Comprehensive Study Notes on Mechanical Systems: Kinematics, Dynamics, Rotational Mechanics, and Simple Harmonic Motion

Scalars and Vectors

• Scalars are physical quantities that possess magnitude (size) only, without any associated direction in space.

  • Examples include:

  • Pressure

  • Temperature (an object cannot be warm in a specific spatial direction)

  • Time (noted as debatable at higher theoretical levels, but treated as a scalar in classical mechanics)

  • Mass (an object cannot possess mass in a specific direction)

  • Energy (energy has no directional component)

• Vectors are physical quantities that possess both magnitude (size) and a specific directional orientation.

  • Examples include:

  • Displacement (displacement from point $A$ to point $B$ involves both a straight-line distance and a direction)

  • Velocity (speed in a specified direction)

  • Acceleration (rate of change of velocity in a direction)

  • Force (a push or pull applied in a specific direction)

  • Torque (rotational influence applied in a rotational direction, such as clockwise or counterclockwise)

Kinematics: Distance, Displacement, Speed, and Acceleration

• Distance vs. Displacement:

  • Distance describes the total length of the continuous path an object has traversed during its motion.

  • Displacement describes the straight-line vector difference between an object's initial start point and final finish point.

  • Distance and displacement share the same numerical value only when an object travels along a strictly straight path.

  • For an object moving along a curved trajectory, the total distance traveled along the path is significantly greater than the magnitude of its net displacement.

• Speed:

  • Speed describes the rate at which an object's position changes over time.

  • The vector counterpart of speed is velocity.

  • Scientific units for speed and velocity are meters per second, expressed as $m/s$ or using negative index notation as $m\,s^{-1}$.

  • The governing linear equation is:     $v = \frac{d}{t}$     where $v$ is speed, $d$ is distance, and $t$ is time elapsed.

  • Worked Example: Calculating the time required for a vehicle traveling at a constant speed of $10\,m/s$ to cover a distance of $25\,m$:     $t = \frac{d}{v}$     $t = \frac{25\,m}{10\,m/s} = 2.5\,s$

• Acceleration:

  • Acceleration describes how rapidly an object's velocity changes over time.

  • Scientific units are meters per second squared, expressed as $m/s/s$, $m/s^2$, or $m\,s^{-2}$.

  • The governing equation is:     $a = \frac{\Delta v}{t}$     where $a$ is acceleration, $\Delta v$ is change in velocity, and $t$ is time.

  • Deceleration occurs when an acceleration vector acts in the direction opposite to the velocity vector, reducing the magnitude of the velocity.

  • Worked Example: Calculating the final speed of a car accelerating from rest ($v_i = 0\,m/s$) at $5\,m/s^2$ for a duration of $7\,s$:     $v = a \times t$     $v = 5\,m/s^2 \times 7\,s = 35\,m/s$

Vector Mathematics and Trigonometry

• Equating Vectors: Vector diagrams can be formulated into mathematical equations using geometric principles:

  • Closed Loops (Circles are Zeros): Vectors that form a continuous closed path return to the starting point, meaning their resultant vector sum equals zero.

  • For four vectors forming a closed loop: $\mathbf{A} + \mathbf{B} + \mathbf{C} + \mathbf{D} = 0$

  • For three vectors forming a closed triangular loop: $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$

  • This principle holds universally regardless of the number of vectors in the closed polygon.

  • Equivalent Paths: Vectors or series of vectors that begin at the same initial point and terminate at the same final point are mathematically equivalent.

  • If two sequential vectors $\mathbf{a} + \mathbf{b}$ share the same start and end points as single vector $\mathbf{c}$, then $\mathbf{a} + \mathbf{b} = \mathbf{c}$.

  • If path $\mathbf{a} + \mathbf{b}$ ends at the same location as path $\mathbf{c} + \mathbf{d}$, then $\mathbf{a} + \mathbf{b} = \mathbf{c} + \mathbf{d}$.

  • If path $\mathbf{a} + \mathbf{b} + \mathbf{c}$ terminates at the end of vector $\mathbf{d}$, then $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{d}$.

• Trigonometric Fundamentals:

  • In a right-angled triangle, the side opposite the right angle is the hypotenuse, the side across from the reference angle $\theta$ is the opposite, and the side next to $\theta$ is the adjacent.

  • Pythagorean Theorem:     $\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$

  • SOH CAH TOA Definitions:     $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$     $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$     $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

• Vector Components: Any vector of magnitude $v$ oriented at an angle $\theta$ relative to the horizontal can be resolved into orthogonal components:

  • Horizontal component ($v_h$ or $v_x$):     $\cos(\theta) = \frac{v_h}{v} \implies v_h = v \cos(\theta)$

  • Vertical component ($v_v$ or $v_y$):     $\sin(\theta) = \frac{v_v}{v} \implies v_v = v \sin(\theta)$

Linear Momentum and Impulse

• Linear Momentum ($p$):

  • Momentum is a measure of motion incorporating both mass and velocity in equal proportion.

  • Defined by the equation:     $p = m v$     where $p$ is momentum in $kg\,m/s$, $m$ is mass in $kg$, and $v$ is velocity in $m/s$.

  • Because momentum is directly proportional to both mass and velocity, objects of wildly different masses traveling at identical velocities possess radically different momentums (e.g., a bumblebee and an automobile each traveling at $10\,m/s$).

• Impulse and Force:

  • Impulse is defined as the change in an object's momentum ($\Delta p$):     $\Delta p = p_f - p_i = m v_f - m v_i$

  • The average force applied to an object is the time rate of change of momentum:     $F = \frac{\Delta p}{\Delta t}$

  • Extending the duration of an impact ($\Delta t$) reduces the peak impact force experienced by the object.

  • On a graph of Force versus Time ($F$ vs. $t$), the area beneath the curve equals the total impulse (change in momentum $\Delta p$).

