Comprehensive Study Notes on Motion in a Plane

VECTORS AND VECTOR ALGEBRA IN TWO DIMENSIONS

• Vectors are mathematical quantities that possess both magnitude and direction, obeying specific laws of vector addition (such as the Triangle Law and Parallelogram Law).

• Basic Vector Terminologies:

  • Null Vector (Zero Vector): A vector having zero magnitude ($|\vec{A}| = 0$) and an arbitrary direction. It represents the result of subtracting a vector from itself.

  • Unit Vector: A dimensionless vector of unit magnitude ($|\hat{A}| = 1$) that specifies direction. It is calculated by dividing a vector by its magnitude:     $\hat{A} = \frac{\vec{A}}{|\vec{A}|}$

  • Equal Vectors: Two or more vectors are equal ($\vec{A} = \vec{B}$) if and only if they possess identical magnitudes and point in the exact same direction.

  • Axial Vectors: Vectors associated with rotational motion (such as angular velocity $\vec{\omega}$, angular acceleration $\vec{\alpha}$, and torque $\vec{\tau}$) whose directions are specified along the axis of rotation using the right-hand screw rule.

• Vector Orientations:

  • Orthogonal Vectors: Two vectors $\vec{A}$ and $\vec{B}$ positioned at an angle of $\theta = 90^\circ$. Their dot product is zero ($\vec{A} \cdot \vec{B} = 0$).

  • Parallel Vectors: Two vectors positioned at an angle of $\theta = 0^\circ$. They satisfy $\vec{A} = n \vec{B}$, where $n$ is a positive scalar.

  • Anti-Parallel Vectors: Two vectors positioned at an angle of $\theta = 180^\circ$. They satisfy $\vec{A} = -n \vec{B}$, where $n$ is a positive scalar.

• Resolution of a Vector in Two Dimensions:

  • Any vector $\vec{V}$ lying in the XY plane can be decomposed into two mutually perpendicular components along the X-axis and Y-axis:     $\vec{V} = V_x \hat{i} + V_y \hat{j}$

  • If $\alpha$ is the angle made by $\vec{V}$ with the X-axis:     $V_x = V \cos(\alpha)$     $V_y = V \sin(\alpha)$

  • The magnitude of the vector is given by:     $|\vec{V}| = \sqrt{V_x^2 + V_y^2}$

  • The direction angle $\alpha$ is determined by:     $\tan(\alpha) = \frac{V_y}{V_x}$

• Arithmetic Vector Operations:

  • Given two vectors $\vec{A} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$ and $\vec{B} = a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}$:

  • Addition:     $\vec{A} + \vec{B} = (a_1 + a_2) \hat{i} + (b_1 + b_2) \hat{j} + (c_1 + c_2) \hat{k}$

  • Subtraction:     $\vec{A} - \vec{B} = (a_1 - a_2) \hat{i} + (b_1 - b_2) \hat{j} + (c_1 - c_2) \hat{k}$

  • Dot Product (Scalar Product):     $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) = a_1 a_2 + b_1 b_2 + c_1 c_2$

  • Fundamental unit vector relationships:       $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$       $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

  • Cross Product (Vector Product):     $\vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin(\theta) \hat{n}$

  • Expressed in determinant form:       $\vec{A} \times \vec{B} = \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{pmatrix} = \hat{i} (b_1 c_2 - c_1 b_2) + \hat{j} (c_1 a_2 - a_1 c_2) + \hat{k} (a_1 b_2 - b_1 a_2)$

  • Fundamental unit vector relationships:       $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$       $\hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{k} = \hat{i}, \quad \hat{k} \times \hat{i} = \hat{j}$       $\hat{j} \times \hat{i} = -\hat{k}, \quad \hat{k} \times \hat{j} = -\hat{i}, \quad \hat{i} \times \hat{k} = -\hat{j}$

• Fundamental Vector Addition Laws:

  • Triangle Law of Vector Addition: If two vectors are represented in magnitude and direction by two sides of a triangle taken in the same order, their resultant is represented by the third side taken in the reverse order.

