Comprehensive Study Notes: Newtonian Mechanics, Torque, Angular Momentum, and Circular Motion

Newton's Laws of Motion

Newton's First Law of Motion (Law of Inertia)

• Definition: Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.

• Physical Concepts and Observations:

  • Sliding a book across a horizontal, rough tabletop causes it to slow down and eventually stop due to the external friction force exerted by the tabletop.

  • Sliding the same book on a slab of ice results in less deceleration because the frictional force is significantly reduced.

  • On an idealized, completely frictionless surface, an object given an initial velocity would continue moving indefinitely in a straight line at constant velocity with no external force required to sustain its motion.

• Concept of Inertia:

  • Inertia is the natural tendency of objects to resist changes in their state of motion (resist acceleration).

  • Because of this fundamental property, Newton's First Law is universally referred to as the Law of Inertia.

• Inertial Frame of Reference:

  • An inertial frame of reference is a non-accelerating perspective used to determine whether an object is at rest or moving with constant velocity.

  • Key properties of an inertial frame of reference:

  • It moves with constant velocity (i.e., it is at rest or moving with constant speed in a straight linear direction).

  • It has zero acceleration ($\vec{a} = 0$).

  • Newton's First Law holds strictly true within this frame.

  • No fictitious or pseudo-forces arise within it.

  • Any object viewed from an inertial frame of reference obeys the First Law and continues moving with constant velocity unless acted upon by a non-zero net force.

• Relative Velocity Between Inertial Reference Frames:

  • If frame $A$ is an inertial reference frame obeying the First Law, another frame $B$ moving relative to frame $A$ at a constant velocity is also an inertial reference frame.

  • Relative velocity mathematical addition formula:     $\vec{v}_{P/A} = \vec{v}_{P/B} + \vec{v}_{B/A}$     where $\vec{v}_{P/A}$ is the velocity of object $P$ relative to frame $A$, $\vec{v}_{P/B}$ is the velocity of object $P$ relative to frame $B$, and $\vec{v}_{B/A}$ is the velocity of frame $B$ relative to frame $A$

• Passenger in a Vehicle Inertia Example:   

  

  • Case 1 (Vehicle Accelerates): Initially, both the passenger and the vehicle are at rest ($t = 0$, $\vec{v} = 0$). When the vehicle accelerates forward ($\vec{a} > 0$), the passenger tends to remain at rest due to inertia, causing the passenger to move backward relative to the accelerating vehicle frame.

  • Case 2 (Vehicle Decelerates): Initially, both the passenger and the vehicle are moving forward at speed $\vec{v}$. When the vehicle brakes and decelerates ($\vec{a} < 0$), the passenger tends to continue moving forward at constant velocity $\vec{v}$, shifting forward relative to the slowing vehicle.

Superposition of Forces and Mechanical Equilibrium

• Principle of Superposition of Forces:

  • Individual forces applied simultaneously at a single point on an object produce the exact same physical effect as a single force equal to their resultant vector sum.   

    

• Mathematical Expressions for Force Superposition:

  • One Dimension:     $\vec{R} = \sum \vec{F} = \vec{F}_1 + \vec{F}_2 + \dots$

  • Two Dimensions (Component Method):     $R_x = \sum F_x$     $R_y = \sum F_y$

  • Resultant Vector Magnitude (via Pythagorean theorem):     $R = \sqrt{R_x^2 + R_y^2}$

  • Resultant Vector Direction / Angle:     $\tan(\theta) = \frac{R_y}{R_x}$

  • Note on Signs: The angle $\theta$ can be located in any of the four quadrants; $\sum F_x$ and $\sum F_y$ may be positive, negative, or zero depending on direction.

• Mechanical Equilibrium:

  • An object is in mechanical equilibrium when no net force acts upon it, causing it to remain at rest or continue moving at constant velocity.   

