Since $\omega > 0$ and $u_y > 0$, we have:

$$x' > 0$$

Alternatively, in the rotating frame of reference, both pebbles experience a Coriolis force in the $+x'$ direction:

$$\vec{F}_{\text{Coriolis}} = -2m(\vec{\omega} \times \vec{v}_{\text{rel}})$$ $$\vec{F}_{\text{Coriolis}} = -2m(\omega \hat{z} \times u_y \hat{y}) = 2m\omega u_y \hat{x}$$

This force deflects both pebbles into the unshaded region.

Answer: Both $P$ and $Q$ land in the unshaded region.