Animated Solution for Physics - Kinematics: Two particles $A$ and $B$ are moving on two concentric circles of radii $R_1$ and $R_2$ with equal angular speed $\omega$. At $t=0$, their positions and direction of motion are shown in the figure. The relative velocity ${\mathbf{v}}_A-{\mathbf{v}}_B$ at $t=\frac{\pi }{2\omega }$ is given by

The problem of finding the relative velocity of two particles moving in circular paths is a classic test of your understanding of kinematics and vector addition. It might look intimidating at first glance, but once we break it down into its core components, the solution reveals itself beautifully.

Imagine you are looking down at two particles, $A$ and $B$, moving on concentric circular tracks. Particle $A$ is on the inner track with radius $R_1$, and particle $B$ is on the outer track with radius $R_2$.

At the very beginning, when $t=0$, both particles are perfectly aligned on the positive x-axis. However, they are moving in opposite directions! Particle $A$ is moving counter-clockwise, while particle $B$ is moving clockwise. Despite moving in opposite directions, they share a crucial similarity: they both have the exact same angular speed, $\omega$.

To find their relative velocity at a specific time, we first need to figure out exactly where they are at that moment. In circular motion, the angular displacement $\theta$ (the angle covered) is simply the product of the angular speed $\omega$ and the time $t$.

This is our master equation for finding their new positions.

The problem asks for the relative velocity at a very specific time: $t=\frac{\pi }{2\omega }$. Let's substitute this time into our master equation to see how much angle each particle has covered.

Notice how elegantly the $\omega$ terms cancel out! We are left with:

Since $\frac{\pi }{2}$ radians is exactly $90^{\circ }$, this means both particles have completed exactly one quarter of their respective circular paths.

Particle $A$, moving counter-clockwise from the positive x-axis, will land squarely on the positive y-axis at the coordinate $(0,R_1)$. Particle $B$, moving clockwise from the positive x-axis, will land on the negative y-axis at the coordinate $(0,-R_2)$.

Now that we know where they are, we need to determine their velocity vectors. In uniform circular motion, the velocity vector is always tangent to the circular path, and its magnitude is given by $v=\omega R$.

For particle $A$ at the top of its circle $(0,R_1)$, the tangent points directly to the left, which is the negative x-direction. Therefore, its velocity vector is:

For particle $B$ at the bottom of its circle $(0,-R_2)$, it is moving clockwise. The tangent at the bottom of a clockwise circle also points directly to the left! Therefore, its velocity vector is:

The final step is to find the relative velocity of $A$ with respect to $B$, which is defined as ${\mathbf{v}}_A-{\mathbf{v}}_B$. Let's substitute the vectors we just found:

Be very careful with the minus signs here. The double negative becomes a positive:

Factoring out the common terms, we arrive at our final, elegant result:

This perfectly matches option (d). By carefully tracking the positions and tangent vectors, a seemingly complex 2D motion problem collapses into a straightforward 1D vector subtraction!