  • Parallelogram Law of Vector Addition: If two vectors $\vec{A}$ and $\vec{B}$ acting simultaneously at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant $\vec{R}$ is given by the diagonal passing through their common point:     $R = \sqrt{A^2 + B^2 + 2 A B \cos(\theta)}$

  • The angle $\alpha$ that the resultant $\vec{R}$ makes with vector $\vec{A}$ is calculated using:       $\tan(\alpha) = \frac{B \sin(\theta)}{A + B \cos(\theta)}$

  • Lami's Theorem: For three coplanar, concurrent, and balanced (equilibrium) forces $\vec{F}_1$, $\vec{F}_2$, and $\vec{F}_3$ making angles $\alpha$, $\beta$, and $\gamma$ opposite to them respectively:     $\frac{F_1}{\sin(\alpha)} = \frac{F_2}{\sin(\beta)} = \frac{F_3}{\sin(\gamma)}$

MOTION WITH UNIFORM ACCELERATION IN A PLANE

• Consider a point object moving in the XY plane with a constant acceleration $\vec{a}$. Let origin 'O' measure time and position.

• At instant $t_1 = 0$, let the initial position vector be $\vec{r}_0$ and initial velocity be $\vec{u}$. At instant $t_2 = t$, let the final position vector be $\vec{r}$ and final velocity be $\vec{v}$.

• Constant Acceleration Definition:   $\vec{a} = \frac{\vec{v} - \vec{u}}{t_2 - t_1} = \frac{\vec{v} - \vec{u}}{t}$   $\vec{v} = \vec{u} + \vec{a} t$

• Rectangular Components of Motion:

  • Initial Velocity: $\vec{u} = u_x \hat{i} + u_y \hat{j}$, with magnitude $u = \sqrt{u_x^2 + u_y^2}$

  • Final Velocity: $\vec{v} = v_x \hat{i} + v_y \hat{j}$, with magnitude $v = \sqrt{v_x^2 + v_y^2}$

  • Acceleration: $\vec{a} = a_x \hat{i} + a_y \hat{j}$, with magnitude $a = \sqrt{a_x^2 + a_y^2}$

• Kinematic Equations per Axis:

  • Velocity components:     $v_x = u_x + a_x t$     $v_y = u_y + a_y t$

  • Position coordinates / Displacements:     $x - x_0 = u_x t + \frac{1}{2} a_x t^2$     $y - y_0 = u_y t + \frac{1}{2} a_y t^2$

HORIZONTAL PROJECTILE MOTION

• A body projected horizontally with an initial velocity $u$ from a height $h$ above ground level describes a parabolic path under the sole influence of gravity.

• Motion Components:

  • Horizontal direction (X-axis): Initial velocity $u_x = u$, Acceleration $a_x = 0$.

  • Vertical direction (Y-axis): Initial velocity $u_y = 0$, Acceleration $a_y = g$ (downwards). sssssssss

• Key Parameters of Horizontal Projectile:

  • Time of descent ($t$):     $h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}}$

  • The time of descent is completely independent of the horizontal initial velocity $u$.

  • Horizontal Range ($R$):     $R = u_x t = u \sqrt{\frac{2h}{g}}$

  • Speed at any time $t$ ($v$):     $v_x = u, \quad v_y = g t = \sqrt{2gh}$     $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + g^2 t^2} = \sqrt{u^2 + 2gh}$

  • Striking angle with the horizontal ($\theta$):     $\tan(\theta) = \frac{v_y}{v_x} = \frac{g t}{u} = \frac{\sqrt{2gh}}{u}$

  • Angle of elevation of projection point from landing point ($\alpha$):     $\tan(\alpha) = \frac{h}{R} = \frac{h}{u \sqrt{\frac{2h}{g}}} = \frac{g t}{2 u} = \frac{\tan(\theta)}{2}$

  • Equation of Trajectory:     $x = u t \implies t = \frac{x}{u}$     $y = \frac{1}{2} g t^2 = \frac{1}{2} g \left(\frac{x}{u}\right)^2 = \frac{g x^2}{2 u^2}$

  • This represents a standard vertical parabola.

OBLIQUE PROJECTILE MOTION

• An oblique projectile is any body projected into the air with an initial velocity $u$ making an angle $\theta$ ($\theta \neq 0^\circ, 90^\circ$) with the horizontal, moving freely under gravity.

• Basic Assumptions:

  • Air resistance is neglected.

  • The curvature and rotation of the earth are neglected.

