Unit 2: Two-Dimensional Motion, Rotational Dynamics, and Universal Gravitation

Overview of Two-Dimensional Motion

• Kinematics Definition: Kinematics is the branch of mechanics concerned with the study of motion without considering its causes (i.e., without considering the forces that cause or alter the motion). An example is analyzing the trajectory of a football without evaluating the forces acting upon it.
• Extension from 1D to 2D: Two-dimensional kinematics extends straight-line (one-dimensional) kinematics developed in Grade 11 to motion taking place across a plane. This extension allows for the physics analysis of real-world curved paths.
• Prevalence in Nature: Most natural motions follow curved paths rather than straight lines. Examples of two-dimensional motion include:
  • A ball kicked by a football player
  • The orbital motion of planets
  • A bicycle rounding a curve
  • The rotation of automobile wheels
• Unit Objectives:
  • Understand the basic physical concepts of two-dimensional motion.
  • Describe object motion in both horizontal and inclined projectile trajectories.
  • Describe uniform rotational motion, rotational dynamics, and Kepler's laws of planetary motion.
  • Describe Newton's law of Universal Gravitation.
  • Develop comprehensive problem-solving skills for two-dimensional physical systems.

Conceptual Foundations & Initial Questions

• Brainstorming Question 2.1 (Part 1): Consider a ball shot horizontally from a very high building at high speed, assuming no gravitational force acts on it. Under zero gravity, the ball experiences no vertical acceleration ($a_y = 0$) and no horizontal acceleration ($a_x = 0$). By Newton's first law, it continues to move in a straight horizontal line at a constant horizontal velocity indefinitely.
• Brainstorming Question 2.1 (Part 2): Consider the same ball projected horizontally from the high building, but with gravity active. Gravity exerts a constant downward force, accelerating the ball downward ($a_y = -g$). Gravity does not affect the horizontal velocity component ($v_x$) because gravity acts purely vertically. The ball follows a curved parabolic path. The horizontal distance traveled is determined by its constant horizontal speed and the total time it takes to fall to the ground.
• Discussion Question 2.1: Identifying the motion that differs from the others among:
  • a) A ball thrown horizontally into the air
  • b) A bullet fired from a gun
  • c) A javelin thrown by an athlete
  • d) A bird flying in the air
  • Analysis: Option (d), a bird flying in the air, is fundamentally different. Options (a), (b), and (c) represent projectile motion, where objects move freely under gravity alone after launch. A bird controls its path using aerodynamic lift and internal mechanical propulsion, self-generating forces rather than moving strictly under gravity.

Projectile Motion Principles

• Definition of a Projectile: A projectile is any object that has been thrown, fired, or released, and continues in motion subject only to the influence of gravitational acceleration ($g = 9.8\,\text{m/s}^2$).
• Examples of Projectile Motion:
  • A football kicked during a game
  • A cannonball fired from a cannon
  • A bullet fired from a gun
  • The flight of a golf ball
  • A jet of water escaping from a hose
• Three Key Simplifying Assumptions:
  1. Constant Acceleration: The free-fall acceleration $g = 9.8\,\text{m/s}^2$ is constant over the entire range of motion and is directed downward.
  2. Negligible Air Resistance: Aerodynamic drag forces are ignored.
  3. Constant Horizontal Velocity: Because gravitational acceleration has no horizontal component, horizontal acceleration is zero ($a_x = 0$), keeping horizontal velocity $v_x$ constant.
• Trajectory: Under these three assumptions, the path of a projectile (called its trajectory) is a parabola.
• Independence of Components: The horizontal ($x$) and vertical ($y$) components of projectile motion are completely independent of each other. They are solved independently using time $t$ as the common variable linking both dimensions.

