Motion in a Plane

MOTION IN A PLANE — COMPREHENSIVE CHAPTER NOTES

INTRODUCTION TO MOTION IN A PLANE

**Definition**: Motion in a plane refers to the movement of an object in two dimensions where position, displacement, velocity, and acceleration require vector representation, unlike one-dimensional motion where ± signs suffice.

**Why Vectors are Essential**:

In 1D motion, direction is represented by + or − signs (only two directions possible)

In 2D/3D motion, infinite directions exist; vectors with magnitude and direction are necessary

Projectile motion, circular motion, and planetary motion occur in planes

**Scope of Chapter**:

Vector algebra (addition, subtraction, multiplication)

Resolution of vectors into components

Motion with constant acceleration in 2D

Projectile motion (detailed analysis)

Uniform circular motion

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SCALARS AND VECTORS

**Definition of Scalar**: A quantity specified completely by magnitude (a number) and proper unit. Direction has no meaning. Examples: distance, mass, temperature, time, speed, volume, density, energy.

**Definition of Vector**: A quantity possessing both magnitude and direction that obeys triangle law or parallelogram law of addition. Examples: displacement, velocity, acceleration, force, momentum, angular momentum.

**Key Difference**:

Scalars follow ordinary algebra (add/subtract/multiply like normal numbers)

Vectors follow special vector algebra (cannot be added/subtracted arbitrarily without considering direction)

**Representation**:

Bold face in printed text: **v**, **A**, **F**

Arrow notation by hand: $\vec{v}$, $\vec{A}$, $\vec{F}$

Magnitude denoted: |**v**| = v (light face, no bold)

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POSITION AND DISPLACEMENT VECTORS

**Position Vector**: A vector drawn from an arbitrary origin O to the point where object is located. Denoted as $\vec{r}$ = OP.

Magnitude: length from O to P

Direction: along OP from O to P

Changes with choice of origin

**Displacement Vector**: Vector joining initial position P (at time t) to final position P′ (at time t′). Denoted as $\vec{PP'}$ = $\vec{r'}$ − $\vec{r}$.

**Critical Property**: Displacement depends only on initial and final positions, NOT on the path taken between them.

**Example**: If object travels from P to Q via path PABCQ, PDQ, or PBEFQ, displacement vector $\vec{PQ}$ remains identical for all paths. Magnitude of displacement ≤ path length (equality only for straight-line motion).

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EQUALITY OF VECTORS

**Definition**: Two vectors **A** and **B** are equal if and only if:

1. They have same magnitude: |**A**| = |**B**|

2. They have same direction

**Test for Equality**: Shift **B** parallel to itself until its tail coincides with tail of **A**. If tips also coincide, vectors are equal (**A** = **B**).

**Important Note**: Vectors having same magnitude but different directions are NOT equal. Vectors are "free vectors" — their location does not matter, only magnitude and direction matter.

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MULTIPLICATION OF VECTORS BY REAL NUMBERS

**Multiplication by Positive Scalar λ**:

Result: λ**A** (where λ > 0)

Magnitude: |λ**A**| = λ|**A**|

Direction: Same as **A**

Example: 2**A** has magnitude doubled, direction unchanged

**Multiplication by Negative Scalar −λ**:

Result: −λ**A**

Magnitude: |−λ**A**| = λ|**A**|

Direction: Opposite to **A**

Example: −1.5**A** points opposite to **A**, magnitude = 1.5|**A**|

**Dimensional Consideration**: If λ has physical dimension (e.g., time) and **A** is a vector (e.g., velocity), then λ**A** has dimension = (dimension of λ) × (dimension of **A**).

**Example**: velocity × time = displacement (dimensionally correct).

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ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD

Triangle Law of Vector Addition

**Statement**: If two vectors **A** and **B** are represented as two sides of a triangle in sequence (head of **A** to tail of **B**), the third side represents their resultant **R** = **A** + **B**.

**Procedure** (Head-to-Tail Method):

1. Draw vector **A** from origin O

2. From head of **A**, draw vector **B**

3. Join origin O to head of **B** — this is resultant **R**

4. **R** = **A** + **B**

Parallelogram Law of Vector Addition

**Statement**: If two vectors **A** and **B** are sides of a parallelogram from common origin, the diagonal represents their resultant.

