Comprehensive Study Guide: Motion in a Plane and Vector Algebra

Fundamentals of Scalars and Vectors

• Motion in One Dimension vs. Multi-Dimensions:

  • In one-dimensional motion, physical quantities can assume only two possible directions, which are specified using positive ($+$) and negative ($-$) signs.
  • In two-dimensional motion (motion in a plane) or three-dimensional motion (motion in space), specifying directions requires vector algebra to describe physical quantities such as position, displacement, velocity, and acceleration.

• Scalar Quantities:

  • A scalar quantity is defined as a physical quantity that possesses magnitude only and has no direction.
  • It is completely specified by a single real number along with an appropriate unit of measurement.
  • Examples of scalar quantities include distance, mass, temperature, and time.

• Vector Quantities:

  • A vector quantity is defined as a physical quantity that possesses both magnitude and direction, and satisfies vector addition rules such as the triangle law of vector addition or the parallelogram law of vector addition.
  • It is specified by stating its magnitude (a positive real number) and its direction in space.
  • Examples of vector quantities include displacement, velocity, acceleration, and force.

Vector Representation and Basic Vector Concepts

• Notation:

  • A vector is represented in printed text using a bold letter such as \n\mathbf{A} or by drawing an arrow over the letter, such as $\vec{\mathbf{A}}$.
  • The magnitude of a vector $\vec{\mathbf{A}}$ is defined as its absolute value, denoted by $|\vec{\mathbf{A}}| = A$.

• Graphical Representation:

  • Graphically, a vector is represented as a directed line segment with an arrow head.
  • The starting point $P$ of the line segment is called the tail (or origin) of the vector.
  • The end point $Q$ containing the arrow head is called the head (or tip) of the vector.
  • The length of the line segment drawn to scale represents the magnitude of the vector, while the arrow head points in the direction of the vector.

• Position Vector:

  • Consider a particle at point $P$ in a coordinate plane relative to an origin $O$.
  • The straight line directed from origin $O$ to point $P$ defines the position vector $\mathbf{r}$ of the object at time $t$.
  • If the particle moves to a new position $P'$ at time $t'$, its position vector becomes $\mathbf{r}' = \mathbf{OP'}$.

• Displacement Vector:

  • The displacement vector $\mathbf{PP'}$ (or $\Delta\mathbf{r}$) is the straight directed line segment joining the initial position $P$ to the final position $P'$.
  • The displacement vector depends solely on the initial and final positions of the particle and is completely independent of the actual path taken by the object between those two points.
  • Mathematically, the displacement vector is given by:     $\Delta\mathbf{r} = \mathbf{r}' - \mathbf{r}$

• Equality of Vectors:

  • Two vectors $\mathbf{A}$ and $\mathbf{B}$ are defined to be equal ($\mathbf{A} = \mathbf{B}$) if and only if they have both the same magnitude and the exact same direction.
  • If two vectors have the same magnitude (same line segment length) but point in different directions, they are not equal ($\mathbf{A}' \neq \mathbf{B}'$).

Vector Operations via Graphical Methods

• Multiplication of a Vector by Real Numbers:
  • Multiplying a vector $\mathbf{A}$ by a positive real scalar number $\lambda > 0$ yields a vector $\lambda\mathbf{A}$ whose magnitude is scaled by the factor $\lambda$ ($|\lambda\mathbf{A}| = \lambda |\mathbf{A}|$) while maintaining the original direction of $\mathbf{A}$.
  • Example: Multiplying $\mathbf{A}$ by $2$ yields $2\mathbf{A}$, which points in the same direction as $\mathbf{A}$ with twice its magnitude.

• Multiplying a vector $\mathbf{A}$ by a negative real scalar number $\lambda < 0$ yields a vector $\lambda\mathbf{A}$ pointing in the direction opposite to $\mathbf{A}$, with a magnitude equal to $-\lambda |\mathbf{A}|$
  • Example: Multiplying $\mathbf{A}$ by $-1$ yields $-\mathbf{A}$, which has the same magnitude as $\mathbf{A}$ but opposite direction.
  • Example: Multiplying $\mathbf{A}$ by $-1.5$ yields $-1.5\mathbf{A}$, pointing opposite to $\mathbf{A}$ with $1.5$ times its magnitude.

