Two identical discs of same radius R are rotating about their axes in opposite directions with the s

Q: Why is the initial relative velocity zero?

At $t=0$, both points $P$ and $Q$ are at the innermost edges of their respective discs. Due to the opposite directions of rotation, both points are moving tangentially downwards with the exact same speed $R\omega$. Since their velocity vectors are identical, their relative velocity is zero.

Q: Why do the vertical components of velocity cancel out?

The discs rotate in opposite directions, but the points $P$ and $Q$ start from opposite sides of their respective centers. This symmetry causes their vertical displacements to be identical at all times, leading to identical vertical velocity components that cancel out when subtracted.

Q: What does the absolute value in the magnitude function do to the graph?

The absolute value function $|\sin(\omega t)|$ ensures the magnitude is always positive. When the sine function crosses zero and becomes negative, the absolute value flips it to positive. This creates a "bouncing" effect on the graph, resulting in sharp, non-differentiable points (cusps) at $v_r = 0$.

Q: Why is graph (b) incorrect?

Graph (b) shows smooth minima at $v_r = 0$. A smooth minimum would imply a function like $1 - \cos(2\omega t)$ or $\sin^2(\omega t)$, where the derivative is zero at the minimum. However, our function is $|\sin(\omega t)|$, which has an undefined derivative at zero, creating sharp cusps as seen in graph (a).

Q: At what times is the relative velocity maximum?

The relative velocity is maximum when $|\sin(\omega t)| = 1$. This occurs at $\omega t = \pi/2, 3\pi/2$, etc., which corresponds to $t = T/4, 3T/4$. At these times, the points are moving in exactly opposite horizontal directions.

Question

Two identical discs of same radius $R$ are rotating about their axes in opposite directions with the same constant angular speed $\omega$. The discs are in the same horizontal plane.

At time $t = 0$, the points $P$ and $Q$ are facing each other as shown in the figure. The relative speed between the two points $P$ and $Q$ is $v_r$. In one time period ($T$) of rotation of the discs, $v_r$ as a function of time is best represented by

Note: Language of the question is wrong. Magnitude of relative velocity should be asked.

Sigma Percentile · Accepted Answer

Imagine standing above two identical spinning discs, watching two specific points, $P$ and $Q$, as they trace their circular paths. This problem is a beautiful exercise in kinematics, challenging us to translate a dynamic physical system into a precise mathematical graph. Let's break down this elegant dance step by step.

### Visualizing the Initial State

Before diving into equations, let's build a strong physical intuition. At time $t = 0$, the points $P$ and $Q$ are facing each other. The left disc is rotating clockwise, which means point $P$, located at the rightmost edge of its disc, is moving straight down.

Simultaneously, the right disc is rotating anti-clockwise. Point $Q$, located at the leftmost edge of its disc, is also moving straight down. Both points are moving in the exact same direction with the exact same speed, $R\omega$.

Because their velocity vectors are identical, their relative velocity at this instant is perfectly zero. If you were sitting on point $P$, point $Q$ would appear completely stationary to you at $t = 0$.

### The Mathematical Formulation

To find the relative velocity at any arbitrary time $t$, we need to write down the position vectors of both points. Let's set up a coordinate system at the center of each disc.

For the left disc, point $P$ starts at an angle of $0^\circ$ and rotates clockwise. Its angular position at time $t$ is $-\omega t$. Therefore, its position vector is:

$\vec{r}_P = R\cos(\omega t)\hat{i} - R\sin(\omega t)\hat{j}$

Taking the derivative with respect to time gives us the velocity of $P$:

$\vec{v}_P = -R\omega\sin(\omega t)\hat{i} - R\omega\cos(\omega t)\hat{j}$

Now, let's look at the right disc. Point $Q$ starts at an angle of $180^\circ$ (or $\pi$ radians) and rotates anti-clockwise. Its angular position is $\pi + \omega t$. Its position vector is:

$\vec{r}_Q = R\cos(\pi + \omega t)\hat{i} + R\sin(\pi + \omega t)\hat{j} = -R\cos(\omega t)\hat{i} - R\sin(\omega t)\hat{j}$

Differentiating this gives the velocity of $Q$:

$\vec{v}_Q = R\omega\sin(\omega t)\hat{i} - R\omega\cos(\omega t)\hat{j}$

### The Magic of Cancellation

Now comes the most satisfying part of the problem. We want the relative velocity, $\vec{v}_r = \vec{v}_P - \vec{v}_Q$.

Notice the vertical ($\hat{j}$) components of both velocities. They are exactly the same: $-R\omega\cos(\omega t)\hat{j}$. When we subtract $\vec{v}_Q$ from $\vec{v}_P$, these vertical components perfectly cancel each other out!

$\vec{v}_r = (-R\omega\sin(\omega t) - R\omega\sin(\omega t))\hat{i}$

$\vec{v}_r = -2R\omega\sin(\omega t)\hat{i}$

The relative velocity is purely horizontal at all times. This is a profound result of the symmetry in the system.

### Decoding the Graph

The question asks for the magnitude of this relative velocity, which is the speed. We take the absolute value of our vector:

$v_r = |-2R\omega\sin(\omega t)| = 2R\omega|\sin(\omega t)|$

This mathematical function is known as a full-wave rectified sine wave. Let's analyze its key features to identify the correct graph:

1. **Zeros:** The function is zero whenever $\sin(\omega t) = 0$. This happens at $t = 0, T/2, T$, etc.
2. **Peaks:** The function reaches its maximum value of $2R\omega$ when $|\sin(\omega t)| = 1$, which occurs at $t = T/4, 3T/4$, etc.
3. **The Cusps:** Because of the absolute value, the graph doesn't smoothly cross the horizontal axis. Instead, it abruptly bounces back up. This creates sharp, non-differentiable points, or "cusps," at the zeros.

Looking at the given options, graph (a) perfectly captures all these features. It starts at zero, peaks at $T/4$, and has sharp cusps at $T/2$ and $T$. Graph (b) is incorrect because it shows smooth minima, which would imply a different mathematical function.

By systematically breaking down the motion into vectors, we transformed a complex visual problem into a simple, undeniable mathematical truth.

Select Answer:

Visualized Solution

$Initial State at t = 0$

$Velocity of P at time t$

$Velocity of Q at time t$

$Relative Velocity Vector$

$Magnitude of Relative Velocity$

$Graph Analysis$

The Sigma Insight: Kinematics of Circular Motion

Visualizing the Initial State

The Mathematical Formulation

The Magic of Cancellation

Decoding the Graph

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State at t=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of P at time t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of Q at time t

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

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

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