Dynamics

Motion of a particle in a plane

Motion of a particle in a plane governs kinematics and dynamics of particles along two-dimensional trajectories using Cartesian and polar coordinates.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Kinematics and Dynamics of Plane Motion

Let the position vector of a moving particle in a plane at time $t$ be given by $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$. The velocity vector is $\vec{v}(t) = \frac{d\vec{r}}{dt} = \dot{x}\hat{i} + \dot{y}\hat{j}$, and the acceleration vector is $\vec{a}(t) = \frac{d\vec{v}}{dt} = \ddot{x}\hat{i} + \ddot{y}\hat{j}$.

📐 Geometric Interpretation

The position vector traces a curve in the $xy$-plane. The velocity vector $\vec{v}$ is always tangent to the trajectory at any instant, while the acceleration vector $\vec{a}$ points towards the concavity of the path, resolvable into tangential and normal components.

⚙️ Algebraic & Structural Meaning

Vector decomposition allows decoupled scalar differential equations along orthogonal axes ($x$ and $y$) or radial and transverse directions ($r$ and $\theta$). Initial value problems reduce to successive integrations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Position Vector Differentiability	`Time interval $t \in [t_1, t_2]$`	Strictly Mandatory / Necessary	If $x(t)$ or $y(t)$ fails to be differentiable, velocity and acceleration become undefined at singular points.
Continuity of Acceleration	`Open interval $(t_1, t_2)$`	Strictly Mandatory	Discontinuous forces lead to impulsive motion where standard kinematic equations fail without Dirac delta formulations.
Coordinate Independence	`Global coordinate domain`	Essential for transformation	Failure to properly account for moving frames introduces fictitious inertial forces (Coriolis and centrifugal).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Radial and Transverse Components of Velocity

$$v_r = \dot{r}, \quad v_\theta = r\dot{\theta}$$

Rule: In polar coordinates, radial velocity is the rate of change of radius, and transverse velocity is radius times angular velocity.

Application: Frequently asked in UP PGT for central orbit and projectile trajectory problems.

Radial and Transverse Components of Acceleration

$$a_r = \ddot{r} - r\dot{\theta}^2, \quad a_\theta = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$$

Rule: Directly memorize the standard acceleration formulas in polar coordinates to bypass tedious coordinate transformations.

Application: Direct formula identification in TGT/PGT vector mechanics.

Tangential and Normal Acceleration

$$a_t = \frac{dv}{dt} = v \frac{dv}{ds}, \quad a_n = \frac{v^2}{\rho}$$

Rule: Use $a_t$ for rate of change of speed and $a_n = \frac{v^2}{\rho}$ where $\rho$ is the radius of curvature for centripetal effects.

Application: Standard numericals on circular motion and path curvature in DSSSB PGT.

Angular Momentum Conservation in Central Orbits

$$h = r^2\dot{\theta} = \text{constant}$$

Rule: When central force acts, transverse acceleration $a_\theta = 0$, implying areal velocity $\frac{1}{2}r^2\dot{\theta}$ is constant.

Projectile Range and Maximum Height

$$R = \frac{u^2 \sin(2\alpha)}{g}, \quad H = \frac{u^2 \sin^2\alpha}{2g}$$

Rule: For complementary angles of projection ($ \alpha $ and $90^\circ - \alpha$), the horizontal range is identical.

Equation of Trajectory of a Projectile

$$y = x \tan\alpha - \frac{g x^2}{2u^2 \cos^2\alpha} = x \tan\alpha \left(1 - \frac{x}{R}\right)$$

Rule: Use the Cartesian form of trajectory to instantly find angle of projection or velocity given a point $(x,y)$ on the path.

Application: Standard problem format in UP PGT.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Total Acceleration with Radial Component

Common Pitfall: Candidates often forget the centrifugal term $-r\dot{\theta}^2$ when calculating radial acceleration.

Counterexample: For uniform circular motion ($r = a = \text{const}$), $\ddot{r} = 0$, but $a_r = -a\dot{\theta}^2 \neq 0$.

Why it fails: Ignoring centripetal acceleration leads to completely incorrect force equations.

🛡️ Exam Defense: Always write full polar acceleration terms: $a_r = \ddot{r} - r\dot{\theta}^2$ and $a_\theta = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta}).$

⚠️ COMMON ERROR #02

Speed vs Velocity Magnitude Misinterpretation

Common Pitfall: Equating the time derivative of speed vector components directly without taking vector magnitude.