  • Acceleration due to gravity ($g$) is constant in magnitude ($9.8\,m/s^2$ or $10\,m/s^2$) and direction at all points of motion.

• Velocity Components and Motion Equations:

  • Horizontal component: $u_x = u \cos(\theta)$, remains constant throughout flight because $a_x = 0$

  • Vertical component: $u_y = u \sin(\theta)$, varies continuously at rate $a_y = -g$

  • Velocity components at time $t$:     $v_x = u \cos(\theta)$     $v_y = u \sin(\theta) - g t$

  • Resultant speed at time $t$:     $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 \cos^2(\theta) + (u \sin(\theta) - g t)^2} = \sqrt{u^2 + g^2 t^2 - 2 u g t \sin(\theta)}$

  • Direction of velocity at time $t$ relative to horizontal:     $\tan(\alpha) = \frac{v_y}{v_x} = \frac{u \sin(\theta) - g t}{u \cos(\theta)}$

  • Displacements at time $t$:     $x = (u \cos(\theta)) t$     $y = (u \sin(\theta)) t - \frac{1}{2} g t^2$     $S = \sqrt{x^2 + y^2}$

• Equation of Trajectory:   $y = x \tan(\theta) - \frac{g x^2}{2 u^2 \cos^2(\theta)} = x \tan(\theta) \left(1 - \frac{x}{R}\right)$

  • Expressed in polynomial form $y = A x - B x^2$ where:     $A = \tan(\theta)$     $B = \frac{g}{2 u^2 \cos^2(\theta)}$

  • Parameters derived from coefficients $A$ and $B$:     $\theta = \tan^{-1}(A)$     $\text{Range } R = \frac{A}{B}$     $\text{Maximum Height } H = \frac{A^2}{4 B}$

• Time Parameters of Flight:

  • Time of ascent ($t_a$) = Time of descent ($t_d$) = $\frac{u \sin(\theta)}{g}$

  • Total Time of Flight ($T$):     $T = t_a + t_d = \frac{2 u \sin(\theta)}{g}$

• Maximum Height ($H$):   $H = \frac{u^2 \sin^2(\theta)}{2g}$

  • At maximum height, vertical component $v_y = 0$

  • Speed is non-zero and at its minimum: $v_{min} = v_x = u \cos(\theta)$

  • Angle between velocity vector $\vec{v}$ and acceleration vector $\vec{g}$ at maximum height is exactly $90^\circ$

• Horizontal Range ($R$):   $R = u_x T = (u \cos(\theta)) \left(\frac{2 u \sin(\theta)}{g}\right) = \frac{u^2 \sin(2\theta)}{g}$

  • Maximum horizontal range occurs at projection angle $\theta = 45^\circ$:     $R_{max} = \frac{u^2}{g}$

  • At maximum range ($\theta = 45^\circ$), maximum height is:     $H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2}{4g} = \frac{R_{max}}{4}$

  • General relationship between Range and Maximum Height:     $\tan(\theta) = \frac{4 H}{R} \implies R = 4 H \cot(\theta)$

  • When Range equals Maximum Height ($R = H$):     $\tan(\theta) = 4 \implies \theta = \tan^{-1}(4) \approx 76^\circ$

• Complementary Angles of Projection:

  • Two bodies thrown with identical speed $u$ at complimentary angles $\theta_1 = \theta$ and $\theta_2 = 90^\circ - \theta$ (or $45^\circ + \theta$ and $45^\circ - \theta$) possess identical horizontal ranges:     $R_1 = R_2 = R$

  • Ratios and relationships for complementary projections:     $\frac{T_1}{T_2} = \tan(\theta)$     $T_1 T_2 = \frac{2 u \sin(\theta)}{g} \cdot \frac{2 u \cos(\theta)}{g} = \frac{2 R}{g} \implies R = \frac{1}{2} g T_1 T_2$     $H_1 + H_2 = \frac{u^2 \sin^2(\theta)}{2g} + \frac{u^2 \cos^2(\theta)}{2g} = \frac{u^2}{2g}$     $\frac{H_1}{H_2} = \tan^2(\theta)$     $H_1 H_2 = \frac{u^4 \sin^2(\theta) \cos^2(\theta)}{4 g^2} = \frac{R^2}{16} \implies R = 4 \sqrt{H_1 H_2}$