**Procedure**:

1. Draw **A** and **B** from common origin O

2. Complete parallelogram: draw line from head of **A** parallel to **B**

3. Draw line from head of **B** parallel to **A**

4. Diagonal OS from O represents **R** = **A** + **B**

**Equivalence**: Both methods give identical resultant; they are equivalent.

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Properties of Vector Addition

**Commutative Law**:

**A** + **B** = **B** + **A**

(Order of addition doesn't matter; resultant is same)

**Associative Law**:

(**A** + **B**) + **C** = **A** + (**B** + **C**)

(Grouping order irrelevant)

**Addition with Null Vector**:

**A** + **0** = **A**

(Null vector or zero vector **0** has zero magnitude, undefined direction)

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Vector Subtraction

**Definition**:

**A** − **B** = **A** + (−**B**)

where −**B** is vector equal in magnitude to **B** but opposite in direction.

**Procedure**:

1. Reverse direction of **B** to get −**B**

2. Add **A** and −**B** using triangle/parallelogram method

3. Result is **R** = **A** − **B**

**Null Vector from Equal Opposite Vectors**:

**A** + (−**A**) = **0**

**Physical Significance**: If object returns to starting point, displacement = **0** (null vector).

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WORKED EXAMPLE: Rain and Wind Problem

**Problem**: Rain falls vertically at 35 m/s. Wind blows east-to-west at 12 m/s. Find umbrella direction.

**Solution**:

Velocity of rain: $\vec{v_r}$ = 35 m/s (vertical downward)

Velocity of wind: $\vec{v_w}$ = 12 m/s (horizontal, east-to-west)

Resultant: $\vec{R}$ = $\sqrt{v_r^2 + v_w^2}$ = $\sqrt{35^2 + 12^2}$ = $\sqrt{1225 + 144}$ = $\sqrt{1369}$ = **37 m/s**

Direction: tan θ = $\frac{v_w}{v_r}$ = $\frac{12}{35}$ = 0.343

θ = tan⁻¹(0.343) = **19° from vertical towards east**

**Answer**: Boy holds umbrella at 19° to vertical in vertical plane towards east.

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RESOLUTION OF VECTORS

**Definition**: Expressing a given vector as sum of component vectors along chosen directions.

**Concept**: Any vector **A** in a plane can be expressed as:

**A** = λ**a** + μ**b**

where **a** and **b** are non-colinear reference vectors; λ and μ are scalars.

**Geometric Method**: Through tail O of **A**, draw line parallel to **a**; through head P, draw line parallel to **b**; these lines meet at Q. Then:

**A** = **OQ** + **QP** = λ**a** + μ**b**

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UNIT VECTORS AND RECTANGULAR COMPONENTS

**Definition of Unit Vector**: A vector of magnitude 1 pointing in specific direction. No dimensions, no units; purely directional.

**Standard Unit Vectors**:

$\hat{i}$ : along positive x-axis, |$\hat{i}$| = 1

$\hat{j}$ : along positive y-axis, |$\hat{j}$| = 1

$\hat{k}$ : along positive z-axis, |$\hat{k}$| = 1

These are mutually perpendicular: $\hat{i}$ ⊥ $\hat{j}$ ⊥ $\hat{k}$

**Resolution in 2D**: For vector **A** in x-y plane making angle θ with x-axis:

**A** = A_x$\hat{i}$ + A_y$\hat{j}$

where:

A_x = A cos θ (x-component)

A_y = A sin θ (y-component)

A = |**A**| = magnitude

**Component Values**: Components can be positive, negative, or zero depending on angle θ.

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CONVERTING BETWEEN FORMS

**Given magnitude A and angle θ → Find A_x and A_y**:

A_x = A cos θ

A_y = A sin θ

**Given A_x and A_y → Find magnitude A and angle θ**:

**Magnitude**:

$$A = \sqrt{A_x^2 + A_y^2} \quad \text{...(3.14)}$$

**Direction**:

$$\tan \theta = \frac{A_y}{A_x} \implies \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \quad \text{...(3.15)}$$

**3D Resolution**: For vector **A** in 3D space making angles α, β, γ with x-, y-, z-axes:

A_x = A cos α

A_y = A cos β

A_z = A cos γ

$$A = \sqrt{A_x^2 + A_y^2 + A_z^2}$$

**Position Vector**:

$$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$

where x, y, z are coordinates of point.