• Triangle Law of Vector Addition (Head-to-Tail Method):
  • Statement: If two vectors are represented in magnitude and direction by two sides of a triangle taken in order (head of the first vector joined to tail of the second vector), then their resultant vector is represented in magnitude and direction by the third side of the triangle drawn from the tail of the first vector to the head of the second vector.
  • Procedure:
  1. Place vector $\mathbf{A}$ such that its tail is at point $O$ and head at point $P$.
  2. Shift vector $\mathbf{B}$ parallel to itself so that its tail connects to the head $P$ of vector $\mathbf{A}$.
  3. Draw the resultant vector $\mathbf{R} = \mathbf{A} + \mathbf{B}$ from the tail $O$ of $\mathbf{A}$ to the head $Q$ of $\mathbf{B}$.

• Properties of Vector Addition:

  • Commutative Law: Vector addition is independent of the order of addition:     $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$
  • Associative Law: Vector addition is independent of the grouping of vectors:     $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$

• Subtraction of Vectors:

  • Subtraction of vector $\mathbf{B}$ from vector $\mathbf{A}$ is defined as the vector addition of $\mathbf{A}$ and the negative vector $-\mathbf{B}$:     $\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$

• Parallelogram Law of Vector Addition:
  • Statement: If two vectors are represented in magnitude and direction by two adjacent sides of a parallelogram drawn from a common initial point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that common point.

• Practical Application: Rain and Wind Problem:
  • Scenario: Rain falls vertically downwards with a speed of $v_r = 35\,\text{m\,s}^{-1}$. Wind begins blowing from East to West with a speed of $v_w = 12\,\text{m\,s}^{-1}$.

• Magnitude of resultant velocity $R$:     $R = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37\,\text{m\,s}^{-1}$
• Direction angle $\theta$ relative to the vertical:     $\tan(\theta) = \frac{v_w}{v_r} = \frac{12}{35} \approx 0.343$$\theta = \tan^{-1}(0.343) \approx 19^\circ$
• Practical Conclusion: A person standing at the bus stop must hold their umbrella tilted at an angle of approximately $19^\circ$ with the vertical towards the East to protect against the rain.

Unit Vectors and Vector Resolution

• Definition of Unit Vector:

  • A unit vector is a vector having a magnitude equal to exactly unit magnitude ($1$) pointing in a specific direction.
  • It is dimensionless and unitless, serving purely to specify spatial orientation.
  • A unit vector in the direction of vector $\mathbf{A}$ is denoted as $\hat{\mathbf{A}}$ and defined as:     $\hat{\mathbf{A}} = \frac{\mathbf{A}}{|\mathbf{A}|}$
  • Consequently, any vector $\mathbf{A}$ can be expressed as the product of its magnitude and its unit vector:     $\mathbf{A} = |\mathbf{A}|\hat{\mathbf{A}}$

• Orthogonal Unit Vectors:

  • Unit vectors along the positive $x$-, $y$-, and $z$-axes of a right-handed rectangular coordinate system are represented by $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$, respectively.
  • Magnitudes of orthogonal unit vectors:     $|\hat{\mathbf{i}}| = |\hat{\mathbf{j}}| = |\hat{\mathbf{k}}| = 1$
  • These three vectors are mutually perpendicular to one another.

• Resolution of a Vector in Two Dimensions (Plane):
  • A vector $\mathbf{A}$ lying in the $x-y$ plane can be resolved into two mutually perpendicular component vectors along the $x$- and $y$-axes:     $\mathbf{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}}$
  • Here, $A_x\hat{\mathbf{i}}$ and $A_y\hat{\mathbf{j}}$ are vector components, whereas scalar quantities $A_x$ and $A_y$ are the scalar components of vector $\mathbf{A}$.