• Angles Between Velocity and Acceleration:

  • During ascent (rising phase): $90^\circ < \theta < 180^\circ$

  • At peak position: $\theta = 90^\circ$

  • During descent (falling phase): $0^\circ < \theta < 90^\circ$

• Energy Relations in Oblique Motion:

  • Initial Kinetic Energy at point of projection: $KE_0 = \frac{1}{2} m u^2$

  • At maximum height $H$:

  • Kinetic Energy: $KE = \frac{1}{2} m v_{min}^2 = \frac{1}{2} m u^2 \cos^2(\theta) = KE_0 \cos^2(\theta)$

  • Potential Energy: $PE = m g H = m g \left(\frac{u^2 \sin^2(\theta)}{2g}\right) = \frac{1}{2} m u^2 \sin^2(\theta) = KE_0 \sin^2(\theta)$

  • Total Energy: $E = KE + PE = KE_0$

  • When $KE = PE$ at highest point: $\cos^2(\theta) = \sin^2(\theta) \implies \theta = 45^\circ$

• Projections From an Elevated Height $h$:

  • Case 1: Projected at angle $\theta$ above horizontal from top of height $h$:

  • Vertical motion equation: $-h = (u \sin(\theta)) T - \frac{1}{2} g T^2$

  • Horizontal range: $x = (u \cos(\theta)) T$

  • Landing speed: $v = \sqrt{u^2 + 2gh}$

  • Landing components: $v_x = u \cos(\theta)$, $v_y = g T - u \sin(\theta) = \sqrt{u^2 \sin^2(\theta) + 2gh}$

  • Landing angle $\alpha$ with horizontal: $\tan(\alpha) = \frac{v_y}{v_x} = \frac{\sqrt{u^2 \sin^2(\theta) + 2gh}}{u \cos(\theta)} = \frac{g T - u \sin(\theta)}{u \cos(\theta)}$

  • Case 2: Projected at angle $\theta$ below horizontal from top of height $h$:

  • Vertical motion equation: $h = (u \sin(\theta)) T + \frac{1}{2} g T^2$

  • Horizontal range: $x = (u \cos(\theta)) T$

  • Landing speed: $v = \sqrt{u^2 + 2gh}$

  • Landing vertical component: $v_y = u \sin(\theta) + g T = \sqrt{u^2 \sin^2(\theta) + 2gh}$

  • Landing angle $\alpha$: $\tan(\alpha) = \frac{v_y}{v_x} = \frac{\sqrt{u^2 \sin^2(\theta) + 2gh}}{u \cos(\theta)} = \frac{u \sin(\theta) + g T}{u \cos(\theta)}$

• Special Projectile Scenarios:

  • Trajectory passing same height $y$ at times $t_1$ and $t_2$:     $t_1 + t_2 = \frac{2 u \sin(\theta)}{g} = T$     $t_1 t_2 = \frac{2 y}{g} \implies y = \frac{1}{2} g t_1 t_2$     $u \sin(\theta) = \frac{g (t_1 + t_2)}{2}$

  • Two bodies projected horizontally in opposite directions from a height $h$ with speeds $u_1$ and $u_2$:

  • Time when their velocity vectors are mutually perpendicular ($\vec{v}_1 \cdot \vec{v}_2 = 0$):       $t = \frac{\sqrt{u_1 u_2}}{g}$

  • Distance separating them at this instant:       $x = (u_1 + u_2) t = \frac{(u_1 + u_2) \sqrt{u_1 u_2}}{g}$

  • Two tall towers of heights $h_1$ and $h_2$ separated by distance $d$:

  • Time taken for ball thrown horizontally at speed $u$ from top of tower 1 to reach tower 2:       $t = \sqrt{\frac{2(h_1 - h_2)}{g}}$

  • Distance condition: $d = u t = u \sqrt{\frac{2(h_1 - h_2)}{g}}$

PROJECTILE MOTION ON AN INCLINED PLANE

• Motion on an inclined plane involves setting up reference axes along and perpendicular to the inclined surface of angle $\beta$.

• Up the Inclined Plane:

  • Plane inclination angle = $\beta$; Projection angle with horizontal = $\alpha$ (Angle with inclined plane = $\alpha - \beta$).

  • Acceleration components:     $a_x = -g \sin(\beta)$     $a_y = -g \cos(\beta)$

  • Initial velocity components:     $u_x = u \cos(\alpha - \beta)$     $u_y = u \sin(\alpha - \beta)$

  • Time of Flight ($T$):     $0 = u_y T + \frac{1}{2} a_y T^2 \implies 0 = (u \sin(\alpha - \beta)) T - \frac{1}{2} (g \cos(\beta)) T^2$     $T = \frac{2 u \sin(\alpha - \beta)}{g \cos(\beta)}$

  • Range along the incline ($R$):     $R = u_x T + \frac{1}{2} a_x T^2 = u \cos(\alpha - \beta) T - \frac{1}{2} g \sin(\beta) T^2$     $R = \frac{u^2}{g \cos^2(\beta)} [\sin(2\alpha - \beta) - \sin(\beta)]$

  • Maximum Range Condition (Up the plane):     $\sin(2\alpha - \beta) = 1 \implies 2\alpha - \beta = \frac{\pi}{2} \implies \alpha - \beta = \frac{\pi}{4} - \frac{\beta}{2} \quad \text{or} \quad \alpha = \frac{\pi}{4} + \frac{\beta}{2}$     $R_{max} = \frac{u^2}{g \cos^2(\beta)} [1 - \sin(\beta)] = \frac{u^2}{g (1 + \sin(\beta))}$

  • Time of flight $T'$ corresponding to maximum range:     $T' = \sqrt{\frac{2 R_{max}}{g}}$

• Down the Inclined Plane:

  • Plane inclination angle = $\beta$; Projection angle relative to plane = $\alpha + \beta$.

  • Acceleration components:     $a_x = g \sin(\beta)$     $a_y = -g \cos(\beta)$

  • Initial velocity components:     $u_x = u \cos(\alpha + \beta)$     $u_y = u \sin(\alpha + \beta)$

  • Time of Flight ($T$):     $T = \frac{2 u \sin(\alpha + \beta)}{g \cos(\beta)}$

  • Range along the incline ($R$):     $R = \frac{u^2}{g \cos^2(\beta)} [\sin(2\alpha + \beta) + \sin(\beta)]$

  • Maximum Range Condition (Down the plane):     $2\alpha + \beta = \frac{\pi}{2} \implies \alpha = \frac{\pi}{4} - \frac{\beta}{2}$     $R_{max} = \frac{u^2}{g (1 - \sin(\beta))}$

  • Time of flight $T'$ corresponding to maximum range:     $T' = \sqrt{\frac{2 R_{max}}{g}}$

RELATIVE MOTION IN TWO DIMENSIONS

• The motion of a particle P relative to a particle Q in a 2D plane is defined by their vector velocity difference:   $\vec{v}_{P/Q} = \vec{v}_P - \vec{v}_Q$   where $\vec{v}_P$ is the velocity of P w.r.t ground and $\vec{v}_Q$ is the velocity of Q w.r.t ground.

• Relative Acceleration:   $\vec{a}_{P/Q} = \vec{a}_P - \vec{a}_Q$

• Relative Displacement:   $\vec{r}_{P/Q} = \vec{r}_P - \vec{r}_Q$

RAIN-MAN RELATIVE MOTION

• When rain is falling downwards with velocity $\vec{v}_R$ and a man moves horizontally with velocity $\vec{v}_M$, the apparent velocity of rain relative to the man is:   $\vec{v}_{RM} = \vec{v}_R - \vec{v}_M$

• Geometry and Calculation:

  • If rain falls vertically w.r.t ground ($\vec{v}_R = -v_R \hat{j}$) and the man moves east ($\vec{v}_M = v_M \hat{i}$):     $\vec{v}_{RM} = -v_M \hat{i} - v_R \hat{j}$     $v_{RM} = \sqrt{v_R^2 + v_M^2}$

  • Direction to hold umbrella (angle $\theta$ with vertical):     $\tan(\theta) = \frac{|\vec{v}_M|}{|\vec{v}_R|} = \frac{v_M}{v_R}$

• Analytical Problem Case Study (Rain-Man):

  • Scenario: A person in a stationary vehicle holds an umbrella at $60^\circ$ with the vertical. Driving at $20\,m/s$, raindrops appear to fall vertically.