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VECTOR ADDITION — ANALYTICAL METHOD

**Principle**: Add corresponding components of vectors separately.

Addition of Two Vectors in 2D

**Given**:

**A** = A_x$\hat{i}$ + A_y$\hat{j}$

**B** = B_x$\hat{i}$ + B_y$\hat{j}$

**Resultant**:

$$\vec{R} = \vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$$

**Component form**:

R_x = A_x + B_x

R_y = A_y + B_y

**Magnitude and Direction of R**:

$$R = \sqrt{R_x^2 + R_y^2} = \sqrt{(A_x + B_x)^2 + (A_y + B_y)^2}$$

$$\tan \phi = \frac{R_y}{R_x} = \frac{A_y + B_y}{A_x + B_x}$$

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Addition in 3D

**Given**:

**A** = A_x$\hat{i}$ + A_y$\hat{j}$ + A_z$\hat{k}$

**B** = B_x$\hat{i}$ + B_y$\hat{j}$ + B_z$\hat{k}$

**Resultant**:

$$\vec{R} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j} + (A_z + B_z)\hat{k}$$

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Multiple Vector Addition

For vectors **a**, **b**, **c**, their sum **T** = **a** + **b** − **c** has components:

T_x = a_x + b_x − c_x

T_y = a_y + b_y − c_y

T_z = a_z + b_z − c_z

**Advantage of Analytical Method**: More accurate than graphical; easily programmable; handles multiple vectors efficiently.

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RESULTANT OF TWO VECTORS — MAGNITUDE AND DIRECTION

Law of Cosines

**Statement**: For two vectors **A** and **B** making angle θ between them, resultant **R**:

$$R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \quad \text{...(3.24a)}$$

**Derivation**: Using parallelogram OQSP where angle between **A** and **B** is θ:

Diagonal OS = R

Drop perpendicular from S to extended OP, meeting at N

ON = A + B cos θ

SN = B sin θ

By Pythagoras: R² = (A + B cos θ)² + (B sin θ)²

Expanding: R² = A² + B² cos² θ + 2AB cos θ + B² sin² θ

Simplifying: **R² = A² + B² + 2AB cos θ**

**Special Cases**:

θ = 0° (parallel): R = A + B (maximum)

θ = 90° (perpendicular): R = √(A² + B²)

θ = 180° (antiparallel): R = |A − B| (minimum)

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Law of Sines

**Statement**: The resultant R and its direction with respect to vector A:

$$\sin\alpha = \frac{B\sin\theta}{R} \quad \text{...(3.24e)}$$

$$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \quad \text{...(3.24f)}$$

where α is angle between **R** and **A**.

**Derivation**: In triangle formed by **A**, **B**, **R**:

From geometry: R sin α = B sin θ (perpendicular distance relation)

Therefore: sin α = (B sin θ)/R

Tangent form: tan α = (perpendicular distance)/(adjacent distance)

**Complete Law of Sines**:

$$\frac{R}{\sin\theta} = \frac{A}{\sin\beta} = \frac{B}{\sin\alpha} \quad \text{...(3.24d)}$$

where α, β, θ are angles opposite to **R**, **A**, **B** respectively.

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WORKED EXAMPLE: Motorboat Problem

**Problem**: Motorboat races north at 25 km/h. Water current 10 km/h at 60° east of south. Find resultant velocity.

**Solution**:

Let north = positive y-axis, east = positive x-axis.

**Boat velocity**: $\vec{v_b}$ = 25$\hat{j}$ km/h (northward)

**Current velocity**: "60° east of south" means 60° from south towards east = 180° − 60° = 120° from north (or −60° from south axis)

Component east: v_c,x = 10 sin(60°) = 10 × (√3/2) = 5√3 km/h

Component south: v_c,y = −10 cos(60°) = −10 × (1/2) = −5 km/h

$\vec{v_c}$ = 5√3$\hat{i}$ − 5$\hat{j}$ km/h

**Resultant velocity**:

$$\vec{v_r} = \vec{v_b} + \vec{v_c} = 5\sqrt{3}\hat{i} + (25-5)\hat{j} = 5\sqrt{3}\hat{i} + 20\hat{j}$$

**Magnitude**:

$$v_r = \sqrt{(5\sqrt{3})^2 + 20^2} = \sqrt{75 + 400} = \sqrt{475} = 5\sqrt{19} \approx 21.8 \text{ km/h}$$

**Direction**: tan φ = $\frac{20}{5\sqrt{3}}$ = $\frac{4}{\sqrt{3}}$ = $\frac{4\sqrt{3}}{3}$ ≈ 2.31

φ = tan⁻¹(2.31) ≈ 67° from east towards north = 23° from north towards east.