  • Braking Comparison: Stopping a car rapidly by slamming on the brakes produces a high force over a short time duration; braking gradually produces a small force over an extended duration. The net change in momentum ($\Delta p$) from traveling velocity down to zero is identical in both scenarios.

• Worked Example: Wall Collision: A $0.050\,kg$ tennis ball traveling horizontally at $6.2\,m/s$ strikes a wall, rebounding at $5.9\,m/s$ in the opposite direction. Contact duration is $0.0045\,s$.

  • Establishing direction: Let rebound direction be positive ($+5.9\,m/s$) and initial approach velocity be negative ($-6.2\,m/s$).

  • Calculate change in momentum:     $\Delta p = m (v_f - v_i) = 0.050\,kg \times (5.9\,m/s - (-6.2\,m/s)) = 0.050 \times 12.1 = 0.61\,kg\,m/s$

  • Calculate average force:     $F = \frac{\Delta p}{\Delta t} = \frac{0.61\,kg\,m/s}{0.0045\,s} = 135\,N$

  • Note on sign conventions: Assigning the initial velocity as positive ($+6.2\,m/s$) and final as negative ($-5.9\,m/s$) yields $\Delta p = -0.61\,kg\,m/s$ and $F = -135\,N$. The magnitude remains unchanged; the negative sign merely indicates the force vector acts in the negative coordinate direction.

Conservation of Linear Momentum

• Principle of Conservation of Momentum: The total linear momentum of an isolated system remains constant unless acted upon by an external net force.

  • 1D Collision Demonstration: Prior to collision, an orange ball has a momentum of $+20\,kg\,m/s$ (directed right) and a green ball has $-30\,kg\,m/s$ (directed left). System total momentum:     $p_{\text{total}} = 20 + (-30) = -10\,kg\,m/s$

  • Following the collision, the orange ball rebounds with $-25\,kg\,m/s$ (left) and the green ball moves with $+15\,kg\,m/s$ (right). System total momentum:     $p_{\text{total}} = -25 + 15 = -10\,kg\,m/s$

  • Total momentum before equals total momentum after; momentum is conserved.

• External Forces vs. Balanced Forces:

  • Momentum remains conserved in the presence of external forces provided those external forces are completely balanced (net external force equals zero).

  • Billiard balls rolling on a horizontal table experience the downward force of gravity and an equal upward support/normal force; because these cancel ($F_{\text{net}} = 0$), horizontal momentum is conserved.

  • A ball flying through the air experiences an unbalanced gravitational force ($F_{\text{net}} \neq 0$), causing its downward momentum to continually increase; momentum of the ball alone is not conserved.

• Nature of External Forces:

  • An external force is simply a force exerted by an entity whose momentum is omitted from system accounting (usually because measuring it is impractical).

  • For a falling apple, Earth exerts a gravitational force downward, increasing the apple's momentum. However, the apple exerts an equal and opposite gravitational force on Earth, imparting an identical upward momentum to Earth.

  • Because Earth's mass ($M_E$) is enormous relative to the apple ($m_a$), its upward velocity ($v_E = \frac{m_a v_a}{M_E}$) is immeasurably small.

  • Apparent non-conservation of momentum arises exclusively when an interacting component of the physical system is excluded.

• Magnetic Propulsion Experiment:

  • A neodymium magnet has steel ball bearings resting on its left face. An identical bearing approaches from the right at a low speed.

  • As the bearing nears the magnet, mutual magnetic attraction accelerates it, increasing its momentum.

  • Upon impact, the bearing on the far left end is ejected with considerably higher velocity and momentum than the incoming bearing possessed initially.

  • Resolution of Conservation: On a standard table, friction holds the magnet assembly stationary. On a frictionless air table, the magnet and attached bearings accelerate toward the approaching bearing with an equal and opposite momentum.

  • Net momentum is strictly conserved:     $p_{\text{initial, approaching}} = p_{\text{propelled}} - p_{\text{magnet assembly}}$

• Worked Example: 2D Momentum Collision:

  • Billiard ball $A$ ($m_A = 0.170\,kg$) moves at $v_{Ai} = 1.80\,m/s$ and collides with stationary billiard ball $B$ ($m_B = 0.120\,kg$, $v_{Bi} = 0\,m/s$).

  • Following collision, the balls depart at right angles ($90^\circ$). Ball $A$ moves with final velocity $v_{Af} = 1.10\,m/s$.

  • Determine initial momentum:     $p_{\text{initial}} = p_{Ai} = m_A v_{Ai} = 0.170\,kg \times 1.80\,m/s = 0.306\,kg\,m/s$

  • Determine final momentum of Ball $A$:     $p_{Af} = m_A v_{Af} = 0.170\,kg \times 1.10\,m/s = 0.187\,kg\,m/s$

  • Form vector right triangle where initial momentum vector is the hypotenuse ($p_i = 0.306\,kg\,m/s$) and final momentum vectors are orthogonal legs ($p_{Af} = 0.187\,kg\,m/s$ and $p_{Bf}$):     $p_i^2 = p_{Af}^2 + p_{Bf}^2$     $p_{Bf} = \sqrt{p_i^2 - p_{Af}^2} = \sqrt{0.306^2 - 0.187^2} = \sqrt{0.093636 - 0.034969} = \sqrt{0.058667} = 0.242\,kg\,m/s$

  • Determine final speed of Ball $B$:     $v_{Bf} = \frac{p_{Bf}}{m_B} = \frac{0.242\,kg\,m/s}{0.120\,kg} = 2.02\,m/s$

Center of Mass in One and Two Dimensions

• Definition: The center of mass (COM) represents the mean position of a distribution of mass in space, weighted by each constituent mass.

  • Two bodies balanced on a pivot will equilibrate precisely at their collective center of mass.

  • For a system of two masses ($m_1$ and $m_2$) measured from an arbitrary reference origin ($x = 0$):     $x_{\text{com}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

  • Choosing the coordinate origin ($x = 0$) directly at the position of mass $m_1$ simplifies calculations by setting $x_1 = 0$:     $x_{\text{com}} = \frac{m_2 x_2}{m_1 + m_2}$     where $x_2$ represents the total separation distance between the two bodies.

• Worked Example: Earth-Moon Center of Mass:

  • Earth mass: $m_E = 5.97 \times 10^{24}\,kg$

  • Moon mass: $m_M = 7.35 \times 10^{22}\,kg$

  • Center-to-center separation distance: $d = 3.84 \times 10^8\,m$

  • Setting Earth's center as $x = 0$ ($x_E = 0$, $x_M = 3.84 \times 10^8\,m$):     $x_{\text{com}} = \frac{(7.35 \times 10^{22}\,kg) \times (3.84 \times 10^8\,m)}{(5.97 \times 10^{24}\,kg) + (7.35 \times 10^{22}\,kg)} = \frac{2.8224 \times 10^{31}}{6.0435 \times 10^{24}} = 4.7 \times 10^6\,m$

  • The center of mass lies $4.7 \times 10^6\,m$ from Earth's center (inside Earth's radius).

• Mechanical Equilibrium Demonstration (Forks and Matchstick):

  • Two interlocking metal forks balanced on a single matchstick rest on the lip of a drinking glass without toppling.

  • While the visible bulk of the forks projects forward, their curved design distributes mass backward such that equal mass lies on either side of the pivot line.

  • The collective center of mass of the fork-matchstick structure resides directly at/below the pivot point.

  • Because the distance from the pivot to the line of action of gravity is zero ($d = 0$), the net gravitational torque is zero ($\tau = F \times d = 0$), preventing rotation or toppling.

• Center of Mass in Two Dimensions:

  • Calculated by resolving coordinates into independent orthogonal axes ($x_{\text{com}}$ and $y_{\text{com}}$):     $x_{\text{com}} = \frac{\sum m_i x_i}{\sum m_i}$     $y_{\text{com}} = \frac{\sum m_i y_i}{\sum m_i}$

• Worked Example: Raft Coordinates:

  • A uniform raft measures $2.0\,m$ by $3.0\,m$. A fish box ($m_1 = 12.3\,kg$) sits at corner $(0, 0)$. A tackle box ($m_2 = 23.7\,kg$) sits at the diagonally opposite corner $(3.0\,m, 2.0\,m)$.

  • Calculate $x_{\text{com}}$:     $x_{\text{com}} = \frac{(12.3 \times 0) + (23.7 \times 3.0)}{12.3 + 23.7} = \frac{71.1}{36.0} = 1.98\,m$

  • Calculate $y_{\text{com}}$:     $y_{\text{com}} = \frac{(12.3 \times 0) + (23.7 \times 2.0)}{12.3 + 23.7} = \frac{47.4}{36.0} = 1.32\,m$

  • The coordinates of the center of mass are $(1.98\,m, 1.32\,m)$.

Analysis of Collisions and Center of Mass Motion

• Center of Mass Dynamics: Provided no net external resultant force acts upon a system, the velocity of the center of mass ($v_{\text{com}}$) remains strictly constant before, during, and after interactions. This represents an alternative phrasing of the law of conservation of momentum.

• Video Analysis of Four Collisions (Tracker Study):

  • Collision 1: Symmetry in Center of Mass Frame:

  • In the lab frame, puck 1 moves at $v = 10\,m/s$ toward identical stationary puck 2 ($v = 0\,m/s$). Post-impact, puck 1 halts ($v = 0\,m/s$) and puck 2 departs at $v = 10\,m/s$.

  • In a frame of reference moving at half the initial velocity ($v_{\text{frame}} = 5\,m/s$), initial velocities appear as $+5\,m/s$ and $-5\,m/s$. Post-impact velocities appear as $-5\,m/s$ and $+5\,m/s$.

  • In this center of mass frame, the collision is visibly symmetric: equal and opposite velocities are mirrored across the interaction.

  • Collision 2: Orthogonal Trajectories in Elastic Collisions:

  • In an elastic collision between two identical masses ($m_1 = m_2 = m$) where one mass is initially stationary, kinetic energy is strictly conserved:       $\frac{1}{2} m v_i^2 = \frac{1}{2} m v_{1f}^2 + \frac{1}{2} m v_{2f}^2$

  • Dividing out the common factor of $\frac{1}{2}m$ yields:       $v_i^2 = v_{1f}^2 + v_{2f}^2$

  • This algebraic relation matches the Pythagorean theorem ($c^2 = a^2 + b^2$), proving that the final velocity vectors must form the legs of a right triangle. The scattering angle between the two identical masses is always $90^\circ$.

  • Collision 3: Quantitative Verification of Momentum Conservation:

  • Analysis of collision tracking data reveals the numerical difference between initial and final system momentum to be $0.76\%$.

  • This deviation is sufficiently small to be dismissed as experimental random error, validating conservation.

  • Collision 4: Constant Velocity of Center of Mass:

  • Tracking the composite center of mass during collision verifies that its trajectory and velocity remain constant along an uninterrupted linear path.

• Explosion Analysis (Melon Demonstration):

  • A stationary melon ($p = 0$) detonates on a rigid surface.

  • In deep space without external forces, the vector sum of all fragment momentums would sum precisely to zero.

  • For a melon resting on a table, two external forces intervene during detonation:

  • Upward normal force from the tabletop acting on fragments in contact.

  • Downward force of gravity acting on all fragments.

  • Because these forces represent net external interactions relative to the melon system, the melon's net linear momentum is not conserved.

Energy, Work, and Power

• Conservation of Energy: Energy can neither be created nor destroyed; it can only be transformed from one form into another.

  • Examples of Energy Conversion:

  • An electric lamp supplied with $10\,J$ of electrical energy converts it into $3\,J$ of light energy and $7\,J$ of thermal (heat) energy ($10 = 3 + 7$).

  • A falling ball starting with $35\,J$ of gravitational potential energy dissipates $2\,J$ through air resistance, converting the remaining $33\,J$ into kinetic energy ($35 = 2 + 33$).