    

  • Equilibrium condition:     $\sum \vec{F} = 0$

  • For an object resting on a flat surface, the upward normal force $\vec{N}$ balances the downward gravitational force (weight) $\vec{W}$:     $\sum \vec{F} = \vec{N} + (-\vec{W}) = 0 \implies N = W$

Worked Examples and Activities: First Law of Motion

• Example 1: Wooden Block on Rough Table:

  • Problem: A $0.020\,\text{kg}$ wooden block slides at constant velocity on a rough horizontal tabletop. An external horizontal force of $2.50\,\text{N}$ is exerted on the block. Calculate the magnitude of the friction force acting on it.

  • Solution: Since the block moves with constant velocity, its acceleration is zero ($\vec{a} = 0$), meaning the net force is zero ($\sum \vec{F} = 0$). Thus, friction opposes the external force directly:     $\sum F_x = F_{\text{ext}} - f_k = 0 \implies f_k = F_{\text{ext}} = 2.50\,\text{N}$

• Example 2: Cargo in Freight Elevator:

  • Problem: A $180\,\text{lb}$ cargo is inside a freight elevator moving upward at constant velocity. The elevator rope snaps and falls. Determine (a) weight of cargo in $\text{N}$, and (b) normal force exerted by the elevator floor before and after the rope snaps.

  • Solution:

  • Weight conversion: $180\,\text{lb} \times 4.44822\,\text{N/lb} \approx 800\,\text{N}$.

  • (a) Weight $W = 800\,\text{N}$.

  • (b) Before the rope snaps ($\vec{a} = 0$), by Law of Inertia, Normal Force $N = W = 800\,\text{N}$. After the rope breaks, both elevator and cargo are in free fall ($\vec{a} = -g$), so Normal Force $N = 0\,\text{N}$.

• Example 3: Metal Slab on Smooth Inclined Plane:

  • Problem: A $250\,\text{lb}$ metal slab is moved up a smooth inclined plane ($l = 0.7\,\text{m}$, $H = 0.3\,\text{m}$) at constant velocity. Ignore friction. Calculate the magnitude of required force.

  • Solution:     $\sin(\theta) = \frac{H}{l} = \frac{0.3}{0.7} = \frac{3}{7}$     $W = 250\,\text{lb} \times 4.44822\,\text{N/lb} = 1112.055\,\text{N}$     $\sum F_{\parallel} = F_{\text{required}} - W \sin(\theta) = 0$     $F_{\text{required}} = W \sin(\theta) = 1112.055\,\text{N} \times \frac{3}{7} = 476.6\,\text{N} \quad (\text{or } 478.29\,\text{N})$

• Activity 2A.1 (True / False Evaluation):

  1. The first law of motion states that an object will continue to be at rest or to move at a constant rate until a net force acts on it: True.

  2. An inertial frame of reference may be accelerating: False.

  3. It is impossible for an object to have zero net force when it is in motion: False.

• Activity 2A.2 (Athlete Suspended from Rope):

  • A $50.0\,\text{kg}$ athlete is suspended motionless from a light rope attached to ceiling.

  • Athlete's weight calculation:     $W = m g = 50.0\,\text{kg} \times 9.80\,\text{m/s}^2 = 490\,\text{N}$

• Activity 2A.3 (Cargo Weight Variation):

  • Recalculate Example 2 for a cargo mass of $215\,\text{lbs}$:     $W = 215\,\text{lbs} \times 4.44822\,\text{N/lb} = 956.37\,\text{N}$

  • Normal force before rope snap = $956.37\,\text{N}$.

  • Normal force after rope snap = $0\,\text{N}$.

Newton's Second Law of Motion (Law of Acceleration)

• Definition: If a net external force acts on an object, acceleration occurs in the direction of the net external force. The net force vector is equal to the product of the object's inertial mass $m$ and acceleration vector $\vec{a}$.   $\sum \vec{F} = m \vec{a}$

• Direct Proportionality and Inertial Mass:

  • Acceleration magnitude $a$ is directly proportional to net external force magnitude $|\sum \vec{F}|$.

