Consider a disc rotating in the horizontal plane with a constant angular speed omega about its centr

Q: Why does pebble P have an initial x-velocity in the ground frame?

Pebble P is located at the edge of the rotating disc. Even before it is thrown, it is moving tangentially with the disc at a speed of $\omega R$. When thrown, it retains this tangential velocity in the ground frame due to inertia.

Q: Why do we need to transform the coordinates back to the disc's frame?

The shaded and unshaded regions are painted on the disc, meaning they rotate with it. To determine where the pebbles land relative to these regions, we must find their final positions in the coordinate system that rotates with the disc.

Q: How do we know pebble Q always lands in the unshaded region?

In the ground frame, Q moves straight along the positive y-axis. Meanwhile, the disc rotates counter-clockwise. From the perspective of the disc, the pebble appears to curve to the right (into the positive x-region), which is the unshaded region.

Q: What does the constraint 'range is less than half the disc radius' tell us?

It tells us that $v_y t < R/2$. However, this is an upper bound. It doesn't give us a lower bound for the range, which is why we cannot determine if the range is large enough to overcome the negative drift caused by the disc's rotation for pebble P.

Q: Why was this question awarded a bonus in the exam?

Because the final landing spot of pebble P depends on the exact magnitude of its initial velocity, which is not provided. Without this data, multiple options could be correct depending on the assumed velocity, making the question ambiguous.

Question

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

Options:

