Kinematics & Constrained Motion: Master Notes

Chapter Overview

Kinematics is the geometric study of motion without regarding the forces that cause it. In IIT-JEE (Advanced) and Olympiads, the key is mastering coordinate systems, calculus relations, relative motion vectors, and string/wedge geometric constraints.

---

1. Core Calculus & Vector Kinematics

1.1 Fundamental Relations

$$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt},\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d^2\overrightarrow{r}}{d{t}^2}=v\frac{dv}{dx}\hat{t}$$

• Position $\to$ Velocity $\to$ Acceleration: Differentiate with respect to time $t$.
• Acceleration $\to$ Velocity $\to$ Position: Integrate:$$v(t)=v_0+{\int }_0^{t}a(t^{\prime })d{t}^{\prime },x(t)=x_0+{\int }_0^{t}v(t^{\prime })d{t}^{\prime }$$
• When acceleration depends on position $a(x)$:$${\int }_{v_0}^v{v}^{\prime }d{v}^{\prime }={\int }_{x_0}^{x}a(x^{\prime })d{x}^{\prime }⟹\frac{1}{2}(v^2-v_0^2)={\int }_{x_0}^{x}a(x^{\prime })d{x}^{\prime }$$

---

1.2 Polar Coordinates $(r,\theta )$ & Curvilinear Motion

When a particle moves in a plane, Cartesian coordinates $(x,y)$ can be tedious. In polar coordinates:

$$\overrightarrow{r}=r{\hat{e}}_r$$$$\overrightarrow{v}=\dot{r}{\hat{e}}_r+r\dot{\theta }{\hat{e}}_{\theta }=v_r{\hat{e}}_r+v_{\theta }{\hat{e}}_{\theta }$$$$\overrightarrow{a}=(\ddot{r}-r{\dot{\theta }}^2){\hat{e}}_r+(r\ddot{\theta }+2\dot{r}\dot{\theta }){\hat{e}}_{\theta }$$

• Radial acceleration: $a_r=\ddot{r}-r{\omega }^2$
• Transverse acceleration: $a_{\theta }=r\alpha +2\dot{r}\omega$
• Coriolis kinematic term: $2\dot{r}\dot{\theta }$ (appears when distance $r$ changes while rotating)

---

1.3 Radius of Curvature ($R_c$)

The radius of curvature of any trajectory curve $y(x)$ or path with tangential speed $v$ and normal acceleration $a_n$:

$$\boxed{{\displaystyle R_c=\frac{v^2}{a_n}=\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}{\left|\frac{d^{2}y}{d{x}^2}\right|}}}$$

::: insight 💡 Projectile Radius of Curvature

• At the apex of projectile motion: $v=u\cos \theta$, $a_n=g⟹R_{\text{apex}}=\frac{u^2{\cos }^2\theta }{g}$.
• At launch: $v=u$, $a_n=g\cos \theta ⟹R_{\text{launch}}=\frac{u^2}{g\cos \theta }$. :::

---

2. Constrained Motion Master Techniques

2.1 String Constraint: The Virtual Work / Differentiating Length Method

Method 1: Total Length Derivative

For any inextensible string of total length $L$:

$$L=x_1+2x_2+x_3+\text{const}⟹\frac{dL}{dt}=0⟹v_1+2v_2+v_3=0⟹a_1+2a_2+a_3=0$$

Method 2: Tension-Dot-Product Method ($\sum \overrightarrow{T}\cdot \overrightarrow{a}=0$)

Since internal tension does no net work in ideal inextensible string systems:

$$\boxed{{\displaystyle \sum _i{\overrightarrow{T}}_i\cdot {\overrightarrow{v}}_i=0\text{and}\sum _i{\overrightarrow{T}}_i\cdot {\overrightarrow{a}}_i=0}}$$

Multi-Mode DiagramKinematics & Movable Pulley Constraint System

Option 1: Publication-Grade Scientific Vector SVG

Movable pulley virtual work and string constraint analysis: Tensions, string acceleration relations, and displacement multipliers.

String Constraint: $x_1 + 2 x_2 = L \implies a_1 = 2 a_2$

Acceleration a₂: $a_2 = \frac{(2m_2 - m_1)g}{4m_1 + m_2}$

String Tension T: $T = \frac{3 m_1 m_2 g}{4m_1 + m_2}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate movable pulley real-time gravity fall under string constraints with mass sliders.

Mass m₁: 3 kg

Mass m₂: 5 kg

Accel a₁ ($a_1$):8.07 m/s²

Accel a₂ ($a_2$):4.04 m/s²

Option 3: Interactive 3D Spatial Viewport

3D Spatial pulley rigging showing vertical displacement coordinates.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Constraint Relation:$a_1 = 2 a_2$

Option 5: Reactive Parametric Physics with Live Calculus

Dynamic mass slider computing instantaneous accelerations and string tensions.

Mass m₁: 3 kg

Mass m₂: 6 kg

a_1 ($a_1$):$9.80 m/s²$

a_2 ($a_2$):$4.90 m/s²$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

---

2.2 Wedge & Surface Contact Constraints

When two bodies remain in physical contact without separating or interpenetrating:

$$\boxed{{\displaystyle {\overrightarrow{v}}_1\cdot \hat{n}={\overrightarrow{v}}_2\cdot \hat{n}}}$$

Rule: The components of velocity of both bodies along their common normal direction must be equal at all times!