  • Mathematical breakdown:

  • When stationary, $\theta = 60^\circ \implies \tan(60^\circ) = \frac{v_M'}{v_R} = \sqrt{3}$.

  • When driving at $v_M = 20\,m/s$, relative rain velocity $\vec{v}_{RM}$ is vertical.

  • From vector triangle: $\sin(60^\circ) = \frac{v_M}{v_R} \implies \frac{\sqrt{3}}{2} = \frac{20}{v_R} \implies v_R = \frac{40}{\sqrt{3}}\,m/s$

  • Relative velocity w.r.t moving vehicle: $\tan(60^\circ) = \frac{v_M}{v_{RM}} \implies \sqrt{3} = \frac{20}{v_{RM}} \implies v_{RM} = \frac{20}{\sqrt{3}}\,m/s$

RIVER-BOAT RELATIVE MOTION

• Velocity Definitions:

  • $\vec{v}_{BW}$ = Velocity of boat w.r.t water (steering velocity in still water)

  • $\vec{v}_W$ = Velocity of water current w.r.t ground

  • $\vec{v}_{BG}$ = Absolute velocity of boat w.r.t ground   $\vec{v}_{BG} = \vec{v}_{BW} + \vec{v}_W$

• Component Equations (River width = $b$, boat steering at angle $\theta$ with normal to banks):

  • Horizontal component (along river flow, X-axis):     $(v_{BG})_x = v_W - v_{BW} \sin(\theta)$

  • Vertical component (across river, Y-axis):     $(v_{BG})_y = v_{BW} \cos(\theta)$

  • Time taken to cross river width $b$:     $t = \frac{b}{(v_{BG})_y} = \frac{b}{v_{BW} \cos(\theta)}$

  • Drift ($d$) (displacement along X-axis upon reaching opposite bank):     $d = (v_{BG})_x t = (v_W - v_{BW} \sin(\theta)) \left(\frac{b}{v_{BW} \cos(\theta)}\right)$

• Special Case 1: Crossing in Shortest Time:

  • For minimum time $t$, $\cos(\theta) = 1 \implies \theta = 0^\circ$ (steering directly perpendicular to bank).     $t_{min} = \frac{b}{v_{BW}}$

  • Drift accumulated during shortest time crossing:     $d = v_W t_{min} = b \left(\frac{v_W}{v_{BW}}\right)$

• Special Case 2: Crossing in Shortest Path / Minimum Distance:

  • Condition A: River velocity less than boat speed ($v_W < v_{BW}$):

  • Zero drift can be achieved ($d = 0$), path length $S_{min} = b$.

  • Steering angle $\theta$ upstream:       $\sin(\theta) = \frac{v_W}{v_{BW}}$       $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \frac{\sqrt{v_{BW}^2 - v_W^2}}{v_{BW}}$

  • Time to cross along shortest path:       $t = \frac{b}{v_{BW} \cos(\theta)} = \frac{b}{\sqrt{v_{BW}^2 - v_W^2}}$

  • Condition B: River velocity greater than boat speed ($v_W > v_{BW}$):

  • Boat cannot reach directly opposite point ($d = 0$ is impossible).

  • To minimize drift $d$, differentiate $d$ w.r.t $\theta$ and set to 0, yielding:       $\sin(\theta) = \frac{v_{BW}}{v_W}$

  • Minimum drift ($d_{min}$):       $d_{min} = b \frac{\sqrt{v_W^2 - v_{BW}^2}}{v_{BW}}$

  • Shortest path length ($S_{min}$):       $S_{min} = \sqrt{b^2 + d_{min}^2} = b \left(\frac{v_W}{v_{BW}}\right)$

UNIFORM AND NON-UNIFORM CIRCULAR MOTION

• Angular Kinematics Parameters:

  • Angular Displacement ($\theta$): Angle turned by the radius vector. Measured in radians ($rad$).

  • Small angular displacements ($d\theta$) are vector quantities; finite angular displacements are scalar quantities as they fail vector addition commutativity.     $\theta = \frac{\text{arc length } (l)}{\text{radius } (r)}$

  • Angular Velocity ($\omega$): Rate of change of angular displacement.     $\omega = \frac{d\theta}{dt}$

  • S.I. Unit: $rad\,s^{-1}$.

