A particle of mass m, moving in a circular path of radius R with a constant speed v2 is located at p

Q: Why is the center of the circle at $(R, 0)$ and not the origin?

The problem states the particle is at $(2R, 0)$ at $t=0$ and the diagram shows the circle touching the origin. For a circle of radius $R$ to pass through $(0,0)$ and $(2R, 0)$, its center must be exactly halfway between them on the X-axis, which is $(R, 0)$.

Q: How do we know the angular velocity $\omega$ is constant?

The problem explicitly states that the particle moves with a "constant speed $v_2$". Since $v = R\omega$ and both $v_2$ and $R$ are constant, the angular velocity $\omega$ must also be constant.

Q: Why is the particle's velocity vector $\vec{v}_p = -v_2 \sin \theta \hat{i} + v_2 \cos \theta \hat{j}$?

The velocity vector is always tangential to the circular path. If the radius vector makes an angle $\theta$ with the X-axis, the tangent makes an angle of $\theta + 90^\circ$. Using trigonometry, $\cos(\theta + 90^\circ) = -\sin \theta$ and $\sin(\theta + 90^\circ) = \cos \theta$, giving us those exact components.

Q: Does the relative momentum vector change its magnitude over time?

Yes! Because the particle's velocity direction is constantly changing while the man's velocity is constant, the vector difference between them changes in both magnitude and direction over time.

Q: Can we solve this by shifting to a rotating frame of reference?

While you could use a rotating frame attached to the particle, it would introduce complex pseudo-forces (Coriolis and Centrifugal). Since we only need the relative velocity, staying in the ground frame and simply subtracting the velocity vectors is much more elegant and less prone to errors.

Question

A particle of mass $m$, moving in a circular path of radius $R$ with a constant speed $v_2$ is located at point $(2R, 0)$ at time $t = 0$ and a man starts moving with a velocity $v_1$ along the positive Y-axis from origin at time $t = 0$. Calculate the linear momentum of the particle w.r.t. man as a function of time.

Sigma Percentile · Accepted Answer

The concept of relative motion often feels intuitive when objects move in straight lines. But what happens when one object moves in a circle while the observer walks in a straight line? This problem is a beautiful exploration of exactly that scenario.

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate system. You start walking straight up the Y-axis with a steady speed $v_1$. At the exact same moment, a particle located at $(2R, 0)$ starts moving along a circular path of radius $R$.

The geometry here is crucial. Since the circle has a radius $R$ and passes through both the origin $(0,0)$ and $(2R, 0)$, its center must be perfectly nestled at $(R, 0)$.

### The Particle's Circular Journey

Let's focus purely on the particle for a moment. It's moving with a constant speed $v_2$. In the world of circular motion, linear speed and angular speed are intimately connected by the relation $v = R\omega$.

Therefore, the angular velocity of our particle is:
$$ \omega = \frac{v_2}{R} $$

Since it moves with a constant angular velocity, the angle $	heta$ it sweeps out from the center $(R, 0)$ at any time $t$ is simply:
$$ 	heta = \omega t = \frac{v_2}{R} t $$

### The Velocity Vectors

To find relative momentum, we first need the absolute velocity vectors of both the particle and the man.

**The Man:**
This is the easy part. The man is strolling along the positive Y-axis with speed $v_1$. His velocity vector is:
$$ \vec{v}_m = v_1 \hat{j} $$

**The Particle:**
This requires a bit more spatial visualization. The particle is at an angle $	heta$ on the circle. Its velocity vector $\vec{v}_p$ is always tangential to the path. If you draw the tangent, you'll see it makes an angle of $	heta + 90^\circ$ with the positive X-axis.

Resolving this into components, we get:
$$ \vec{v}_p = -v_2 \sin 	heta \hat{i} + v_2 \cos 	heta \hat{j} $$

Substituting our expression for $	heta$, the particle's velocity as a function of time becomes:
$$ \vec{v}_p = -v_2 \sin(\omega t) \hat{i} + v_2 \cos(\omega t) \hat{j} $$

### The Relative Perspective

Now, we bring the two motions together. The velocity of the particle *with respect to* the man is the vector difference of their absolute velocities:
$$ \vec{v}_{pm} = \vec{v}_p - \vec{v}_m $$

Substituting our vectors:
$$ \vec{v}_{pm} = \left( -v_2 \sin(\omega t) \hat{i} + v_2 \cos(\omega t) \hat{j} ight) - v_1 \hat{j} $$

Grouping the $\hat{j}$ components together, we get the relative velocity:
$$ \vec{v}_{pm} = -v_2 \sin(\omega t) \hat{i} + (v_2 \cos(\omega t) - v_1) \hat{j} $$

### Final Calculation: Relative Momentum

Momentum is simply mass times velocity. To find the linear momentum of the particle with respect to the man, we multiply our relative velocity vector by the mass $m$:

$$ \vec{p}_{pm} = m \vec{v}_{pm} $$

$$ \vec{p}_{pm} = m \left[ -v_2 \sin(\omega t) \hat{i} + (v_2 \cos(\omega t) - v_1) \hat{j} ight] $$

And there we have it! A dynamic, time-varying vector that perfectly describes how the particle's momentum appears to the moving observer. The beauty of this result lies in how it elegantly captures the interplay between linear and circular kinematics.

Visualized Solution

The Physical Setup

Angular Kinematics

Velocity of the Particle $(v_{p})$

Velocity of the Man $(v_{m})$

Relative Velocity $(v_{p m})$

Relative Linear Momentum $(p_{p m})$

The Sigma Insight: Relative Velocity

Analyzing the Setup

The Particle's Circular Journey

The Velocity Vectors

The Relative Perspective

Final Calculation

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