(A) $P$ lands in the shaded region and $Q$ in the unshaded region

(B) $P$ lands in the unshaded region and $Q$ in the shaded region

(C) both $P$ and $Q$ land in the unshaded region

(D) both $P$ and $Q$ land in the shaded region

Sigma Percentile · Accepted Answer

The problem we are looking at today is a fascinating one from JEE Advanced 2012. It is a beautiful mix of kinematics and rotating reference frames, but it also comes with a twist—a twist that made it one of the most debated questions of its year! Let's dive deep into the physics of this rotating disc and uncover why this question left both students and professors scratching their heads. ### Analyzing the Setup Imagine a large disc rotating counter-clockwise in a horizontal plane with a constant angular speed $\omega$. The disc is divided into two halves by a diameter along the y-axis. The left half (where $x < 0$) is shaded, and the right half (where $x > 0$) is unshaded. We have two pebbles, $P$ and $Q$. Pebble $Q$ is sitting right at the center of the disc (the origin), while pebble $P$ is at the bottom edge of the disc, at coordinates $(0, -R)$. Both pebbles are projected simultaneously towards a point $R$ located at the top edge of the disc, at $(0, R)$. The problem states that the velocity of projection is in the y-z plane and is the same for both pebbles **with respect to the disc**. This is a crucial detail! It means that to an observer standing on the rotating disc, both pebbles are thrown exactly the same way. However, to understand where they will land, we need to analyze their motion from the perspective of a stationary observer standing on the ground. ### The Master Equation: Ground Frame Velocity To track the pebbles, we must first find their absolute velocities in the stationary ground frame. The absolute velocity of any object thrown from a moving platform is the vector sum of its velocity relative to the platform and the velocity of the platform itself at the point of projection. Mathematically, this is written as: $\vec{v}_{abs} = \vec{v}_{disc} + \vec{v}_{rel}$ Let the relative velocity of projection be $\vec{v}_{rel} = v_y \hat{j} + v_z \hat{k}$. Since the pebbles are projected towards $R$ (which is along the positive y-axis), we know that $v_y > 0$. Now, let's look at the velocity of the disc at the projection points. The velocity of any point on the disc is given by $\vec{v}_{disc} = \vec{\omega} imes \vec{r}$. For pebble $Q$ at the center ($\vec{r} = 0$), the disc is not moving! Therefore, its absolute velocity is simply its relative velocity: $\vec{v}_{Q, abs} = v_y \hat{j} + v_z \hat{k}$ For pebble $P$ at the bottom edge ($\vec{r} = -R \hat{j}$), the disc has a tangential velocity. Using the cross product, $\vec{\omega} imes \vec{r} = (\omega \hat{k}) imes (-R \hat{j}) = \omega R \hat{i}$. So, pebble $P$ gets an extra horizontal push in the positive x-direction! Its absolute velocity is: $\vec{v}_{P, abs} = \omega R \hat{i} + v_y \hat{j} + v_z \hat{k}$ ### Tracking Pebble Q Let's follow pebble $Q$ first. In the ground frame, it has no x-velocity. It simply moves straight along the y-axis. At any time $t$, its coordinates in the ground frame are $x = 0$ and $y = v_y t$. But remember, while pebble $Q$ is flying through the air, the disc is rotating underneath it! By the time the pebble lands, the disc has rotated by an angle $ heta = \omega t$. To find where $Q$ lands on the disc, we need to transform its ground coordinates back to the disc's rotating frame. The x-coordinate on the disc, let's call it $x'$, is given by the rotation matrix formula: $x' = x \cos heta + y \sin heta$ Substituting $x = 0$ and $y = v_y t$, we get: $x' = v_y t \sin(\omega t)$ Since $v_y$ is positive and the angle $\omega t$ is small (less than $\pi/4$ as per the problem), $\sin(\omega t)$ is also positive. Therefore, $x'$ is strictly positive. This means pebble $Q$ will always land in the region where $x > 0$, which is the **unshaded region**. ### Tracking Pebble P Now comes the tricky part—pebble $P$. Because of the disc's rotation, $P$ was launched with an initial x-velocity of $\omega R$. In the ground frame, its position at time $t$ will be: $x = \omega R t$ $y = -R + v_y t$ Again, we transform these coordinates back to the rotating disc to find its landing spot: $x' = x \cos heta + y \sin heta$ $x' = (\omega R t) \cos(\omega t) + (-R + v_y t) \sin(\omega t)$ Let's rearrange this to group the terms with $R$ and the terms with $v_y t$: $x' = R (\omega t \cos(\omega t) - \sin(\omega t)) + v_y t \sin(\omega t)$ To make sense of this, we can use small-angle approximations (since $ heta = \omega t$ is small). We know that $\sin heta \approx heta - heta^3/6$ and $\cos heta \approx 1 - heta^2/2$. Substituting these in, the term $( heta \cos heta - \sin heta)$ simplifies to approximately $- heta^3/3$. So, the x-coordinate on the disc becomes: $x' \approx -R \frac{ heta^3}{3} + v_y t heta$ $x' \approx heta \left( v_y t - R \frac{ heta^2}{3} ight)$ ### The Ambiguity Revealed Look closely at that final expression for $x'$. The sign of $x'$ dictates whether pebble $P$ lands in the shaded or unshaded region. And this sign depends entirely on the term inside the bracket: $\left( v_y t - R \frac{ heta^2}{3} ight)$. Here is where the problem falls apart. The term $v_y t$ represents the horizontal range of the pebble relative to the disc. The problem states that this range is "less than half the disc radius" ($v_y t < R/2$). However, this constraint is not tight enough! - If the range $v_y t$ is relatively large (say, close to $R/2$), the bracket evaluates to a positive number. In this case, $x' > 0$, and pebble $P$ lands in the **unshaded region**. - But what if the pebble was thrown with a very small velocity? If the range $v_y t$ is tiny (smaller than $R heta^2/3$), the bracket evaluates to a negative number. In this case, $x' < 0$, and pebble $P$ lands in the **shaded region**. Because the exact initial velocity of projection is not provided, we cannot definitively say where pebble $P$ will land. It could be either! ### Conclusion This question is a brilliant test of relative motion and coordinate transformations, but it lacks the sufficient data required to reach a single, unambiguous answer. Depending on the exact speed of the throw, both option (a) and option (c) could be correct. Recognizing this flaw, the exam authorities ultimately awarded a bonus star (*) for this question. It serves as a great reminder that in physics, the exact values of parameters can sometimes completely change the qualitative outcome of a system!

Select Answer:

Visualized Solution

$Analyzing the Rotating Disc$

$Velocity in Ground Frame$

$Trajectory of Pebble Q$

$Landing Position of Q$

$Trajectory of Pebble P$

$Position of P in Ground Frame$

$Landing Position of P$

$The Ambiguity$

$Conclusion$

The Sigma Insight: Kinematics of Rotational Motion

Analyzing the Setup

The Master Equation

Tracking Pebble Q

Tracking Pebble P

The Ambiguity Revealed

Conclusion

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A body rotating with an angular speed of $600 rpm$ is uniformly accelerated to $1800 rpm$ in $10 s$ . The number of rotations made in the process is …… .

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Rotating Disc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity in Ground Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Trajectory of Pebble Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Landing Position of Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Trajectory of Pebble P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position of P in Ground Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Landing Position of P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Ambiguity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion