Starting from the centre of a circular path of radius R, a particle P chases another particle Q that

Q: Why is the angular velocity of P same as Q?

Because P is constrained to always lie on the straight line connecting the center to Q. If Q moves by an angle $\theta$, the line moves by $\theta$, so P must also move by $\theta$ in the same time.

Q: Why did we use polar coordinates?

The problem involves circular motion and a radial chase. Cartesian coordinates $(x,y)$ would make the math incredibly messy. Polar coordinates $(r, \theta)$ naturally align with the radial and tangential velocity components.

Q: What happens if $u < v$?

If the chaser is slower, the maximum radius it can reach is $\frac{u}{v}R$, which is less than $R$. It will never catch Q and will end up orbiting on a smaller inner circle.

Q: Why does the radial velocity become zero?

As P moves outwards, its radius $r$ increases. To maintain the same angular velocity $\omega$, its tangential speed $v_\theta = r\omega$ must increase. Since its total speed $u$ is constant, the radial speed must decrease to compensate, eventually hitting zero.

Q: Could P oscillate back and forth along the line?

Mathematically, the differential equation allows for oscillation. However, physically, P is 'chasing' Q. Once it reaches the maximum radius where its velocity is purely tangential, any inward movement would mean it's actively moving away from its target's path, which contradicts the nature of a chase. It settles into the circular orbit.

Question

Starting from the centre of a circular path of radius $R$, a particle P chases another particle Q that is moving with a uniform speed $v$ on the circular path. The chaser P moves with a constant speed $u$ and always remains collinear with the centre and the location of the chased Q.
(a) On which path will P eventually move and how long will it take to reach on this path? Consider the cases $u < v$, $u = v$ and $u > v$.
(b) If speeds of the particles are $v = 4 	ext{ m/s}$ and $u = 8 	ext{ m/s}$ and radius of the circular path is $R = 84 	ext{ m}$, how long P will take to reach Q. Use $\pi = 22/7$.

Sigma Percentile · Accepted Answer

### The Art of the Chase

Imagine a thrilling pursuit. A target, particle $Q$, is moving steadily along a circular track of radius $R$ with a constant speed $v$. Our chaser, particle $P$, starts right from the center of this circle. But $P$ isn't just running wildly; it follows a strict, almost robotic rule: it must always stay perfectly aligned on the straight line connecting the center to the target $Q$.

This single constraint changes everything. It means that as $Q$ sweeps out an angle $	heta$ along its circular path, $P$ must sweep out the exact same angle $	heta$ to stay collinear.

### Decoding the Kinematic Constraint

Because both particles cover the same angle in the same amount of time, their angular velocities must be identical. The angular velocity of the target $Q$ is simply its speed divided by its radius, $\omega_Q = \frac{v}{R}$. Therefore, the angular velocity of our chaser $P$ is also locked at $\omega_P = \frac{v}{R}$.

To understand $P$'s motion, we need to look at it through the lens of polar coordinates $(r, 	heta)$. In this system, any velocity vector is broken down into two perpendicular components:
1.  **Radial Velocity ($v_r$):** The speed at which the particle moves directly away from the center, given by $\frac{dr}{dt}$.
2.  **Tangential Velocity ($v_	heta$):** The speed at which the particle sweeps around the center, given by $r\omega$.

We are told that $P$ moves with a constant total speed $u$. According to the Pythagorean theorem, the sum of the squares of its velocity components must equal the square of its total speed:

$$v_r^2 + v_	heta^2 = u^2$$

Substituting our known expressions, we get the master equation for the chase:

$$\left(\frac{dr}{dt}ight)^2 + \left(r\frac{v}{R}ight)^2 = u^2$$

### The Master Differential Equation

This equation is the heart of the problem. Let's isolate the radial velocity, $\frac{dr}{dt}$, to see how $P$'s distance from the center changes over time:

$$\frac{dr}{dt} = \sqrt{u^2 - \left(\frac{rv}{R}ight)^2}$$

To find the actual path $r(t)$, we separate the variables and integrate:

$$\int \frac{dr}{\sqrt{u^2 - \left(\frac{rv}{R}ight)^2}} = \int dt$$

This might look intimidating, but it's a standard integral that yields an inverse sine function. By using the substitution $r\frac{v}{R} = u \sin\phi$, the integral beautifully simplifies, leading us to the trajectory equation:

$$r(t) = \frac{uR}{v} \sin\left(\frac{v}{R}tight)$$

### The Ultimate Fate of the Chaser

The equation $r(t)$ reveals that the chaser's distance from the center increases sinusoidally. But a sine function maxes out at $1$. This means the maximum radius $P$ can ever reach is $r_{max} = \frac{u}{v}R$.

This maximum occurs when the sine term is $1$, which happens at time $t = \frac{\pi R}{2v}$. At this exact instant, a fascinating physical shift occurs. The radial velocity $\frac{dr}{dt}$ drops to zero. All of $P$'s constant speed $u$ is now directed purely tangentially ($v_	heta = u$). Because it must maintain the angular velocity $\omega = \frac{v}{R}$ to stay collinear, and its tangential speed perfectly matches this requirement at this radius ($u = r_{max} \omega$), $P$ will stop moving outwards and simply orbit forever on this new circular path of radius $\frac{u}{v}R$.

### Catching the Target

Now, what if we actually want $P$ to catch $Q$? For a successful interception, $P$'s radius must reach $Q$'s radius, meaning $r(t) = R$.

Setting this up in our trajectory equation:

$$\frac{uR}{v} \sin\left(\frac{v}{R}tight) = R \implies \sin\left(\frac{v}{R}tight) = \frac{v}{u}$$

Notice that this is only mathematically possible if $u \geq v$. If the chaser is slower than the target, the sine value would need to be greater than $1$, which is impossible.

In part (b) of our problem, we are given $v = 4 	ext{ m/s}$, $u = 8 	ext{ m/s}$, and $R = 84 	ext{ m}$. Since $u > v$, a catch is guaranteed! Let's plug in the numbers:

$$\sin\left(\frac{4}{84}tight) = \frac{4}{8} = \frac{1}{2}$$

The angle whose sine is $\frac{1}{2}$ is $\frac{\pi}{6}$. So:

$$\frac{t}{21} = \frac{\pi}{6}$$

Solving for $t$ and using the approximation $\pi = \frac{22}{7}$:

$$t = 21 	imes \frac{22/7}{6} = 11 	ext{ seconds}$$

And there we have it. In exactly 11 seconds, the relentless chaser finally catches its prey.

Visualized Solution

$Kinematic Setup$

$Angular Velocity Constraint$

$Velocity Components of P$

$Constant Speed Condition$

$Differential Equation for Path$

$Integrating the Equation$

$Ultimate Path of P$

$Catching Particle Q$

$Calculating the Time$

The Sigma Insight: Kinematics of Circular Motion

The Art of the Chase

Decoding the Kinematic Constraint

The Master Differential Equation

The Ultimate Fate of the Chaser

Catching the Target

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Visualized Solution

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

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

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

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

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

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

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

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

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