• Trigonometric Relationships in Two Dimensions:

  • If vector $\mathbf{A}$ makes an angle $\theta$ with the positive $x$--axis:     $\cos(\theta) = \frac{A_x}{A} \implies A_x = A\cos(\theta)$$\sin(\theta) = \frac{A_y}{A} \implies A_y = A\sin(\theta)$
  • Magnitude $A$ from component values:     $A_x^2 + A_y^2 = A^2\cos^2(\theta) + A^2\sin^2(\theta) = A^2(\cos^2(\theta) + \sin^2(\theta)) = A^2$$A = \sqrt{A_x^2 + A_y^2}$
  • Direction angle $\theta$:     $\tan(\theta) = \frac{A_y}{A_x} \implies \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$

• Resolution of a Vector in Three Dimensions (Space):

  • If vector $\mathbf{A}$ exists in three-dimensional space, it expands as:     $\mathbf{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}} + A_z\hat{\mathbf{k}}$
  • The magnitude of vector $\mathbf{A}$ in 3D is given by:     $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$

Analytical Method of Vector Addition

• Geometric Derivation for Resultant Magnitude and Direction:
  • Consider two vectors $\mathbf{A}$ and $\mathbf{B}$ acting simultaneously on a body, inclined at an angle $\theta$ to each other.

• Extend line $OP$ to point $N$ and drop perpendicular $SN$ to $OP$ extended such that $SN \perp OP$.

• In right-angled triangle $\triangle SNP$:     $\cos(\theta) = \frac{PN}{PS} = \frac{PN}{B} \implies PN = B\cos(\theta)$$\sin(\theta) = \frac{SN}{PS} = \frac{SN}{B} \implies SN = B\sin(\theta)$

• In right-angled triangle $\triangle OSN$:     $OS^2 = ON^2 + SN^2$$ON = OP + PN = A + B\cos(\theta)$$OS^2 = (A + B\cos(\theta))^2 + (B\sin(\theta))^2$$R^2 = A^2 + 2AB\cos(\theta) + B^2\cos^2(\theta) + B^2\sin^2(\theta)$$R^2 = A^2 + B^2(\cos^2(\theta) + \sin^2(\theta)) + 2AB\cos(\theta)$$R^2 = A^2 + B^2 + 2AB\cos(\theta)$$R = \sqrt{A^2 + B^2 + 2AB\cos(\theta)}$

• Direction angle $\alpha$ of resultant vector $\mathbf{R}$ relative to vector $\mathbf{A}$:     $\tan(\alpha) = \frac{SN}{ON} = \frac{SN}{OP + PN} = \frac{B\sin(\theta)}{A + B\cos(\theta)}$$\alpha = \tan^{-1}\left(\frac{B\sin(\theta)}{A + B\cos(\theta)}\right)$

• Alternatively, applying Law of Sines in $\triangle OPS$:     $\frac{R}{\sin(\theta)} = \frac{B}{\sin(\alpha)} \implies \sin(\alpha) = \frac{B}{R}\sin(\theta)$

  • Worked Example: Motorboat and Water Current Velocity:

• Problem: A motorboat races North at a speed of $v_b = 25\,\text{km/h}$. Water current flows at $v_c = 10\,\text{km/h}$ in the direction $60^\circ$ East of South. Find the resultant velocity of the boat.

• Calculation steps:
  • The angle between North (direction of $\mathbf{v}_b$) and South is $180^\circ$. Since current is $60^\circ$ East of South, the angle $\theta$ between $\mathbf{v}_b$ and $\mathbf{v}_c$ is:       $\theta = 180^\circ - 60^\circ = 120^\circ$
  • Magnitude of resultant velocity $R$:       $R = \sqrt{v_b^2 + v_c^2 + 2v_b v_c \cos(120^\circ)}$       Given $\cos(120^\circ) = -\frac{1}{2}$:       $R = \sqrt{25^2 + 10^2 + 2 \times 25 \times 10 \times \left(-\frac{1}{2}\right)}$$R = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 21.8\,\text{km/h} \approx 22\,\text{km/h}$
  • Direction angle $\phi$ relative to North:       Applying Law of Sines:       $\frac{R}{\sin(\theta)} = \frac{v_c}{\sin(\phi)} \implies \sin(\phi) = \frac{v_c}{R}\sin(\theta)$$\sin(\phi) = \frac{10 \times \sin(120^\circ)}{21.8} = \frac{10 \times \frac{\sqrt{3}}{2}}{21.8} = \frac{10 \times 0.866}{21.8} \approx 0.397$$\phi = \sin^{-1}(0.397) \approx 23.4^\circ$
  • Result: The boat travels at an effective speed of $22\,\text{km/h}$ in a direction $23.4^\circ$ East of North.