  • A burning log containing $100\,J$ of chemical energy releases $10\,J$ as radiant light and $90\,J$ as thermal energy ($100 = 10 + 90$).

• Worked Example: Tuatara Freefall:

  • A tuatara falls $2.0\,m$ from a tree branch. Calculate impact velocity assuming complete conservation of mechanical energy ($E_p \rightarrow E_k$):     $E_k = E_p$     $\frac{1}{2} m v^2 = m g h$     $\frac{1}{2} v^2 = g h$     $v = \sqrt{2 g h} = \sqrt{2 \times 9.8\,m/s^2 \times 2.0\,m} = \sqrt{39.2} = 6.3\,m/s$

• Work ($W$):

  • Work is performed when a force acts upon an object over a displacement in the direction of the force:     $W = F d$     where $W$ is work in Joules ($J$), $F$ is applied force in Newtons ($N$), and $d$ is displacement in meters ($m$).

  • When lifting an object against gravity, the required force equals weight ($F_g = m g$) and displacement equals height ($h$):     $W = (m g) h = m g h$     This reproduces the equation for gravitational potential energy ($E_p$).

  • Worked Example: Dirt Bike on a Ramp: A $95\,kg$ dirt bike is pushed up an inclined ramp onto a trailer bed $0.80\,m$ high.

  • Force against gravity:       $F_g = m g = 95\,kg \times 9.8\,m/s^2 = 931\,N \approx 930\,N$

  • Work performed:       $W = F_g d = 931\,N \times 0.80\,m = 744.8\,J \approx 740\,J$

  • The work performed against gravity depends solely on the vertical displacement ($0.80\,m$). The extended distance traveled along the inclined path does not alter the work done against gravity.

• Power ($P$):

  • Power is the rate of energy consumption or work performed per unit time:     $P = \frac{\Delta E}{t}$

  • Measured in Joules per second ($J/s$), designated as Watts ($W$).

  • A $25\,W$ stereo consumes $25\,J$ of electrical energy per second ($\frac{25\,J}{1\,s} = 25\,W$); a $10\,W$ fan consumes $10\,J/s$.

Newton's Laws of Motion

• Newton's First Law (Law of Inertia): An object will maintain a constant velocity (or remain at rest) unless acted upon by a non-zero net external force.

  • A kicked soccer ball initiates motion from rest solely because an external contact force is applied.

  • The ball decelerates and stops because the external force of friction acts against its motion.

  • Any change in velocity (speed or direction) requires a net force.

• Newton's Second Law: When a net force acts upon an object, the object undergoes an acceleration directly proportional to the magnitude of the force and inversely proportional to its mass:   $F = m a$

  • Applying an identical force to an object with larger mass results in a correspondingly smaller acceleration.

• Newton's Third Law: Forces occur in matched, equal, and opposite interaction pairs. When body $A$ exerts a force on body $B$, body $B$ simultaneously exerts an equal and opposite force on body $A$:   $F_{A \rightarrow B} = -F_{B \rightarrow A}$

  • A brick resting on a table exerts a downward contact force on the tabletop; the table exerts an equal, upward reaction/support force on the brick.

  • Earth exerts a downward gravitational pull on the Moon; the Moon exerts an equal, upward gravitational pull on Earth.

Circular Motion Fundamentals and Dynamics

• Geometric Terminology:

  • Radius ($r$): The distance from the center to any point on the circular path (half the circle width).

  • Diameter ($d$): The full width across the circle through its center ($d = 2r$).

  • Circumference ($C$): The linear perimeter length of one complete lap ($C = 2 \pi r$).

  • Revolution: Exactly one complete lap around the circular path.

  • Period ($T$): The time required in seconds to complete one revolution ($s/\text{rev}$).

  • Frequency ($f$): The number of revolutions completed per second ($\text{rev}/s$ or $Hz$).

  • Mathematical reciprocal relation:     $f = \frac{1}{T} \quad \text{and} \quad T = \frac{1}{f}$

  • Orbital Speed ($v$): Distance divided by time for one lap:     $v = \frac{2 \pi r}{T} = 2 \pi r f$

• Centripetal Acceleration ($a_c$):

  • Because velocity is a vector, an object traveling in a circle continuously alters its direction and therefore undergoes constant acceleration, even if speed remains uniform.

  • Centripetal translates literally as "center-seeking"; centripetal acceleration vectors always point radially inward toward the circle center.

  • Velocity vectors are directed tangentially to the path, oriented at a perpendicular angle ($90^\circ$) relative to acceleration.

  • Mathematical formula:     $a_c = \frac{v^2}{r}$

• Centripetal Force ($F_c$):

  • Centripetal force is not an independent fundamental force; it is a mechanical role fulfilled by real physical forces.

  • Examples of fulfilling forces:

  • Gravity provides centripetal force for Earth orbiting the Sun.

  • Mechanical tension provides centripetal force for a mass whirled on a string.

  • Friction between tires and asphalt provides centripetal force for a vehicle rounding a flat curve.

  • Mathematical expression:     $F_c = m a_c = \frac{m v^2}{r}$

• Worked Example: Car on a Flat Circular Track:

  • Vehicle mass: $m = 1500\,kg$

  • Path radius: $r = 20\,m$

  • Lap period: $T = 60\,s$

  • Calculate tangential speed:     $v = \frac{2 \pi r}{T} = \frac{2 \pi (20\,m)}{60\,s} = 2.094\,m/s \approx 2.1\,m/s$

  • Calculate minimum required friction force ($F_f = F_c$):     $F_c = \frac{m v^2}{r} = \frac{1500\,kg \times (2.094\,m/s)^2}{20\,m} = 328.9\,N \approx 330\,N$

• Centrifugal Force Mechanism (Spinning Fluid Demonstration):

  • A cylindrical container filled with green water is rotated on a turntable. The liquid climbs the perimeter walls, leaving a central depression.

  • Inertial Frame Analysis: Individual fluid particles possess tangential velocity. In the absence of force, Newton's first law dictates they continue in a straight line. The solid container wall blocks this path, exerting an inward radial normal force (centripetal force) on the water. Unable to compress or move outward, the constrained liquid is forced upward.

  • Rotating Frame Analysis: An observer rotating with the container perceives the water as stationary but subject to an apparent outward force pushing it toward the walls. This perceived effect is a fictitious, perspective-dependent artifact called centrifugal force.

Banked Corners and Centripetal Force

• Mechanics of Banked Roads:

  • Banking a roadway at an angle $\theta$ allows vehicles to navigate curves safely without relying on tire friction.

  • The support/reaction force ($F_r$ or $F_N$) acts perpendicular to the inclined pavement.

  • The downward gravitational force ($F_g = m g$) acts vertically.

  • Resolving the reaction force into components:

  • Vertical component balances gravity (since vertical acceleration is zero):       $F_r \cos(\theta) = m g$

  • Horizontal component acts radially inward toward the curve center, providing centripetal force:       $F_r \sin(\theta) = F_c = \frac{m v^2}{r}$

  • Dividing the horizontal equation by the vertical equation eliminates $F_r$:     $\tan(\theta) = \frac{F_c}{F_g} = \frac{\frac{m v^2}{r}}{m g} = \frac{v^2}{r g}$

• Worked Example: Frictionless Icy Banked Turn:

  • Car mass: $m = 1500\,kg$

  • Curve radius: $r = 80.0\,m$

  • Banking angle: $\theta = 25.4^\circ$

  • Calculate required centripetal force:     $\tan(\theta) = \frac{F_c}{F_g} \implies F_c = m g \tan(\theta)$     $F_c = 1500\,kg \times 9.8\,m/s^2 \times \tan(25.4^\circ) = 14700 \times 0.4748 = 6980\,N \approx 6990\,N$

  • Calculate designated design speed:     $F_c = \frac{m v^2}{r} \implies v^2 = \frac{F_c r}{m} = \frac{6980\,N \times 80.0\,m}{1500\,kg} = 372.27$     $v = \sqrt{372.27} = 19.3\,m/s$

Gravitation, Orbital Mechanics, and Satellites

• Newton's Universal Law of Gravitation: All masses in the universe exert an attractive force on each other. For two masses $m_1$ and $m_2$ separated by distance $r$ between their centers of mass:   $F_g = \frac{G m_1 m_2}{r^2}$   where $G = 6.67 \times 10^{-11}\,N\,m^2\,kg^{-2}$ is the universal gravitational constant.

  • By Newton's third law, both masses experience the exact same gravitational attraction magnitude, regardless of mass disparity.

• Worked Example: Earth-Moon Gravitational Attraction:

  • Mass of Earth: $m_E = 5.97 \times 10^{24}\,kg$

  • Mass of Moon: $m_M = 7.35 \times 10^{22}\,kg$

  • Orbital radius: $r = 3.84 \times 10^8\,m$

  • Force calculation:     $F_g = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (7.35 \times 10^{22})}{(3.84 \times 10^8)^2} = \frac{2.9267 \times 10^{37}}{1.4746 \times 10^{17}} = 1.98 \times 10^{20}\,N$

• Orbital Velocity Derivation:

  • In orbital motion, gravitational attraction supplies the necessary centripetal force:     $F_c = F_g$     $\frac{m_{\text{sat}} v^2}{r} = \frac{G M_{\text{central}} m_{\text{sat}}}{r^2}$

  • Dividing out satellite mass $m_{\text{sat}}$ and multiplying by $r$:     $v^2 = \frac{G M_{\text{central}}}{r}$     $v = \sqrt{\frac{G M_{\text{central}}}{r}}$

• Worked Example: Moon's Orbital Velocity:

  • $v = \sqrt{\frac{(6.67 \times 10^{-11}\,N\,m^2\,kg^{-2}) \times (5.97 \times 10^{24}\,kg)}{3.84 \times 10^8\,m}} = \sqrt{\frac{3.982 \times 10^{14}}{3.84 \times 10^8}} = \sqrt{1.037 \times 10^6} = 1018\,m/s \approx 1020\,m/s$

• Geostationary Orbit Requirements:

  • A geostationary satellite remains hovering over an identical, fixed geographic position on the Earth's surface.

  • Two strict conditions must be satisfied:

  1. The orbit must reside directly above Earth's equator (an inclined orbit would oscillate north-south across latitudes because orbits must center on Earth's center of mass).

  2. The orbital period must precisely match the rotational period of the planet ($T = 24\,\text{hours} = 1\,\text{day}$).

  • Because orbital period depends uniquely upon orbital radius/altitude, every celestial body has a single unique geostationary altitude.

  • For Earth, this orbit requires an altitude of approximately $35,800\,km$ and a orbital velocity of approximately $3,000\,m/s$.

Vertical Circular Motion

• Forces on a Mass Swung in a Vertical Circle (Bucket on a String):

  • Downward gravitational force ($F_g = m g$) remains constant throughout the loop.

  • Tension ($F_T$) continually varies to supply centripetal force and manage weight components.

• Top of the Loop:

  • Both tension and gravity act downward toward the center:     $F_c = F_T + F_g$

  • At critical minimum velocity to complete the circle, string tension drops to zero ($F_T = 0$):     $F_c = F_g \implies \frac{m v_{\text{min}}^2}{r} = m g \implies v_{\text{min}} = \sqrt{g r}$

  • At speeds below $v_{\text{min}}$, the string slackens and the object leaves the circular path.

• Bottom of the Loop:

  • Tension acts upward toward the center, while gravity acts downward away from the center:     $F_c = F_T - F_g \implies F_T = F_c + F_g$

  • Tension must counteract the object's weight while simultaneously providing centripetal acceleration.

  • Tension reaches its absolute maximum at the bottom of the swing.

• Sides of the Loop (Ascent / Descent):

  • Gravity acts perpendicular ($90^\circ$) to the radial string line, exerting zero radial influence.