Counterexample: Speed $v = \sqrt{\dot{x}^2 + \dot{y}^2}$, hence $\frac{dv}{dt} \neq \sqrt{\ddot{x}^2 + \ddot{y}^2}$.

Why it fails: Magnitude of acceleration vector $|\vec{a}| = \sqrt{\ddot{x}^2 + \ddot{y}^2}$, which equals tangential acceleration only in straight-line motion.

🛡️ Exam Defense: Differentiate scalar speed for tangential acceleration; use vector magnitude for total acceleration.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Cartesian Kinematics	Independent orthogonal axes	$v = \dot{x}\hat{i} + \dot{y}\hat{j}$	Motion resolved along fixed $X$ and $Y$ grids	Given $x(t)$ and $y(t)$ explicitly as functions of time
Polar Kinematics	Central or angular coordinates	$v_r = \dot{r}, v_\theta = r\dot{\theta}$	Motion resolved along radius vector and perpendicular	Given $r(t)$ and $\theta(t)$ or central forces
Tangential & Normal	Curvilinear path dynamics	$a_t = \dot{v}, a_n = \frac{v^2}{\rho}$	Decomposition along tangent and curvature center	Given radius of curvature $\rho$ or speed along curve
Projectile Motion	Constant downward gravity $g$	$y = x\tan\alpha - \frac{gx^2}{2u^2\cos^2\alpha}$	Parabolic trajectory under uniform gravity	Words like 'projection', 'range', 'maximum height'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the position of a particle at time $t$ is given by $\vec{r} = a \cos t \hat{i} + a \sin t \hat{j}$, find the magnitude of its acceleration.

(A) $a$(B) $at$(C) $a \sin t$(D) $0$

📚 Method A: Formal Textbook Derivation

Step 1: Find velocity $\vec{v} = \frac{d\vec{r}}{dt} = -a \sin t \hat{i} + a \cos t \hat{j}$. Step 2: Find acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -a \cos t \hat{i} - a \sin t \hat{j}$. Step 3: Magnitude $|\vec{a}| = \sqrt{(-a\cos t)^2 + (-a\sin t)^2} = a$.

⚡ Method B: 25-Second Exam Shortcut

Recognize that $\vec{r} = a \cos t \hat{i} + a \sin t \hat{j}$ represents uniform circular motion of radius $a$ with unit angular speed, hence $a_n = \frac{v^2}{\rho} = \frac{a^2}{a} = a$.

💡 Tip: Whenever position components involve $\sin t$ and $\cos t$ with equal amplitudes, it is uniform circular motion.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The radial velocity of a particle moving in a plane is proportional to its transverse velocity, and its angular velocity is constant. The path of the particle is:

(A) Circle(B) Parabola(C) Equiangular Spiral(D) Straight Line

📚 Method A: Formal Textbook Derivation

Step 1: Given $\dot{r} = k (r\dot{\theta})$, where $\dot{\theta} = \omega$ (constant). Step 2: $\frac{dr}{dt} = k r \omega \implies \frac{dr}{r} = k\omega dt$. Step 3: Integrate to get $\ln r = k\omega t + c \implies r = Ae^{m\theta}$. This is the equation of an equiangular spiral.

⚡ Method B: 25-Second Exam Shortcut

Proportionality between radial and transverse velocity with constant angular velocity directly characterizes logarithmic spirals in polar mechanics.

💡 Tip: Memorize standard curves in polar coordinates: $r = a(1+\cos\theta)$ (cardioid), $r\theta = a$ (hyperbolic spiral), $r = ae^{m\theta}$ (equiangular spiral).

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If the position vector of a particle is $\vec{r}(t) = t^3\hat{i} + t^2\hat{j}$, the acceleration vector at $t = 1$ is:

A $6\hat{i} + 2\hat{j}$

B $3\hat{i} + 2\hat{j}$

C $6\hat{i} + t\hat{j}$

D $3\hat{i} + t^2\hat{j}$

Derivation & Analysis:

Velocity $\vec{v} = 3t^2\hat{i} + 2t\hat{j}$. Acceleration $\vec{a} = 6t\hat{i} + 2\hat{j}$. At $t=1$, $\vec{a} = 6\hat{i} + 2\hat{j}$.