Kinematics in Two Dimensions (Motion in a Plane)

• Position Vector and Displacement Vector:
  • The position vector $\mathbf{r}$ of a particle $P$ at time $t$ in a 2D plane is:     $\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}$
  • At a later time $t'$, the position vector $\mathbf{r}'$ is:     $\mathbf{r}' = x'\hat{\mathbf{i}} + y'\hat{\mathbf{j}}$
  • Displacement vector $\Delta\mathbf{r}$ over time interval $\Delta t = t' - t$:     $\Delta\mathbf{r} = \mathbf{r}' - \mathbf{r} = (x'\hat{\mathbf{i}} + y'\hat{\mathbf{j}}) - (x\hat{\mathbf{i}} + y\hat{\mathbf{j}}) = (x' - x)\hat{\mathbf{i}} + (y' - y)\hat{\mathbf{j}}$$\Delta\mathbf{r} = \Delta x\hat{\mathbf{i}} + \Delta y\hat{\mathbf{j}}$

• Velocity Vector:

  • Average velocity $\bar{\mathbf{v}}$ over time interval $\Delta t$:     $\bar{\mathbf{v}} = \frac{\Delta\mathbf{r}}{\Delta t} = \frac{\Delta x\hat{\mathbf{i}} + \Delta y\hat{\mathbf{j}}}{\Delta t} = \left(\frac{\Delta x}{\Delta t}\right)\hat{\mathbf{i}} + \left(\frac{\Delta y}{\Delta t}\right)\hat{\mathbf{j}}$
  • Instantaneous velocity $\mathbf{v}$ as $\Delta t \to 0$:     $\mathbf{v} = \lim_{\Delta t \to 0} \frac{\Delta\mathbf{r}}{\Delta t} = \frac{d\mathbf{r}}{dt}$$\mathbf{v} = v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}}$     Where scalar velocity components are:     $v_x = \frac{dx}{dt}, \quad v_y = \frac{dy}{dt}$
  • Magnitude of instantaneous velocity $v$:     $v = \sqrt{v_x^2 + v_y^2}$
  • Direction angle $\theta$ of velocity vector $\mathbf{v}$ relative to the $x$-axis:     $\tan(\theta) = \frac{v_y}{v_x} \implies \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$
  • Direction property: At any point on a curved path, the direction of instantaneous velocity is strictly tangential to the path at that point.

• Acceleration Vector:

  • Average acceleration $\bar{\mathbf{a}}$ over time interval $\Delta t$:     $\bar{\mathbf{a}} = \frac{\Delta\mathbf{v}}{\Delta t} = \frac{\Delta (v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}})}{\Delta t} = \left(\frac{\Delta v_x}{\Delta t}\right)\hat{\mathbf{i}} + \left(\frac{\Delta v_y}{\Delta t}\right)\hat{\mathbf{j}}$
  • Instantaneous acceleration $\mathbf{a}$ as $\Delta t \to 0$:     $\mathbf{a} = \lim_{\Delta t \to 0} \frac{\Delta\mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}$$\mathbf{a} = a_x\hat{\mathbf{i}} + a_y\hat{\mathbf{j}}$     Where scalar acceleration components are:     $a_x = \frac{dv_x}{dt}, \quad a_y = \frac{dv_y}{dt}$

• Worked Example: Kinematic Derivation (Example 4.4):