---

3. Relative Velocity & Shortest Distance of Approach

$${\overrightarrow{v}}_{A/B}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$$

Shortest Distance Between Two Moving Particles

1. Freeze particle $B$ by giving $-{\overrightarrow{v}}_B$ to both particles.
2. Particle $A$ now travels in a straight line with relative velocity ${\overrightarrow{v}}_{\text{rel}}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$.
3. Drop a perpendicular from stationary point $B$ to the line of motion of ${\overrightarrow{v}}_{\text{rel}}$:$$\boxed{{\displaystyle d_{min}=\frac{|{\overrightarrow{r}}_{A/B}\times {\overrightarrow{v}}_{\text{rel}}|}{|{\overrightarrow{v}}_{\text{rel}}|}}}$$$$t_{min}=-\frac{{\overrightarrow{r}}_{A/B}\cdot {\overrightarrow{v}}_{\text{rel}}}{|{\overrightarrow{v}}_{\text{rel}}{|}^2}$$

---

4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2021 (Paper 1) — Variable Acceleration & Trajectory

Question:
A particle of mass $m$ moves in the $xy$-plane such that its velocity vector is given by $\overrightarrow{v}(t)=\alpha \hat{i}+\beta t\hat{j}$, where $\alpha$ and $\beta$ are positive constants. At $t=0$, the particle is at the origin $(0,0)$.

1. Find the Cartesian equation of the trajectory $y(x)$.
2. Determine the radius of curvature $R_c$ of the trajectory at $x=\alpha$.

Step-by-Step Solution:

• Part 1:

  $$\frac{dx}{dt}=\alpha ⟹x(t)=\alpha t⟹t=\frac{x}{\alpha }$$$$\frac{dy}{dt}=\beta t⟹y(t)={\int }_0^t\beta t^{\prime }d{t}^{\prime }=\frac{1}{2}\beta t^2$$

  Substitute $t=x/\alpha$:

  $$y(x)=\frac{\beta }{2{\alpha }^2}{x}^2\text{(A Parabola)}$$

• Part 2:

  $$\frac{dy}{dx}=\frac{\beta }{{\alpha }^2}x,\frac{d^{2}y}{d{x}^2}=\frac{\beta }{{\alpha }^2}$$

  At $x=\alpha$:

  $$\frac{dy}{dx}=\frac{\beta }{\alpha }$$

  Using the radius of curvature formula:

  $$R_c=\frac{{[1+{\left(\frac{\beta }{\alpha }\right)}^2]}^{3/2}}{\frac{\beta }{{\alpha }^2}}=\frac{({\alpha }^2+{\beta }^2{)}^{3/2}}{{\alpha }^3\cdot \frac{\beta }{{\alpha }^2}}=\boxed{{\displaystyle \frac{({\alpha }^2+{\beta }^2{)}^{3/2}}{\alpha \beta }}}$$

---

PYQ 2: JEE Advanced 2019 — River Swimmer Optimization

Question:
A swimmer can swim at speed $v_s=5\text{ km/h}$ relative to still water. The river flows east at speed $v_r=3\text{ km/h}$. The width of the river is $d=1\text{ km}$.

• (a) What heading angle $\theta$ relative to the flow direction should the swimmer choose to cross the river in the minimum possible time?
• (b) What angle should the swimmer choose to cross with minimum drift?

Step-by-Step Solution:

• (a) Minimum Time: The vertical crossing velocity is $v_y=v_s\sin \theta$. Time $t=\frac{d}{v_s\sin \theta }$.
  To minimize $t$, $\sin \theta =1⟹\theta ={90}^{\circ }$ (perpendicular to river bank).

  $$t_{min}=\frac{1\text{ km}}{5\text{ km/h}}=0.2\text{ h}=12\text{ minutes}$$

• (b) Minimum Drift: Since $v_s>v_r$ ($5>3$), zero drift is possible. The swimmer must head upstream at angle $\varphi$ to normal:

  $$\sin \varphi =\frac{v_r}{v_s}=\frac{3}{5}⟹\varphi ={37}^{\circ }\text{ upstream}(\theta ={127}^{\circ }\text{ to flow})$$

---

5. Common Exam Pitfalls

::: pitfall ❌ 1. Misinterpreting Average Speed vs Average Velocity $\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}\ge |\text{Average Velocity}|=\frac{|\mathrm{\Delta }\overrightarrow{r}|}{\mathrm{\Delta }t}$. They are equal only for strictly unidirectional 1D motion without turning back. :::

::: pitfall ❌ 2. Projectile on an Incline Angle Mistakes On an incline of angle $\beta$ with projection angle $\alpha$ relative to the incline:

• Replace $g\to g\cos \beta$ in the perpendicular ($y^{\prime }$) direction.
• Replace $0\to -g\sin \beta$ in the parallel ($x^{\prime }$) direction.
• Time of flight: $T=\frac{2u\sin \alpha }{g\cos \beta }$. :::

---

6. High-Yield Formula Sheet

Quantity	Formula	Condition
Equation of Trajectory	$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }=x\tan \theta (1-\frac{x}{R})$	Ground-to-ground projectile
Max Range on Flat Ground	$R_{max}=\frac{u^2}{g}$	at $\theta ={45}^{\circ }$
Max Range on Incline $\beta$	$R_{max}=\frac{u^2}{g(1+\sin \beta )}$ (Up), $\frac{u^2}{g(1-\sin \beta )}$ (Down)	Launch at $\frac{\pi }{4}\pm \frac{\beta }{2}$
Tension Constraint	$\sum {\overrightarrow{T}}_i\cdot {\overrightarrow{a}}_i=0$	Light inextensible strings
Wedge Constraint	$v_{1,\perp }=v_{2,\perp }$	Solid surface contact