  • Tension alone supplies the centripetal force: $F_T = F_c$.

• Energy Distribution in Vertical Circles:

  • Gravitational Potential Energy ($E_p$): Reaches maximum at top of loop (greatest elevation) and minimum at bottom of loop.

  • Kinetic Energy ($E_k$): Reaches maximum at bottom of loop (greatest speed) and minimum at top of loop.

Rotational Kinematics and Angle Measurement

• Angular Measure (Radians):

  • One radian ($1\,\text{rad}$) is equivalent to $57.3^\circ$.

  • Expressed as fractions and multiples of $\pi$:

  • One complete revolution ($360^\circ$) $= 2 \pi\,\text{rad}$

  • Half revolution ($180^\circ$) $= \pi\,\text{rad}$

  • Quarter revolution ($90^\circ$) $= \frac{\pi}{2}\,\text{rad}$

  • One-eighth revolution ($45^\circ$) $= \frac{\pi}{4}\,\text{rad}$

• Translational vs. Rotational Motion:

  • Pure Translational Motion: Produced when an applied force vector passes directly through the object's center of mass (the object accelerates linearly without rotating).

  • Pure Rotational Motion: Produced when net force equals zero ($F_{\text{net}} = 0$) but net torque is non-zero ($\tau_{\text{net}} \neq 0$), such as when equal and opposite forces are applied at offset positions (a force couple).

  • Combined Motion: Occurs in all intermediate conditions where a net force is applied off-center from the center of mass.

• Translational and Rotational Equivalents:

  • Linear Displacement ($d$ in $m$) $\longleftrightarrow$ Angular Displacement ($\theta$ in $\text{rad}$)

  • Linear Velocity ($v$ in $m/s$) $\longleftrightarrow$ Angular Velocity ($\omega$ in $\text{rad}/s$):     $\omega = \frac{\Delta \theta}{t}$

  • Linear Acceleration ($a$ in $m/s^2$) $\longleftrightarrow$ Angular Acceleration ($\alpha$ in $\text{rad}/s^2$):     $\alpha = \frac{\Delta \omega}{t}$

• Connecting Linear and Rotational Quantities:

  • Linear distance: $d = r \theta$

  • Tangential speed: $v = r \omega$

  • Tangential acceleration: $a = r \alpha$

• Rotational Kinematics Equations:   $\omega_f = \omega_i + \alpha t$   $\theta = \omega_i t + \frac{1}{2} \alpha t^2$   $\omega_f^2 = \omega_i^2 + 2 \alpha \theta$   $\theta = \frac{(\omega_i + \omega_f) t}{2}$

• Worked Example: Lawn Mower Blade Acceleration:

  • Accelerates from rest ($\omega_i = 0$) at $\alpha = 90\,\text{rad}/s^2$ to $\omega_f = 300\,\text{rad}/s$.

  • Select kinematic equation independent of time:     $\omega_f^2 = \omega_i^2 + 2 \alpha \theta \implies \theta = \frac{\omega_f^2}{2 \alpha}$     $\theta = \frac{(300\,\text{rad}/s)^2}{2 \times 90\,\text{rad}/s^2} = \frac{90000}{180} = 500\,\text{rad}$

  • Convert radians to revolutions:     $\text{Revolutions} = \frac{\theta}{2 \pi} = \frac{500}{2 \pi} = 79.58 \approx 80\,\text{rotations}$

Torque and Rotational Dynamics

• Torque ($\tau$):

  • Torque describes the rotational effectiveness of a force in causing angular acceleration.

  • Measured in Newton-meters ($N\,m$).

  • Basic equation:     $\tau = F d_\perp$     where $d_\perp$ is the perpendicular distance from the axis of rotation to the line of action of the force.

• Rotational Equivalent of Newton's Second Law:

  • In linear mechanics: $F = m a$ (force equals mass times linear acceleration).

  • In rotational mechanics: $\tau = I \alpha$     where $\tau$ is torque ($N\,m$), $I$ is rotational inertia ($kg\,m^2$), and $\alpha$ is angular acceleration ($\text{rad}/s^2$).

• Worked Example: Merry-Go-Round:

  • Radius: $r = 2.0\,m$

  • Rotational inertia: $I = 500\,kg\,m^2$

  • Tangential pushing force: $F = 150\,N$

  • Torque exerted:     $\tau = F r = 150\,N \times 2.0\,m = 300\,N\,m$

  • Calculate angular acceleration:     $\alpha = \frac{\tau}{I} = \frac{300\,N\,m}{500\,kg\,m^2} = 0.6\,\text{rad}/s^2$

• Air Rifle Pellet Impact Demonstration:

  • A wooden dowel mounted to a disc on a rotary sensor is struck by a lead pellet fired from an air rifle.

  • Shot 1 (Perpendicular): Velocity is perpendicular to the radius arm ($90^\circ$). The entire impact force acts tangentially, maximizing torque ($\tau = r F$), yielding rapid angular acceleration and high final angular velocity.

  • Shot 2 (Angled): The pellet strikes at an oblique angle. Only the perpendicular component ($F_\perp = F \sin(\phi)$) generates torque. This reduced torque produces lower angular acceleration and lower angular velocity. The parallel force component produces linear force on the axle, manifested as pronounced disc wobble.

Rotational Inertia

• Definition: Rotational inertia ($I$) represents an object's resistance to changes in its rotational state (angular acceleration).

  • While mass represents linear inertia (resistance to linear acceleration), rotational inertia depends on both total mass and the spatial distribution of that mass relative to the axis of rotation.

  • Standard units: $kg\,m^2$.