Horizontal Projection

• Kinematic Setup: An object is launched horizontally from a height $\Delta y$ with an initial horizontal velocity $v_{0x}$. Its initial vertical velocity component is zero ($v_{0y} = 0$).
• Horizontal Motion Equations:
  • Horizontal acceleration: $a_x = 0$
  • Horizontal velocity at any time $t$: $v_x = v_{0x} = \text{constant}$
  • Horizontal displacement (Distance):         $\Delta x = v_{0x} t$
• Vertical Motion Equations:
  • Vertical acceleration: $a_y = g$ (directed downward)
  • Vertical velocity at any time $t$:         $v_y = v_{0y} + g t$         Since $v_{0y} = 0$:         $v_y = g t$
  • Vertical displacement at any time $t$:         $\Delta y = v_{0y} t + \frac{1}{2} g t^2$         Since $v_{0y} = 0$:         $\Delta y = \frac{1}{2} g t^2$
• Sign Conventions: Upward vectors (displacement, velocity) are positive ($+$), and downward vectors are negative ($-$).
• Time of Flight ($t$): The total time taken for a horizontally projected body to strike the ground:     $t = \sqrt{\frac{2\Delta y}{g}}$
• Horizontal Range ($R$): The maximum horizontal distance traveled prior to impact:     $R = v_{0x} \sqrt{\frac{2\Delta y}{g}}$
• Discussion Question 2.2: An airplane flying horizontally drops a package to a remote village.
  • Motion: The package undergoes horizontal projectile motion with initial velocity equal to the horizontal speed of the aircraft.
  • Trajectory: Parabolic curve.
  • Aircraft Position at Impact: Assuming the airplane maintains constant speed and direction, it will be located directly above the package when the package hits the ground because both have identical horizontal velocities.

Horizontal Projection Demonstrations & Quantitative Examples

• Activity 2.1 (Tabletop Tennis Ball Experiment):
  • Procedure: Place two tennis balls at the edge of a tabletop. Snap one ball horizontally off the table while simultaneously tapping the second ball so it drops vertically straight down.
  • Observation & Analysis: Both balls strike the floor at the exact same instant because vertical motion is completely independent of horizontal motion. The time of flight for both is $t = \sqrt{\frac{2y}{g}}$. Initial horizontal velocity of the launched ball is determined by measuring range $R$ and dividing by $t$ ($v_{0x} = \frac{R}{t}$).
• Example 2.1 (Rifle Aimed Horizontally):
  • Problem: A rifle is aimed horizontally at a target $30\,\text{m}$ away. The bullet hits the target $2\,\text{cm}$ below the aiming point. Using $g = 10\,\text{m/s}^2$, calculate (a) time of flight, (b) initial velocity.
  • Given: $\Delta x = 30\,\text{m}$, $\Delta y = -2\,\text{cm} = -0.02\,\text{m}$, $g = -10\,\text{m/s}^2$.
  • Solution (a):         $\Delta y = \frac{1}{2} g t^2$$-0.02\,\text{m} = \frac{1}{2} (-10\,\text{m/s}^2) t^2$$-0.02 = -5 t^2$$t^2 = 0.004 \implies t = 0.06\,\text{s}$
  • Solution (b):         $\Delta x = v_{0x} t$$v_{0x} = \frac{\Delta x}{t} = \frac{30\,\text{m}}{0.06\,\text{s}} = 500\,\text{m/s}$
• Example 2.2 (Rescue Airplane Package Drop):
  • Problem: A rescue plane travelling horizontally at $360\,\text{km/h}$ drops a package from a height of $300\,\text{m}$. Assuming $g = 10\,\text{m/s}^2$, find (a) time taken to reach the ground, (b) horizontal distance from the stranded driver where the drop occurs.
  • Given: $v_{0x} = 360\,\text{km/h} = \frac{360 \times 1000}{3600} = 100\,\text{m/s}$, $\Delta y = -300\,\text{m}$, $g = -10\,\text{m/s}^2$.
  • Solution (a):         $\Delta y = \frac{1}{2} g t^2$$-300\,\text{m} = \frac{1}{2} (-10\,\text{m/s}^2) t^2$$t^2 = 60 \implies t = 7.74\,\text{s}$
  • Solution (b):         $\Delta x = v_{0x} t = 100\,\text{m/s} \times 7.74\,\text{s} = 774\,\text{m}$
• Activity 2.2 (Laboratory Tube Launcher):
  • Procedure: Use a paper tube, rubber band, and foil ball. Keep rubber band stretch identical to maintain launch velocity. Calculate flight time from table height $y$ via $t = \sqrt{\frac{2y}{g}}$, measure distance $R$, and solve for velocity $v_{0x} = \frac{R}{t}$.