  • Problem: The position of a particle as a function of time $t$ is given by:     $\mathbf{r}(t) = 3.0t\hat{\mathbf{i}} + 2.0t^2\hat{\mathbf{j}} + 5.0\hat{\mathbf{k}}$     where $t$ is in seconds and coefficients produce $\mathbf{r}$ in meters.
  • Part (a): Find $\mathbf{v}(t)$ and $\mathbf{a}(t)$.     $\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}\left(3.0t\hat{\mathbf{i}} + 2.0t^2\hat{\mathbf{j}} + 5.0\hat{\mathbf{k}}\right) = 3.0\hat{\mathbf{i}} + 4.0t\hat{\mathbf{j}}$$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}\left(3.0\hat{\mathbf{i}} + 4.0t\hat{\mathbf{j}}\right) = +4.0\hat{\mathbf{j}}\,\text{m\,s}^{-2}$     (Acceleration is constant with magnitude $4.0\,\text{m\,s}^{-2}$ directed entirely along the positive $y$-axis).
  • Part (b): Find the magnitude and direction of $\mathbf{v}(t)$ at $t = 1.0\,\text{s}$.     $\mathbf{v}(1.0) = 3.0\hat{\mathbf{i}} + 4.0(1.0)\hat{\mathbf{j}} = 3.0\hat{\mathbf{i}} + 4.0\hat{\mathbf{j}}$
  • Magnitude $v$:       $v = \sqrt{3.0^2 + 4.0^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\,\text{m\,s}^{-1}$
  • Direction angle $\theta$ with positive $x$-axis:       $\tan(\theta) = \frac{v_y}{v_x} = \frac{4.0}{3.0} \approx 1.333$$\theta = \tan^{-1}(1.333) \approx 53^\circ$

Projectile Motion

• Definition:

  • An object launched into space and allowed to move freely under the influence of gravity alone is defined as a projectile.
  • The path traversed by a projectile is called its trajectory, which is parabolic in shape.

• Initial Velocity Components:

  • If an object is projected with initial speed $u$ at an angle $\theta$ (or $\theta_0$) with the horizontal:
  • Horizontal initial component: $u_x = u\cos(\theta)$
  • Vertical initial component: $u_y = u\sin(\theta)$

• Acceleration Characteristics:

  • Horizontal acceleration $a_x = 0$ (neglecting air resistance), meaning the horizontal velocity component $u_x = u\cos(\theta)$ remains constant throughout the entire flight.
  • Vertical acceleration $a_y = -g$ (directed vertically downward).
  • At the peak point of maximum height, vertical velocity component becomes zero ($v_y = 0$), while horizontal velocity component remains unchanged ($v_x = u\cos(\theta)$).

• Equation of the Path (Trajectory Derivation):

  • Displacement along $x$-axis at time $t$:     $x = u_x t = u\cos(\theta) t \implies t = \frac{x}{u\cos(\theta)}$
  • Displacement along $y$-axis at time $t$:     $y = u_y t - \frac{1}{2}gt^2 = u\sin(\theta) t - \frac{1}{2}gt^2$
  • Substituting $t = \frac{x}{u\cos(\theta)}$ into vertical displacement equation:     $y = u\sin(\theta)\left(\frac{x}{u\cos(\theta)}\right) - \frac{1}{2}g\left(\frac{x}{u\cos(\theta)}\right)^2$$y = \tan(\theta) x - \left(\frac{g}{2u^2\cos^2(\theta)}\right)x^2$
  • Analysis: This equation is of the standard mathematical form $y = ax + bx^2$ (where $a = \tan(\theta)$ and $b = -\frac{g}{2u^2\cos^2(\theta)}$ are constants), proving that the path of a projectile is a parabola.

• Time of Flight ($T$):

  • Definition: The total duration $T$ during which the projectile remains in flight.
  • Derivation: Considering vertical motion from launch until landing back at the initial horizontal level ($s_y = 0$):     $s_y = u_y T + \frac{1}{2}a_y T^2$$0 = u\sin(\theta) T - \frac{1}{2}g T^2$$u\sin(\theta) T = \frac{1}{2}g T^2$$T = \frac{2u\sin(\theta)}{g}$

• Maximum Height ($H$):

  • Definition: The maximum vertical distance reached by the projectile above its projection plane.