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MOTION IN A PLANE

**Definition**: Displacement, velocity, and acceleration in two dimensions where all quantities are vectors.

**Position Vector**:

$$\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$$

**Displacement**:

$$\Delta\vec{r} = \vec{r_2} - \vec{r_1} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} = \Delta x\hat{i} + \Delta y\hat{j}$$

**Velocity Vector**: Rate of change of position

$$\vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = v_x\hat{i} + v_y\hat{j}$$

**Average Velocity**:

$$\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t}$$

**Instantaneous Velocity**:

$$\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta\vec{r}}{\Delta t}$$

**Speed**: Magnitude of velocity

$$|\vec{v}| = \sqrt{v_x^2 + v_y^2}$$

**Acceleration Vector**: Rate of change of velocity

$$\vec{a} = \frac{d\vec{v}}{dt} = \frac{dv_x}{dt}\hat{i} + \frac{dv_y}{dt}\hat{j} = a_x\hat{i} + a_y\hat{j}$$

**Average Acceleration**:

$$\vec{a}_{avg} = \frac{\Delta\vec{v}}{\Delta t}$$

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MOTION IN A PLANE WITH CONSTANT ACCELERATION

**Assumption**: Acceleration **a** is constant (magnitude and direction both constant).

**Equations of Motion** (Vector Form):

$$\vec{v} = \vec{v_0} + \vec{a}t \quad \text{...(3.25)}$$

$$\vec{r} = \vec{r_0} + \vec{v_0}t + \frac{1}{2}\vec{a}t^2 \quad \text{...(3.26)}$$

$$|\vec{v}|^2 = |\vec{v_0}|^2 + 2\vec{a} \cdot (\vec{r} - \vec{r_0}) \quad \text{...(3.27)}$$

**Component Form** (Most Practical):

**x-direction**:

v_x = v₀ₓ + a_x t

x = x₀ + v₀ₓ t + ½a_x t²

v_x² = v₀ₓ² + 2a_x(x − x₀)

**y-direction**:

v_y = v₀ᵧ + a_y t

y = y₀ + v₀ᵧ t + ½a_y t²

v_y² = v₀ᵧ² + 2a_y(y − y₀)

**Advantage of Component Method**: x and y motions are independent; solve separately using 1D equations.

**Key Principle**: Acceleration in one direction does not affect motion in perpendicular direction. This principle is foundation of projectile motion.

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PROJECTILE MOTION

**Definition**: Motion of an object launched into air, moving under gravity alone (air resistance negligible), following a parabolic path.

**Types of Projectile Motion**:

1. Horizontal projection (launched horizontally)

2. Oblique projection (launched at angle θ to horizontal)

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HORIZONTAL PROJECTILE MOTION

**Initial Conditions**:

Initial velocity $\vec{v_0}$ = v₀$\hat{i}$ (horizontal)

Initial height = h

Acceleration = −g$\hat{j}$ (gravity, downward)

**Equations**:

**x-direction** (no acceleration):

v_x = v₀ (constant)

x = v₀t

**y-direction** (constant acceleration −g):

v_y = 0 − gt = −gt

y = h − ½gt²

v_y² = 2gh

**Time of Flight** (when y = 0):

$$0 = h - \frac{1}{2}gt^2 \implies T = \sqrt{\frac{2h}{g}}$$

**Horizontal Range** (distance traveled horizontally):

$$R = v_0 T = v_0\sqrt{\frac{2h}{g}}$$

**Velocity at Impact**:

Horizontal component: v_x = v₀ (unchanged)

Vertical component: v_y = gT = g√(2h/g) = √(2gh)

Total velocity: v = √(v₀² + 2gh)

Angle with horizontal: tan β = v_y/v_x = √(2gh)/v₀

**Trajectory Equation** (Parabolic Path):

Eliminating t from x = v₀t and y = h − ½gt²:

$$t = \frac{x}{v_0} \implies y = h - \frac{gx^2}{2v_0^2}$$

This is equation of parabola (y ∝ −x²).