  • Doubling net force doubles acceleration ($\sum \vec{F} = 2\vec{F}_1 \implies 2\vec{a}$); halving force halves acceleration ($\sum \vec{F} = \frac{1}{2}\vec{F}_1 \implies \frac{\vec{a}}{2}$).   

    

  • The constant ratio between net force magnitude and acceleration magnitude is defined as inertial mass $m$:     $m = \frac{|\sum \vec{F}|}{a}$

  • When two objects with masses $m_1$ and $m_2$ are fastened together, their composite mass under the same method is additive ($m_3 = m_1 + m_2$):     

    

• Units of Mass and Force:

  • Standard SI unit of mass: Kilogram ($\text{kg}$).

  • Standard SI unit of force: Newton ($\text{N}$).

  • Definition of 1 Newton: Amount of net force yielding an acceleration of $1\,\text{m/s}^2$ to an object with a mass of $1\,\text{kg}$:     $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$

• Limitations of Newton's Second Law:

  1. Applies strictly to external forces (forces exerted from outside the defined system of interest).

  2. Valid only when the object's mass $m$ remains constant.

  3. Valid only within inertial frames of reference.

• Crucial Conceptual Distinction:

  • Even though $\sum \vec{F} = m\vec{a}$, the quantity $m\vec{a}$ is NOT a force and must never be drawn as a force vector on free-body diagrams. Acceleration $\vec{a}$ is the kinematic result of net force.

Worked Examples and Activities: Second Law of Motion

• Example 1: Vehicle Net Force:

  • Problem: What net external force must be applied to make a vehicle with a mass of $3250\,\text{kg}$ accelerate at $2.5\,\text{m/s}^2$?

  • Solution:     $\sum F = m a = 3250\,\text{kg} \times 2.5\,\text{m/s}^2 = 8125\,\text{N}$

• Example 2: Displacement Under Net Force:

  • Problem: A net external force of $7.5\,\text{N}$ is exerted upon a $4.65\,\text{kg}$ object with initial velocity $2.3\,\text{m/s}$. Determine its displacement after $6\,\text{seconds}$.

  • Solution:     $a = \frac{\sum F}{m} = \frac{7.5\,\text{N}}{4.65\,\text{kg}} \approx 1.613\,\text{m/s}^2$     $\Delta x = v_i t + \frac{1}{2} a t^2 = (2.3\,\text{m/s})(6\,\text{s}) + \frac{1}{2}(1.613\,\text{m/s}^2)(6\,\text{s})^2 = 13.8 + 29.033 = 42.83\,\text{m} \quad (\text{or } 42.78\,\text{m})$

• Example 3: Wind-driven Iceboat Acceleration:

  • Problem: An iceboat with rider (combined mass $215.50\,\text{kg}$) starts from rest on horizontal ice and reaches $20\,\text{kph}$ eastward in $3.5\,\text{seconds}$. Calculate horizontal wind force F_w$.\n - *Solution*:\n    v_f = 20\,\text{kph} = \frac{20 \times 1000\,\text{m}}{3600\,\text{s}} \approx 5.556\,\text{m/s}\n    a = \frac{v_f - v_i}{\Delta t} = \frac{5.556\,\text{m/s} - 0}{3.5\,\text{s}} \approx 1.587\,\text{m/s}^2\n    F_w = m a = 215.50\,\text{kg} \times 1.587\,\text{m/s}^2 = 398.68\,\text{N}\n\n- **Activity 2B.1 (Terminology Verification)**:\n 1. A change in velocity is referred to as **acceleration**.\n 2. **Net force / Resultant force** is the combination of individual forces exerted on an object.\n 3. There is a **constant** ratio between the magnitudes of net force and acceleration.\n\n- **Activity 2B.2 (Two-Block Pulley System)**:\n - Rope connects a 4\,\text{kg}$block on a frictionless surface to hanging block$m_2$over a frictionless pulley. Tension force$T = 15\,\text{N}.\n - For horizontal block m_1:\n    T = m_1 a \implies 15\,\text{N} = (4\,\text{kg}) a \implies a = 3.75\,\text{m/s}^2\n - For vertical hanging block m_2:\n    m_2 g - T = m_2 a \implies m_2(g - a) = T\n    m_2(9.80\,\text{m/s}^2 - 3.75\,\text{m/s}^2) = 15\,\text{N} \implies m_2(6.05) = 15 \implies m_2 \approx 2.48\,\text{kg}\n\n- **Activity 2B.3 (Iceboat Extended Duration)**:\n - Iceboat (m = 215.50\,\text{kg}$) reaches$35\,\text{kph}$($9.722\,\text{m/s}$) eastward in$6.0\,\text{seconds}:\n    a = \frac{9.722\,\text{m/s}}{6.0\,\text{s}} = 1.620\,\text{m/s}^2\n    F_w = 215.50\,\text{kg} \times 1.620\,\text{m/s}^2 = 349.19\,\text{N}\n\n\n## Newton's Third Law of Motion (Law of Action-Reaction)\n\n- **Definition**: If object A$applies a force on object$B$($\vec{F}{A \text{ on } B}$), object$B$simultaneously applies a force on object$A$($\vec{F}{B \text{ on } A}), forming an action-reaction pair. The two forces are equal in magnitude and opposite in direction.\n  \vec{F}{A \text{ on } B} = -\vec{F}{B \text{ on } A}\n  \n\n- **Mandatory Criterion for Action-Reaction Pairs**:\n - The two forces MUST act on **different objects**. Forces acting on the *same object* can never form an action-reaction pair.\n - Applicable to both contact forces and non-contact (field) forces.\n\n- **Detailed Case Study: Apple Resting on Table**:\n - Forces exerted on the apple:\n 1. Downward gravitational force by Earth: \vec{F}{\text{Earth on apple}}\n 2. Upward normal force by Table: \vec{F}{\text{table on apple}}\n - *Critical Insight*: These two forces act on the same body (the apple) and cancel out to produce equilibrium (\sum \vec{F} = 0), but they are **NOT** an action-reaction pair!\n - The true action-reaction pairs in this setup are:\n - **Pair 1**: Earth pulls downward on apple (\vec{F}_{\text{Earth on apple}}$); apple pulls upward on Earth ($\vec{F}_{\text{apple on Earth}}).\n      \vec{F}{\text{Earth on apple}} = -\vec{F}{\text{apple on Earth}}\n - **Pair 2**: Table pushes upward on apple (\vec{F}_{\text{table on apple}}$); apple pushes downward on table ($\vec{F}_{\text{apple on table}}).\n      \vec{F}{\text{table on apple}} = -\vec{F}{\text{apple on table}}\n - **Effect of Table Removal**:\n    \n - Removing the table eliminates \vec{F}{\text{table on apple}}$. The downward gravitational force$\vec{F}{\text{Earth on apple}}$causes the apple to accelerate downward. The pair$\vec{F}{\text{Earth on apple}} = -\vec{F}{\text{apple on Earth}} remains active during free fall.\n\n- **Case Study: Mason Pulling Block with Rope**:\n  \n - Let B$= Block,$R$= Rope,$M = Mason.\n - **Pair 1 (Rope and Mason)**: \vec{F}{R \text{ on } M} = -\vec{F}{M \text{ on } R}\n - **Pair 2 (Block and Rope)**: \vec{F}{B \text{ on } R} = -\vec{F}{R \text{ on } B}\n\n- **Activity 2C.2 Action-Reaction Pair Identification**:\n  \n - Identification of five action-reaction pairs from the illustration:\n 1. Rope point 1 and Tree T$:$\vec{F}{R \text{ on } T} = -\vec{F}{T \text{ on } R}\n 2. Monkey hand 2 and Rope point 2: \vec{F}{M \text{ on } R2} = -\vec{F}{R2 \text{ on } M}\n 3. Monkey hand 3 and Rope point 3: \vec{F}{M \text{ on } R3} = -\vec{F}{R3 \text{ on } M}\n 4. Rope point 4 and Car C$:$\vec{F}{R \text{ on } C} = -\vec{F}{C \text{ on } R}\n 5. Monkey feet and Ground G$:$\vec{F}{M \text{ on } G} = -\vec{F}{G \text{ on } M}\n\n\n# Rotational Dynamics, Torque, and Angular Momentum\n\n## Torque-Angular Momentum Relation\n\n- **Translational vs. Rotational Analogy**:\n - Linear force \vec{F}$changes linear momentum$\vec{p}$; twisting force (torque$\vec{\tau}$) changes angular momentum$\vec{L}.\n - A Ferris wheel spins and acquires angular momentum when its engine supplies sufficient torque.\n  \n\n- **Derivation of Torque and Angular Momentum Relation**:\n - Second Law alternative expression in linear mechanics:\n    \sum \vec{F} = \frac{\Delta \vec{p}}{\Delta t}\n - Rotational counterpart: Summation of all torques acting on an object equals time rate of change of angular momentum:\n    \sum \tau = \frac{\Delta L}{\Delta t}\n - Mathematical Proof:\n    \sum \tau = I \alpha\n    \alpha = \frac{\omega_f - \omega_i}{\Delta t}\n    \sum \tau = \frac{I \omega_f - I \omega_i}{\Delta t}\n    Since angular momentum L = I \omega:\n    \sum \tau = \frac{L_f - L_i}{\Delta t} = \frac{\Delta L}{\Delta t}\n\n- **Physical Effects**:\n  \n - Applying net torque on a wheel causes angular acceleration \alpha, increasing angular momentum.\n - Opposing net torque slows down rotation, decreasing angular momentum.\n\n- **Law of Conservation of Angular Momentum**:\n - When zero net external torque acts on an object (\sum \tau = 0$), no change in angular momentum occurs ($\Delta L = 0 \implies L_i = L_f).\n\n- **Mathematical Relation to Linear Momentum**:\n - Time derivative cross product relation:\n    \frac{\Delta \vec{L}}{\Delta t} = \vec{r} \times \vec{F}\n - Scalar magnitude equation:\n    L = r p \sin(\theta)\n    where r$is radial distance from axis of rotation,$p$is linear momentum ($p = m v$), and$\theta$is angle between$\vec{r}$and$\vec{p}.\n\n\n## Worked Examples: Torque and Angular Momentum\n\n- **Example 1: Space Station Stick**:\n - *Problem*: An engineer at ISS plays with a stick initially not spinning at zero gravity. Angular momentum increases from 0$to$10\,\text{kg}\cdot\text{m}^2/\text{s}$in$\Delta t = 0.5\,\text{seconds}. Calculate given torque.\n - *Solution*:\n    \tau = \frac{\Delta L}{\Delta t} = \frac{10\,\text{kg}\cdot\text{m}^2/\text{s} - 0}{0.5\,\text{s}} = 20\,\text{N}\cdot\text{m}\n\n- **Example 2: Ferris Wheel Engine Startup**:\n - *Problem*: A Ferris wheel engine provides torque \tau = 7 \times 10^6\,\text{N}\cdot\text{m}$during a$20\,\text{second} startup period from rest. How much angular momentum did it gain?