• Standard Geometric Formulas for Rotational Inertia:

  • Thin ring / hoop of radius $r$ (all mass located at perimeter):     $I = m r^2$

  • Thick ring with inner radius $r_1$ and outer radius $r_2$:     $I = \frac{1}{2} m (r_1^2 + r_2^2)$

  • Solid uniform sphere of radius $r$:     $I = \frac{2}{5} m r^2$

  • Hollow thin-walled sphere of radius $r$:     $I = \frac{2}{3} m r^2$

  • Solid cylinder / uniform disc of radius $r$:     $I = \frac{1}{2} m r^2$

  • Thin rod rotated about its long axis:     $I = \frac{1}{12} m r^2$

• Rotating Platform Experiment (Mass Distribution Verification):

  • A turntable driven by a constant torque applied via a hanging mass and pulley is tested with two movable masses.

  • Configuration 1: Masses positioned $10\,cm$ ($0.10\,m$) from center. After accelerating for $30\,s$, rotational period is measured as $T = 0.53\,s$.     $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.53} = 11.85\,\text{rad}/s$     $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{11.85}{30} = 0.40\,\text{rad}/s^2$

  • Configuration 2: Masses repositioned at double the distance ($20\,cm = 0.20\,m$). After accelerating for $42\,s$, period is measured as $T = 1.5\,s$.     $\omega = \frac{2 \pi}{1.5} = 4.19\,\text{rad}/s$     $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{4.19}{42} = 0.10\,\text{rad}/s^2$

  • Theoretical Analysis: Rotational inertia scales as $I = k m r^2$. Doubling the radial position of the masses ($2 \times r$) increases their rotational inertia by a factor of $2^2 = 4$. Under constant torque, fourfold greater inertia reduces angular acceleration by a factor of four ($0.40\,\text{rad}/s^2 \rightarrow 0.10\,\text{rad}/s^2$).

Rotational Kinetic Energy and Angular Momentum

• Rotational Kinetic Energy ($E_{k,\text{rot}}$):

  • The kinetic energy stored in an object's rotation is given by:     $E_{k,\text{rot}} = \frac{1}{2} I \omega^2$

  • Perfectly analogous to linear kinetic energy ($E_{k,\text{lin}} = \frac{1}{2} m v^2$).

• Rolling Without Slipping:

  • An object rolling along a surface possesses both translational and rotational kinetic energy:     $E_{k,\text{total}} = E_{k,\text{lin}} + E_{k,\text{rot}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$

  • When a ball rolls down an incline, initial gravitational potential energy converts into both linear and rotational kinetic energy:     $E_p = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$

• Angular Momentum ($L$):

  • Rotational equivalent of linear momentum ($p = m v$).

  • Defined for an extended rigid body as:     $L = I \omega$

  • For a point mass $m$ moving with velocity $v$ at a perpendicular distance $r$ from an axis:     $L = m v r = p r$

  • Units are $kg\,m^2\,s^{-1}$.

• Worked Example: Astronaut Catching a Satellite Ball:

  • Floating astronaut with rotational inertia $I = 12.0\,kg\,m^2$ catches a $0.060\,kg$ tennis ball moving at $v = 15.5\,m/s$.

  • The impact occurs at radial distance $r = 0.820\,m$ from her center of mass.

  • Angular momentum delivered by ball:     $L = m v r = 0.060\,kg \times 15.5\,m/s \times 0.820\,m = 0.7626\,kg\,m^2\,s^{-1}$

  • Calculate resulting angular velocity:     $L = I \omega \implies \omega = \frac{L}{I} = \frac{0.7626}{12.0\,kg\,m^2} = 0.0636\,\text{rad}/s$

• Gyroscopic Precession Demonstration (Suspended Wheel):

  • A bicycle wheel equipped with axle handles is suspended from a ceiling cord.

  • When stationary, attempting to balance the wheel horizontally fails; it flops downward immediately under gravitational torque.

  • When spinning rapidly, the wheel remains upright horizontally, processing slowly rather than falling.

  • Large angular momentum resists changes to its rotational orientation. Angular momentum is a vector; altering the axle's orientation requires an external torque acting over time ($\tau = \frac{\Delta L}{\Delta t}$).

Conservation of Angular Momentum

• Principle of Conservation: The total angular momentum of a system remains strictly constant unless acted upon by an external net torque:   $L_i = L_f \implies I_i \omega_i = I_f \omega_f$

• Astrophysical Demonstration (Stellar Core Collapse):

  • A massive main-sequence star rotates slowly (e.g., once every 30 days). Due to its immense radius, it possesses huge rotational inertia ($I$) and low angular velocity ($\omega$).

  • Upon exhausting its nuclear fuel, gravitational collapse shrinks its radius into a dense neutron star.

  • Drastic reduction in radius collapses rotational inertia ($I \propto r^2$). To conserve angular momentum, angular velocity increases dramatically.

  • Observed neutron stars (pulsars) rotate as rapidly as 700 times per second.

• Worked Example: Satellite Deploying Solar Arrays:

  • Satellite rotates initially once every $15\,\text{minutes}$ ($T_i = 15 \times 60\,s = 900\,s$).

  • Initial rotational inertia: $I_i = 1.5 \times 10^3\,kg\,m^2$

  • Initial angular velocity:     $\omega_i = \frac{2 \pi}{T_i} = \frac{2 \pi}{900\,s} = 0.698\,\text{rad}/s$

  • Deploying solar panels extends mass outward, increasing rotational inertia to $I_f = 6.0 \times 10^4\,kg\,m^2$.

  • Calculate new angular velocity:     $I_i \omega_i = I_f \omega_f \implies \omega_f = \frac{I_i \omega_i}{I_f} = \frac{(1.5 \times 10^3) \times 0.698}{6.0 \times 10^4} = 0.175\,\text{rad}/s$

  • Calculate new rotational period:     $T_f = \frac{2 \pi}{\omega_f} = \frac{2 \pi}{0.175\,\text{rad}/s} = 35900\,s \approx 598\,\text{minutes}$

• Figure Skater Spin Dynamics:

  • A spinning figure skater pulls her outstretched arms and free leg inward tight to her body's rotational axis.

  • Radial distance of limb mass decreases, significantly reducing total rotational inertia ($I$).

  • By conservation of angular momentum ($I_i \omega_i = I_f \omega_f$), angular velocity ($\omega$) increases, accelerating the spin.