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OBLIQUE PROJECTILE MOTION

**Initial Conditions**:

Initial velocity v₀ at angle θ to horizontal

v₀ₓ = v₀ cos θ

v₀ᵧ = v₀ sin θ

Acceleration: a = −g$\hat{j}$

**Key Equations**:

**Velocity Components**:

v_x = v₀ cos θ (constant)

v_y = v₀ sin θ − gt

**Position**:

x = v₀ cos θ · t

y = v₀ sin θ · t − ½gt²

**Time of Flight** (when y = 0):

$$T = \frac{2v_0\sin\theta}{g} \quad \text{...(3.28)}$$

**Horizontal Range**:

$$R = v_0\cos\theta \cdot T = v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g} = \frac{2v_0^2\sin\theta\cos\theta}{g}$$

Using sin 2θ = 2 sin θ cos θ:

$$R = \frac{v_0^2\sin2\theta}{g} \quad \text{...(3.29)}$$

**Maximum Height** (when v_y = 0):

Time to max height: t = $\frac{v_0\sin\theta}{g}$

$$H = v_0\sin\theta \cdot t - \frac{1}{2}gt^2 = v_0\sin\theta \cdot \frac{v_0\sin\theta}{g} - \frac{1}{2}g\left(\frac{v_0\sin\theta}{g}\right)^2$$

$$H = \frac{v_0^2\sin^2\theta}{2g} \quad \text{...(3.30)}$$

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IMPORTANT RESULTS FOR PROJECTILE MOTION

**Maximum Range**: R_max occurs when sin 2θ = 1, i.e., θ = 45°

$$R_{max} = \frac{v_0^2}{g}$$

**Complementary Angles**: Angles θ and (90° − θ) give same range but different heights and times.

**Trajectory Equation** (Path):

$$y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta} \quad \text{...(3.31)}$$

This is parabolic path — characteristic equation of projectile.

**Velocity at Any Time**:

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(v_0\cos\theta)^2 + (v_0\sin\theta - gt)^2}$$

**Angle of Velocity with Horizontal**:

$$\tan\alpha = \frac{v_y}{v_x} = \frac{v_0\sin\theta - gt}{v_0\cos\theta}$$

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WORKED EXAMPLE: OBLIQUE PROJECTILE

**Problem**: Ball projected at 30° to horizontal with v₀ = 20 m/s. Find: (a) time of flight, (b) range, (c) maximum height. (g = 10 m/s²)

**Solution**:

**(a) Time of Flight**:

$$T = \frac{2v_0\sin\theta}{g} = \frac{2 \times 20 \times \sin30°}{10} = \frac{2 \times 20 \times 0.5}{10} = \frac{20}{10} = 2 \text{ s}$$

**(b) Range**:

$$R = \frac{v_0^2\sin2\theta}{g} = \frac{20^2 \times \sin60°}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = \frac{200\sqrt{3}}{10} = 20\sqrt{3} \approx 34.6 \text{ m}$$

**(c) Maximum Height**:

$$H = \frac{v_0^2\sin^2\theta}{2g} = \frac{20^2 \times \sin^230°}{2 \times 10} = \frac{400 \times (0.5)^2}{20} = \frac{400 \times 0.25}{20} = \frac{100}{20} = 5 \text{ m}$$

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UNIFORM CIRCULAR MOTION

**Definition**: Motion of an object along a circular path with constant speed (v = constant, but velocity direction changes continuously).

**Important Note**: Although speed is constant, velocity is NOT constant (direction changes); hence acceleration exists despite constant speed.