\n - *Solution*:\n    \Delta L = \tau \Delta t = (7 \times 10^6\,\text{N}\cdot\text{m})(20\,\text{s}) = 1.4 \times 10^7\,\text{kg}\cdot\text{m}^2/\text{s}\n\n- **Example 3: Food Tray Rotation**:\n - *Problem*: A man exerts a perpendicular force F = 3.0\,\text{N}$at radius$r = 0.3\,\text{m}$on a food tray ($L_i = 0$) for$\Delta t = 0.7\,\text{seconds}. Calculate final angular momentum.\n - *Solution*:\n    \tau = r F \sin(90^\circ) = (0.3\,\text{m})(3.0\,\text{N}) = 0.9\,\text{N}\cdot\text{m}\n    L_f = \tau \Delta t = (0.9\,\text{N}\cdot\text{m})(0.7\,\text{s}) = 0.63\,\text{kg}\cdot\text{m}^2/\text{s}\n\n\n# Uniform Circular Motion and Tangential Velocity\n\n## Uniform vs. Non-Uniform Circular Motion\n\n- **Centripetal Force Requirement**:\n - An object moving with constant speed along a curved path accelerates because its direction changes. Without a centripetal force directed toward the center of curvature, the object moves off in a straight tangent line.\n\n- **Uniform Circular Motion Definition**:\n - Occurs when an object traverses a circular path at constant speed.\n - Acceleration vector (\vec{a}{\text{rad}} or centripetal acceleration) is perpendicular to the circular path and directed radially inward toward the circle's center.\n  \n - Acceleration magnitude formula:\n    a{\text{rad}} = \frac{v^2}{R}\n - In terms of revolution period T (time for one complete revolution):\n    a_{\text{rad}} = \frac{4\pi^2 R}{T^2}\n - *Note*: Acceleration \vec{a}{\text{rad}}$and velocity$\vec{v} are two distinct vectors that are always mutually perpendicular in uniform circular motion.\n\n- **Comparison with Non-Uniform Circular Motion**:\n  \n - **Uniform Circular Motion**: Acceleration is strictly perpendicular to velocity; no parallel component (a{\parallel} = 0).\n - **Car Speeding Up on Circular Path**: Acceleration has a component parallel to velocity (a_{\parallel} > 0) which increases speed, plus a radial component changing direction.\n - **Car Slowing Down on Circular Path**: Acceleration has a component antiparallel to velocity (a_{\parallel} < 0) which decreases speed, plus a radial component changing direction.\n\n\n## Rotational Speed and Tangential Velocity\n\n- **Rotational / Angular Speed (\omega)**:\n - Number of complete revolutions around the circular path per unit time.\n - Common units: RPM (revolutions per minute), rev/s, or radians per second (\text{rad/s}).\n\n- **Tangential Speed (v_t)**:\n - Linear speed of an object along its circular path.\n - "Tangential" signifies that the velocity vector touches the circular circumference at that exact instant.\n - Mathematical relation:\n    v_t = r \omega\n    where r$is radial distance from rotation axis, and$\omega$is rotational speed in$\text{rad/s}.\n\n- **Ladybugs on Turntable Conceptual Comparison**:\n - Ladybug A near center vs. Ladybug B at outer rim on same spinning turntable:\n - Both ladybugs have identical rotational speed \omega.\n - Ladybug B at outer radius (r_B > r_A$) travels greater linear distance per revolution and has twice the tangential speed ($v_{t,B} = 2 v_{t,A}).\n\n\n## Worked Examples and Activities: Circular Motion\n\n- **Example 1: Electric Drill Bit**:\n - *Problem*: An electric drill whirls at 60\,\text{rad/s}$. Radius$r = 0.002\,\text{m}. Calculate tangential speed at surface.