• Derivation: Considering vertical motion up to the highest point where vertical speed $v_y = 0$:     $v_y^2 - u_y^2 = 2a_y s_y$$0 - (u\sin(\theta))^2 = 2(-g) H$$-u^2\sin^2(\theta) = -2gH$$H = \frac{u^2\sin^2(\theta)}{2g}$

  • Horizontal Range ($R$):

• Definition: The total horizontal distance covered by the projectile during its time of flight $T$

• Derivation:     $R = \text{Horizontal Component of Velocity} \times \text{Time of Flight}$$R = (u\cos(\theta)) \times \left(\frac{2u\sin(\theta)}{g}\right)$$R = \frac{u^2 (2\sin(\theta)\cos(\theta))}{g}$     Utilizing trigonometric identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$:     $R = \frac{u^2\sin(2\theta)}{g}$

• Condition for Maximum Horizontal Range ($R_{\max}$):

  • $R$ is maximum when $\sin(2\theta) = 1$, which occurs when $2\theta = 90^\circ \implies \theta = 45^\circ$$R_{\max} = \frac{u^2}{g}$

  • Equality of Ranges for Complementary Angles ($\theta$ and $90^\circ - \theta$):

• For angle of projection $\theta$:     $R_1 = \frac{u^2\sin(2\theta)}{g}$

• For angle of projection $90^\circ - \theta$:     $R_2 = \frac{u^2\sin(2(90^\circ - \theta))}{g} = \frac{u^2\sin(180^\circ - 2\theta)}{g}$

• Since $\sin(180^\circ - 2\theta) = \sin(2\theta)$:     $R_2 = \frac{u^2\sin(2\theta)}{g} = R_1$

• Conclusion: For a given initial speed $u$, the horizontal range is identical for launch angles $\theta$ and $\left(90^\circ - \theta\right)$.

  • Worked Example: Cricket Ball Trajectory Calculations:

• Problem: A cricket ball is thrown with speed $u = 28\,\text{m\,s}^{-1}$ at an angle $\theta = 30^\circ$ above the horizontal. Calculate (a) maximum height $H$, (b) time of flight $T$, and (c) horizontal range $R$ (take $g = 9.8\,\text{m\,s}^{-2}$).

• Solutions:

  • (a) Maximum Height $H$:       $H = \frac{u^2\sin^2(\theta)}{2g} = \frac{28^2 \times \sin^2(30^\circ)}{2 \times 9.8} = \frac{784 \times (0.5)^2}{19.6} = \frac{784 \times 0.25}{19.6} = 10\,\text{m}$
  • (b) Time of Flight $T$:       $T = \frac{2u\sin(\theta)}{g} = \frac{2 \times 28 \times \sin(30^\circ)}{9.8} = \frac{56 \times 0.5}{9.8} = \frac{28}{9.8} \approx 2.9\,\text{s}$
  • (c) Horizontal Range $R$:       $R = \frac{u^2\sin(2\theta)}{g} = \frac{28^2 \times \sin(60^\circ)}{9.8} = \frac{784 \times 0.866}{9.8} \approx 69\,\text{m}$

Uniform Circular Motion

• Definition:

  • When an object traverses a circular path of fixed radius at constant speed, its motion is classified as uniform circular motion.
  • The term "uniform" refers specifically to constant speed; velocity itself changes continuously because its direction varies at every instant.

• Time Period ($T$) and Frequency ($\nu$):

  • Time Period ($T$): The time required for an object to complete one full circular revolution.
  • Frequency ($\nu$): The number of complete revolutions executed per unit time (1 second).     $\nu = \frac{1}{T}$
  • Unit of frequency: hertz (Hz) or $\text{s}^{-1}$.