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PARAMETERS OF CIRCULAR MOTION

**Radius of Circle**: r = constant

**Period (T)**: Time for one complete revolution

$$T = \frac{2\pi r}{v} \quad \text{(in seconds)}$$

**Frequency (f)**: Number of revolutions per unit time

$$f = \frac{1}{T} = \frac{v}{2\pi r} \quad \text{(in Hz or rev/s)}$$

**Angular Velocity (ω)**: Rate of change of angular position (in rad/s)

$$\omega = \frac{2\pi}{T} = 2\pi f = \frac{v}{r} \quad \text{...(3.32)}$$

**Relation between v and ω**:

$$v = \omega r \quad \text{...(3.33)}$$

(Linear speed = angular velocity × radius)

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CENTRIPETAL ACCELERATION

**Concept**: Although speed is constant, direction of velocity changes continuously. This requires centripetal (center-seeking) acceleration perpendicular to velocity.

**Derivation**: Consider object at positions P and Q on circle, separated by angle Δθ and time Δt.

Velocity at P: $\vec{v_P}$ with magnitude v, direction tangent at P

Velocity at Q: $\vec{v_Q}$ with magnitude v (same), but tangent at Q (different direction)

Change in velocity: Δ$\vec{v}$ = $\vec{v_Q}$ − $\vec{v_P}$

For small Δθ, |Δ$\vec{v}$| = v sin(Δθ) ≈ v·Δθ (in radians)

Acceleration: a = |Δ$\vec{v}$|/Δt = v·Δθ/Δt = v·ω = v²/r

$$a_c = \frac{v^2}{r} = \omega^2 r = v\omega \quad \text{...(3.34)}$$

**Direction**: Always towards center (centripetal).

**Characteristics of Centripetal Acceleration**:

Perpendicular to velocity

Directed towards center

Magnitude = v²/r = ω²r

Changes continuously in direction (tangent to circle at each point)

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CENTRIPETAL FORCE

**Definition**: Net force towards center required to maintain circular motion.

$$F_c = ma_c = m\frac{v^2}{r} = m\omega^2 r \quad \text{...(3.35)}$$

**SI Unit**: Newton (N)

**Source of Centripetal Force**: Depends on situation:

String tension: F = T

Normal force: F = N

Friction: F = f

Gravity: F = mg (for planetary motion)

Component of applied force

**Important**: Centripetal force is NOT a new force; it's NET force in centripetal direction.

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WORKED EXAMPLE: CIRCULAR MOTION

**Problem**: Car of mass 1000 kg moves in circular path of radius 100 m at constant speed 20 m/s. Find: (a) centripetal acceleration, (b) centripetal force.

**Solution**:

**(a) Centripetal Acceleration**:

$$a_c = \frac{v^2}{r} = \frac{(20)^2}{100} = \frac{400}{100} = 4 \text{ m/s}^2$$

**(b) Centripetal Force**:

$$F_c = ma_c = 1000 \times 4 = 4000 \text{ N} = 4 \text{ kN}$$

Alternatively:

$$F_c = m\omega^2 r \text{ where } \omega = \frac{v}{r} = \frac{20}{100} = 0.2 \text{ rad/s}$$

$$F_c = 1000 \times (0.2)^2 \times 100 = 1000 \times 0.04 \times 100 = 4000 \text{ N}$$

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ANGULAR KINEMATICS IN CIRCULAR MOTION

**Angular Displacement**: θ = angle swept (in radians)

$$\theta = \frac{s}{r}$$

where s = arc length.

**Angular Velocity**:

$$\omega = \frac{d\theta}{dt}$$

**Angular Acceleration**: α = dω/dt (only if speed changes; α = 0 for uniform circular motion)

**Kinematic Equations** (Uniform Circular Motion, α = 0):

θ = ωt (starting from θ = 0)

v = ωr (tangential speed)

a_c = ω²r = v²/r (centripetal acceleration)

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ISSUES AND MISCONCEPTIONS CLARIFIED

**Issue 1**: "If speed is constant, acceleration is zero."

**Clarification**: Speed constant means magnitude of velocity constant. But direction changes, so velocity changes. Hence, acceleration ≠ 0. Centripetal acceleration = v²/r.

**Issue 2**: "Centripetal force is a new force."

**Clarification**: NO. Centripetal force is NET force in radial direction (towards center). It could be tension, normal force, friction, or component of any existing force.

**Issue 3**: "An object in circular motion will move outward (centrifugal force)."

**Clarification**: Centrifugal force is pseudo/fictitious force that appears only in rotating reference frame. In inertial frame,