\n - *Solution*:\n    v_t = r \omega = (0.002\,\text{m})(60\,\text{rad/s}) = 0.12\,\text{m/s}\n\n- **Example 2: Spinning Propeller**:\n - *Problem*: Propeller diameter d = 2.5\,\text{m} \implies r = 1.25\,\text{m}$, spinning at$1.91\,\text{rev/s}$. Calculate rotational speed in$\text{rad/s} and tangential speed.\n - *Solution*:\n    \omega = 1.91\,\text{rev/s} \times \frac{2\pi\,\text{rad}}{1\,\text{rev}} \approx 12.0\,\text{rad/s}\n    v_t = r \omega = (1.25\,\text{m})(12.0\,\text{rad/s}) = 15.0\,\text{m/s}\n\n- **Example 3: Beetle on Spinning Car Wheel**:\n - *Problem*: Beetle at r_1 = 0.075\,\text{m}$has tangential speed$v_1 = 7.50\,\text{m/s}$. Calculate tangential speed at wheel outer rim ($r_2 = 0.225\,\text{m}).\n - *Solution*:\n    \omega = \frac{v_1}{r_1} = \frac{7.50\,\text{m/s}}{0.075\,\text{m}} = 100\,\text{rad/s}\n    v_2 = r_2 \omega = (0.225\,\text{m})(100\,\text{rad/s}) = 22.5\,\text{m/s}\n\n- **Activity 1 (True / False Evaluation)**:\n 1. Uniform circular motion entails that the object in motion has constant speed: **True**.\n 2. Uniform circular motion is also called centrifugal acceleration: **False** (It is called centripetal acceleration).\n 3. One radian per second is equivalent to 0.105\,\text{RPM}$: **False** ($1\,\text{rad/s} \approx 9.55\,\text{RPM}).\n\n- **Activity 1 Rotational Speed Conversions to \text{rad/s}**:\n 1. 2.5\,\text{rev/s} \times 2\pi = 5\pi \approx 15.71\,\text{rad/s}\n 2. 12.15\,\text{rev/s} \times 2\pi \approx 76.34\,\text{rad/s}\n 3. 1.2\,\text{RPM} \times \frac{2\pi\,\text{rad}}{60\,\text{s}} \approx 0.126\,\text{rad/s}\n 4. 5.0\,\text{RPM} \times \frac{2\pi\,\text{rad}}{60\,\text{s}} \approx 0.524\,\text{rad/s}\n 5. 1\,\text{rev/h} = \frac{2\pi\,\text{rad}}{3600\,\text{s}} \approx 0.00175\,\text{rad/s}\n\n\n# Reference Formula Summary Tables\n\n## Table 1: Force Superposition and Linear Dynamics Formulas\n\n| Concept | Formula | Description |\n| :--- | :--- | :--- |\n| **Superposition of Forces (1D)** | \vec{R} = \sum \vec{F} = \vec{F}1 + \vec{F}_2 + \dots | Net force calculation when all constituent forces lie along a single linear dimension. |\n| **Superposition of Forces (2D)** | R = \sqrt{R_x^2 + R_y^2}$| Uses Pythagorean theorem to compute net force magnitude from components$R_x = \sum F_x$and$R_y = \sum F_y. |\n| **Newton's Second Law** | \sum \vec{F} = m \vec{a}$| Calculates net external force exerted on an object given inertial mass$m$and acceleration$\vec{a}. |\n\n## Table 2: Torque, Rotational Dynamics, and Circular Motion Formulas\n\n| Concept | Formula | Description |\n| :--- | :--- | :--- |\n| **Torque-Angular Momentum Relation** | \tau = \frac{\Delta L}{\Delta t}$| Computes net torque$\tau$given angular momentum change$\Delta L$over time interval$\Delta t. |\n| **Angular & Linear Momentum Relation** | L = r p \sin(\theta)$| Relates angular momentum$L$to linear momentum$p$, radial distance$r$, and angle$\theta. |\n| **Centripetal Acceleration** | a{\text{rad}} = \frac{v^2}{R} = \frac{4\pi^2 R}{T^2}$| Solves radially inward acceleration for uniform circular motion given path speed$v$or revolution period$T. |\n| **Tangential Speed** | v_t = r \omega$| Computes linear tangential speed$v_t$for an object at radius$r$rotating at angular speed$\omega$(in$\text{rad/s}$$). |