• Angular Velocity ($\omega$):

  • Definition: The time rate of change of angular displacement $\Delta\theta$.     $\omega = \frac{\Delta\theta}{\Delta t}$
  • Instantaneous angular velocity:     $\omega = \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} = \frac{d\theta}{dt}$
  • Unit of angular velocity: radians per second (rad/s).
  • Relation to time period $T$ and frequency $\nu$:
  • In one full period $T$, angular displacement is $2\pi$ radians:       $\omega = \frac{2\pi}{T} = 2\pi\nu$

• Relationship Between Linear Velocity ($v$) and Angular Velocity ($\omega$):

  • Consider motion along circular arc of radius $r$ through angle $\Delta\theta$ with arc length $\Delta r$:

• Angular relation:     $\Delta\theta = \frac{\Delta r}{r} \implies \Delta r = r \Delta\theta$

• Dividing both sides by $\Delta t$:     $\frac{\Delta r}{\Delta t} = r \left(\frac{\Delta\theta}{\Delta t}\right)$

• Taking limit as $\Delta t \to 0$:     $v = r\omega$

  • Angular Acceleration ($\alpha$):

• Definition: The rate of change of angular velocity with respect to time.     $\alpha = \frac{d\omega}{dt} = \frac{d}{dt}\left(\frac{d\theta}{dt}\right) = \frac{d^2\theta}{dt^2}$

  • Centripetal Acceleration ($a_c$):

• Definition: An object undergoing uniform circular motion experiences an acceleration directed along the radius towards the center of the circular path, termed centripetal acceleration.

• Derivation via Similar Triangles:

  • Let position vectors be $\mathbf{r}$ and $\mathbf{r}'$ and velocity vectors be $\mathbf{v}$ and $\mathbf{v}'$ at points $P$ and $P'$ separated by angle $\Delta\theta$.
  • The isosceles triangle formed by $\mathbf{r}$, $\mathbf{r}'$, and $\Delta\mathbf{r}$ is geometrically similar to the triangle formed by $\mathbf{v}$, $\mathbf{v}'$, and $\Delta\mathbf{v}$:       $\frac{|\Delta\mathbf{v}|}{v} = \frac{|\Delta\mathbf{r}|}{r} \implies |\Delta\mathbf{v}| = \frac{v}{r}|\Delta\mathbf{r}|$
  • Dividing by $\Delta t$ and taking limit $\Delta t \to 0$:       $a_c = \lim_{\Delta t \to 0} \frac{|\Delta\mathbf{v}|}{\Delta t} = \frac{v}{r} \left(\lim_{\Delta t \to 0} \frac{|\Delta\mathbf{r}|}{\Delta t}\right) = \frac{v}{r} \times v = \frac{v^2}{r}$
  • Replacing radius $r$ with radius $R$:       $a_c = \frac{v^2}{R}$

• Equivalent Formulations for Centripetal Acceleration:

  • Using $v = R\omega$:       $a_c = \frac{(R\omega)^2}{R} = \omega^2 R$

  • Using $R = \frac{v}{\omega}$:       $a_c = v\omega$

  • Worked Example: Insect Trapped in a Circular Groove:

• Problem: An insect trapped in a circular groove of radius $R = 12\,\text{cm} = 0.12\,\text{m}$ moves steadily, completing $7$ revolutions in $100\,\text{s}$.

  • (a) Calculate the angular speed $\omega$ and linear speed $v$.
  • (b) Is the acceleration vector constant? Find its magnitude.

• Solutions:

  • Period $T = \frac{100}{7}\,\text{s}$.
  • (a) Angular Speed $\omega$:       $\omega = \frac{2\pi}{T} = \frac{2\pi}{\left(\frac{100}{7}\right)} = \frac{2\pi \times 7}{100} = \frac{14\pi}{100} \approx 0.44\,\text{rad/s}$       Linear Speed $v$:       $v = \omega R = 0.44 \times 0.12 = 5.3 \times 10^{-2}\,\text{m\,s}^{-1}$
  • (b) Acceleration Analysis:
    • The velocity vector $\mathbf{v}$ is everywhere tangent to the circle, while centripetal acceleration $\mathbf{a}_c$ points towards the center.
    • Because the direction of motion continuously rotates, the acceleration vector direction changes continuously; hence, the acceleration vector is not constant.
    • Magnitude of centripetal acceleration:         $a_c = \omega^2 R = (0.44)^2 \times 0.12 \approx 0.1936 \times 0.12 = 2.3 \times 10^{-2}\,\text{m\,s}^{-2}$