  • Period ($T$) and Frequency ($f$) relations:       $\omega = \frac{2\pi}{T} = 2\pi f$

  • Unit Conversions:       $1\,\text{rps} = 60\,\text{rpm} = 2\pi\,rad\,s^{-1}$       $1\,\text{rpm} = \frac{\pi}{30}\,rad\,s^{-1}$

  • Angular Acceleration ($\alpha$): Rate of change of angular velocity.     $\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

  • S.I. Unit: $rad\,s^{-2}$.

• Vector Relations between Linear and Angular Quantities:   $\vec{v} = \vec{\omega} \times \vec{r}$   $v = \omega r$

  • The directions of $\vec{v}$, $\vec{\omega}$, and $\vec{r}$ are mutually perpendicular:

  • $\vec{v}$ lies along the tangent in the plane of rotation.

  • $\vec{\omega}$ lies along the axis of rotation perpendicular to the orbital plane.

  • $\vec{r}$ lies along the radius vector radially outwards from center.

• Equations of Rotational Motion (Constant Angular Acceleration $\alpha$):   $\omega = \omega_0 + \alpha t$   $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$   $\omega^2 - \omega_0^2 = 2 \alpha \theta$

• Uniform Circular Motion (UCM):

  • Motion of a particle along a circular path at constant speed ($v$ = constant).

  • Scalar quantities that remain constant: Speed, Kinetic Energy, Time Period, Frequency, Angular Speed ($\omega$), Magnitude of Centripetal Acceleration, Magnitude of Centripetal Force.

  • Vector quantities that continuously change (due to changing direction): Linear Velocity, Centripetal Acceleration, Centripetal Force, Linear Momentum.

• Accelerations in Circular Motion:

  • Centripetal (Radial) Acceleration ($a_r$):

  • Arises from continuous change in the direction of velocity vector. Directed radially inwards toward the center.     $\vec{a}_r = \vec{\omega} \times \vec{v} = \vec{\omega} \times (\vec{\omega} \times \vec{r})$     $a_r = \frac{v^2}{r} = \omega^2 r = v \omega$

  • Tangential Acceleration ($a_t$):

  • Arises from change in magnitude of linear speed ($v$). Directed along tangent to circle.     $a_t = \frac{dv}{dt} = \alpha r$

  • Total / Net Acceleration ($a_{net}$):     $\vec{a}_{net} = \vec{a}_r + \vec{a}_t$     $a_{net} = \sqrt{a_r^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + (\alpha r)^2}$

  • Classification based on acceleration components:

  • If $a_t = 0, a_r = 0$: Uniform linear motion.

  • If $a_t = 0, a_r \neq 0$: Uniform circular motion.

  • If $a_t \neq 0, a_r = 0$: Accelerated linear motion.

  • If $a_t \neq 0, a_r \neq 0$: Non-uniform circular motion.

• Dynamics of Circular Motion:

  • Centripetal Force ($F_c$): Real inward force required to maintain circular motion.     $F_c = m a_r = \frac{m v^2}{r} = m \omega^2 r$

  • Work done by centripetal force is strictly zero ($W = 0$) because instantaneous force is perpendicular to velocity ($\vec{F}_c \cdot \vec{v} = 0$).

  • Centripetal force is not an additional force; it is provided by physical forces (tension, gravity, normal force, friction).

  • Example: Mass $m$ attached to string of length $l$ revolved in vertical circle: At lowest position, net force $T - m g = \frac{m v^2}{l} \implies T = m g + \frac{m v^2}{l}$.

  • Centrifugal Force:

  • A pseudo force felt in a rotating (non-inertial) frame of reference. Directed radially outwards.

  • Magnitude $F_{cf} = m \omega^2 r = \frac{m v^2}{r}$.

  • Centripetal and centrifugal forces do not form an action-reaction pair because they are defined in different reference frames.

ANALYTICAL PROBLEMS AND DETAILED DERIVATIONS

• Problem 1: Projection at $45^\circ$ with initial speed $u = 20 \sqrt{2}\,m/s$ ($g = 10\,m/s^2$).