MCQ 2: In polar coordinates, the transverse component of acceleration is given by:

A $\ddot{r} - r\dot{\theta}^2$

B $\frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$

C $r\ddot{\theta} + \dot{r}\dot{\theta}$

D $\ddot{\theta}$

Derivation & Analysis:

By standard derivation of polar kinematics, transverse acceleration is $a_\theta = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta}) = 2\dot{r}\dot{\theta} + r\ddot{\theta}$.

MCQ 3: If a particle describes a central orbit under a force $F$, then which of the following is constant?

A Radial velocity

B Angular momentum

C Kinetic energy

D Radial acceleration

Derivation & Analysis:

Since central force has zero torque about the center, angular momentum $h = r^2\dot{\theta}$ is conserved.

MCQ 4: The maximum horizontal range of a projectile with initial velocity $u$ under gravity $g$ is:

A $\frac{u^2}{g}$

B $\frac{u^2}{2g}$

C $\frac{2u^2}{g}$

D $\frac{u^4}{g}$

Derivation & Analysis:

Maximum range occurs when angle of projection $\alpha = 45^\circ$, giving $R_{\max} = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g}$.

MCQ 5: If the velocity of a particle moving along a curve of radius $\rho$ is $v$ and is increasing at rate $\frac{dv}{dt}$, the total acceleration is:

A $\sqrt{\left(\frac{dv}{dt}\right)^2 + \frac{v^4}{\rho^2}}$

B $\frac{dv}{dt} + \frac{v^2}{\rho}$

C $\frac{v^2}{\rho}$

D $\frac{dv}{dt}$

Derivation & Analysis:

Tangential acceleration $a_t = \frac{dv}{dt}$ and normal acceleration $a_n = \frac{v^2}{\rho}$ are orthogonal, so total acceleration magnitude is $\sqrt{a_t^2 + a_n^2}$.

MCQ 6: The path of a projectile as seen from another projectile moving with constant velocity is:

A Parabola

B Ellipse

C Straight line

D Circle

Derivation & Analysis:

Relative acceleration between two projectiles in uniform gravity is zero ($\vec{g} - \vec{g} = 0$). Thus, relative motion is uniform rectilinear motion (straight line).

MCQ 7: If areal velocity of a particle moving in a central orbit is constant, the force is:

A Zero

B Directed towards the center

C Transverse only

D Constant in magnitude and direction

Derivation & Analysis:

Constant areal velocity $\frac{1}{2}r^2\dot{\theta} = h$ implies transverse acceleration $a_\theta = 0$, meaning the resultant force acts entirely along the radius vector (central force).

MCQ 8: Assertion (A): In uniform circular motion, tangential acceleration is zero. Reason (R): Speed remains constant in uniform circular motion.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (C)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Tangential acceleration is $a_t = \frac{dv}{dt}$. Since speed $v$ is constant in uniform circular motion, $\frac{dv}{dt} = 0$.

MCQ 9: If the time of flight of a projectile is $T$, the maximum height $H$ reached is given by:

A $\frac{1}{8} g T^2$

B $\frac{1}{2} g T^2$

C $\frac{1}{4} g T^2$

D $\frac{1}{16} g T^2$

Derivation & Analysis:

Time of flight $T = \frac{2u \sin\alpha}{g}$. Maximum height $H = \frac{u^2 \sin^2\alpha}{2g} = \frac{(gT/2)^2}{2g} = \frac{1}{8}gT^2$.

Q10

MCQ 10: If $r = a(1 + \cos\theta)$ is the path of a particle, the radial velocity $\dot{r}$ when $\dot{\theta} = \omega$ is:

A $-a\omega \sin\theta$

B $a\omega \cos\theta$

C $a\omega(1+\sin\theta)$

D $0$

Derivation & Analysis:

Given $r = a(1+\cos\theta)$, differentiating with respect to $t$ gives $\dot{r} = -a\sin\theta \cdot \dot{\theta} = -a\omega\sin\theta$.

Common questions

Central forces are directed towards a fixed pole. Expressing motion in polar coordinates ($r, \theta$) naturally simplifies the equations of motion because the force acts solely along the radial direction, eliminating unnecessary $x$ and $y$ cross-coupling.

Confusing maximum height time ($T/2$) with total time of flight ($T$), or making sign convention errors when the point of projection is at a different elevation than the landing point.

Memorize standard formulas for $v_r, v_\theta, a_r, a_\theta$ and standard curve derivatives (e.g., spirals, cardioids) to avoid deriving vector unit vectors $\hat{r}$ and $\hat{\theta}$ during the exam.

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