Simple Harmonic Motion Principles

• Defining Criteria of Simple Harmonic Motion (SHM):

  1. The system experiences an acceleration directed toward its equilibrium position at all times.

  2. The magnitude of this acceleration is directly proportional to its displacement from equilibrium.

  • Expressed mathematically:     $a \propto -y$

• Kinematic Trajectory: The displacement, velocity, and acceleration profiles of a simple harmonic oscillator follow sinusoidal functions over time.

  • Amplitude ($A$): The maximum linear displacement from the central equilibrium position.

The Simple Pendulum

• Physical System: A bob suspended from a pivot by a string of negligible mass.

  • The effective pendulum length ($l$) is measured strictly from the suspension pivot to the center of mass of the bob.

• Small Angle Approximation:

  • The theoretical period formula is:     $T = 2 \pi \sqrt{\frac{l}{g}}$

  • This formula requires that $\sin(\theta) \approx \theta$ (with $\theta$ measured in radians).

  • The restoring force equation requires the restoring force to act horizontally toward equilibrium. Because a swinging pendulum traces an arc, the restoring force curves upward at higher angles.

  • At angles greater than $10^\circ$, deviations between $\sin(\theta)$ and $\theta$ induce errors exceeding three significant figures. In standard physics analysis, a small angle is defined as $\theta < 10^\circ$.

• Force Decomposition on a Pendulum:

  • Resolving gravitational force ($F_g = m g$) into coordinates:

  • Tangential Force component:       $F_{\text{tangential}} = F_g \sin(\theta) = m g \sin(\theta)$       This represents the uncountered restoring force accelerating the bob toward equilibrium.

  • Radial Force component:       $F_{\text{radial}} = -F_g \cos(\theta) = -m g \cos(\theta)$

  • The net radial force supplies centripetal force ($F_c$):     $F_c = F_T - F_g \cos(\theta)$     $F_T = F_c + F_g \cos(\theta) = \frac{m v^2}{l} + m g \cos(\theta)$

• Tension and Velocity Variations:

  • Maximum tension occurs at equilibrium (lowest point): velocity is maximum, maximizing centripetal force ($\frac{m v^2}{l}$), and $\cos(0^\circ) = 1$, meaning string tension must fully balance gravity.

  • Minimum tension occurs at maximum amplitude ($\theta_{\max}$): speed drops to zero ($v = 0$, so $F_c = 0$), leaving tension equal only to the reduced gravity component ($F_T = m g \cos(\theta_{\max})$).

• Mass Invariance and Effective Length Demonstration:

  • Pendulum period ($T = 2 \pi \sqrt{\frac{l}{g}}$) is independent of mass.

  • Adding slotted weights to the top of an existing pendulum bob systematically decreases its period.

  • Explanation: Stacking weights on top of the bob shifts the composite center of mass upward toward the pivot, reducing effective pendulum length $l$, which decreases period $T$.

• Velocity at Half-Height Derivation:

  • Let maximum release height be $h$. At this point, total mechanical energy is purely potential:     $E_{\text{total}} = m g h$

  • At half maximum height ($\frac{h}{2}$):     $E_{\text{total}} = m g \left(\frac{h}{2}\right) + \frac{1}{2} m v_{\text{half}}^2$

  • Equating energy states:     $m g h = \frac{1}{2} m g h + \frac{1}{2} m v_{\text{half}}^2$     $\frac{1}{2} g h = \frac{1}{2} v_{\text{half}}^2 \implies v_{\text{half}}^2 = g h$

  • At equilibrium ($h = 0$), all energy is kinetic:     $m g h = \frac{1}{2} m v_{\text{eq}}^2 \implies g h = \frac{1}{2} v_{\text{eq}}^2$

  • Substituting $g h$:     $v_{\text{half}}^2 = \frac{1}{2} v_{\text{eq}}^2$     $v_{\text{half}} = \sqrt{\frac{1}{2}} v_{\text{eq}} = 0.707 v_{\text{eq}}$

  • Velocity at half height is not half of equilibrium velocity; it is over $70.7\%$ of the maximum equilibrium velocity.

Mass on a Spring Systems

• Oscillation Period:

  • An oscillating mass on a spring exhibits simple harmonic motion with period:     $T = 2 \pi \sqrt{\frac{m}{k}}$     where $m$ is oscillating mass ($kg$) and $k$ is the spring constant ($N/m$).

• Spring Combinations:

  • Springs in Series: End-to-end configuration. Compliance adds, reducing total stiffness:     $\frac{1}{k_{\text{total}}} = \frac{1}{k_1} + \frac{1}{k_2} + \dots = \sum \frac{1}{k_i}$

  • Springs in Parallel: Side-by-side configuration. Forces share the load, increasing total stiffness:     $k_{\text{total}} = k_1 + k_2 + \dots = \sum k_i$

• Experimental Verification:

  • Oscillating load: $m = 403\,g = 0.403\,kg$.

  • Individual spring constants measured via period: $k_1 = 21.6\,N/m$ and $k_2 = 19.3\,N/m$.

  • Parallel Configuration: Combined spring constant $k_p = 21.6 + 19.3 = 40.9\,N/m$.

  • Predicted period:       $T = 2 \pi \sqrt{\frac{0.403\,kg}{40.9\,N/m}} = 0.62\,s$

  • Experimentally measured period: $0.64\,s$ (a deviation of $1.6\%$ from theory).

  • Series Configuration: Combined spring constant:     $\frac{1}{k_s} = \frac{1}{21.6} + \frac{1}{19.3} \implies k_s = 10.19\,N/m$

  • Predicted period:       $T = 2 \pi \sqrt{\frac{0.403\,kg}{10.19\,N/m}} = 1.25\,s$

  • Experimentally measured period: $1.28\,s$ (a deviation of $2.3\%$ from theory).

  • Plotting period against variable masses confirms the theoretical square-root relationship ($T \propto \sqrt{m}$).