  • Range ($R$):     $R = \frac{u^2 \sin(2\theta)}{g} = \frac{(20 \sqrt{2})^2 \sin(90^\circ)}{10} = \frac{800 \times 1}{10} = 80\,m$

  • Maximum Height ($H$):     $H = \frac{u^2 \sin^2(\theta)}{2g} = \frac{800 \times (1/\sqrt{2})^2}{20} = \frac{400}{20} = 20\,m$

  • Time of Flight ($T$):     $T = \frac{2 u \sin(\theta)}{g} = \frac{2 \times 20 \sqrt{2} \times (1/\sqrt{2})}{10} = 4\,s$

• Problem 2: Particle thrown from tower of height $40\,m$ at angle $45^\circ$ with initial speed $u = 20 \sqrt{2}\,m/s$ ($g = 10\,m/s^2$).

  • Vertical parameters: $u_y = 20 \sqrt{2} \sin(45^\circ) = 20\,m/s$, $a_y = -10\,m/s^2$, $s_y = -40\,m$

  • Equation: $-40 = 20 T - 5 T^2 \implies T^2 - 4 T - 8 = 0$

  • Time root: $T = \frac{4 + \sqrt{16 + 32}}{2} = 2 + 2 \sqrt{3} \approx 5.46\,s$

  • Horizontal distance: $R = u_x T = (20 \sqrt{2} \cos(45^\circ)) \times 5.46 = 20 \times 5.46 = 109.2\,m$

• Problem 3: Stone whirled horizontally at height $2.9\,m$ on radius $1.5\,m$. String breaks, stone travels horizontal distance $10\,m$ as horizontal projectile ($g = 9.8\,m/s^2$).

  • Flight time: $h = \frac{1}{2} g t^2 \implies 2.9 = 4.9 t^2 \implies t = \sqrt{\frac{2.9}{4.9}} \approx 0.769\,s$

  • Horizontal launch speed: $u = \frac{R}{t} = \frac{10}{\sqrt{5.8 / 9.8}} \approx 13\,m/s$

  • Centripetal acceleration in circular track:     $a_r = \frac{u^2}{r} = \frac{13^2}{1.5} = \frac{169}{1.5} \approx 43\,m/s^2$

• Problem 4: Relative rotation of concentric particles A and B ($T_A = 3\,\text{min}$, $T_B = 1\,\text{min}$).

  • Angular speeds: $\omega_A = \frac{2\pi}{3}$, $\omega_B = \frac{2\pi}{1}$

  • Relative angular speed: $\omega_{AB} = \omega_B - \omega_A = 2\pi \left(1 - \frac{1}{3}\right) = \frac{4\pi}{3}\,rad/min$

  • Time for one relative revolution:     $T_{rel} = \frac{2\pi}{\omega_{AB}} = \frac{2\pi}{4\pi / 3} = 1.5\,\text{min}$

• Problem 5: Collision of horizontal projectile A from height $H$ and surface particle B moving toward tower.

  • Height $H = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2H}{g}}$

  • Horizontal distance traveled by A: $R = u t = u \sqrt{\frac{2H}{g}}$

  • Distance traveled by B moving at speed $v$: $x_B = v t = v \sqrt{\frac{2H}{g}}$

  • Separation $d = R + x_B = (u + v) \sqrt{\frac{2H}{g}}$

  • Solving for $v$:     $u + v = d \sqrt{\frac{g}{2H}} \implies v = d \sqrt{\frac{g}{2H}} - u$

• Problem 6: Elastic collision of projectile with smooth wall at highest point $H$.

  • Vertical component unaltered by elastic impact; flight time unchanged $T = \frac{2 u_y}{g}$.

  • Horizontal velocity reverses direction without magnitude change.

  • Path splits into segments: time from launch to wall = $2t$, time from wall to ground landing = $t$. Total flight time = $3t$.

  • Height $H = \frac{1}{2} g (2t)^2 = 2 g t^2$

  • Wall impact height $h_{wall} = H - \frac{1}{2} g t^2 = H - \frac{H}{4} = \frac{3H}{4}$

• Problem 7: Total contact force on block revolving on horizontal smooth plane.

  • Vertical normal reaction balance: $N_2 = m g$

  • Horizontal centripetal force requirement: $N_1 = \frac{m v^2}{R}$

  • Total net contact force $N_{net}$:     $N_{net} = \sqrt{N_1^2 + N_2^2} = \sqrt{\left(\frac{m v^2}{R}\right)^2 + (m g)^2} = m \sqrt{g^2 + \frac{v^4}{R^2}}$