Inclined Projectile Motion

• Kinematic Setup: An object is projected with an initial velocity $v_0$ at an angle $\theta$ with respect to the horizontal.
• Initial Velocity Component Resolution:
  • Horizontal component: $v_{0x} = v_0 \cos(\theta)$
  • Vertical component: $v_{0y} = v_0 \sin(\theta)$
• Velocity Components at Time $t$:
  • Horizontal velocity: $v_x = v_0 \cos(\theta) = \text{constant}$
  • Vertical velocity:         $v_y = v_0 \sin(\theta) + g t$
• Displacement Components at Time $t$:
  • Horizontal displacement:         $\Delta x = v_0 \cos(\theta) t$
  • Vertical displacement:         $\Delta y = v_0 \sin(\theta) t + \frac{1}{2} g t^2$
• Key Trajectory Characteristics:
  • At peak height ($H$), vertical velocity becomes zero ($v_y = 0$).
  • At the peak, the object possesses non-zero total velocity equal strictly to its horizontal velocity ($v = v_x = v_0 \cos(\theta)$).
  • After the peak, vertical velocity reverses direction and increases downward under gravity.
• Time to Reach Maximum Height ($t_{\text{peak}}$):     $v_y = v_0 \sin(\theta) + g t$     Setting $v_y = 0$ and using $g = -g_{\text{magnitude}}$:     $t_{\text{peak}} = \frac{v_0 \sin(\theta)}{g}$
• Total Time of Flight ($t_{\text{total}}$):
  • For launch and landing at equal elevation ($\Delta y = 0$):         $\Delta y = v_0 \sin(\theta) t + \frac{1}{2} g t^2 = 0$$t \left( v_0 \sin(\theta) + \frac{1}{2} g t \right) = 0$$t_{\text{total}} = \frac{2 v_0 \sin(\theta)}{g}$
  • Restriction: This expression applies strictly when launch and landing elevations are equal.
• Horizontal Range ($R$):
  • Substitute $t_{\text{total}}$ into horizontal displacement equation:         $R = v_0 \cos(\theta) \left( \frac{2 v_0 \sin(\theta)}{g} \right) = \frac{v_0^2 (2 \sin(\theta) \cos(\theta))}{g}$
  • Using trigonometric identity $2 \sin(\theta) \cos(\theta) = \sin(2\theta)$:         $R = \frac{v_0^2 \sin(2\theta)}{g}$
  • Properties of Range:
    • Range is directly proportional to $v_0^2$ and $\sin(2\theta)$.
    • Maximum range occurs at $\theta = 45^\circ$, because $\sin(2 \times 45^\circ) = \sin(90^\circ) = 1$.
    • Complementary launch angles (angles that sum to $90^\circ$, e.g., $30^\circ$ and $60^\circ$, or $37^\circ$ and $53^\circ$) yield identical horizontal ranges for equal launch speeds. The larger angle yields a higher peak height.
• Maximum Height ($H$):
  • Substitute $t_{\text{peak}} = \frac{v_0 \sin(\theta)}{g}$ into vertical displacement equation:         $H = v_0 \sin(\theta) \left( \frac{v_0 \sin(\theta)}{g} \right) + \frac{1}{2} (-g) \left( \frac{v_0 \sin(\theta)}{g} \right)^2$$H = \frac{v_0^2 \sin^2(\theta)}{2g}$
• Mathematical Relation Between Range and Maximum Height:     $\frac{H}{R} = \frac{\frac{v_0^2 \sin^2(\theta)}{2g}}{\frac{v_0^2 \sin(2\theta)}{g}} = \frac{\sin^2(\theta)}{2 \times 2 \sin(\theta) \cos(\theta)} = \frac{\sin(\theta)}{4 \cos(\theta)} = \frac{\tan(\theta)}{4}$$H = \frac{R \tan(\theta)}{4}$
• Discussion Question 2.3: Balls A and B kicked at $37^\circ$ and $53^\circ$ respectively with identical initial speeds v_0$.\n * a) *Maximum horizontal displacement*: Both have equal horizontal range since 37^\circ + 53^\circ = 90^\circ.\n * b) *Maximum height*: Ball B (53^\circ$) reaches greater maximum height because$\sin(53^\circ) > \sin(37^\circ).\n* **Discussion Question 2.4**:\n * 1. If horizontal range equals three times maximum height (R = 3H):\n        \frac{H}{3H} = \frac{\tan(\theta)}{4} \implies \frac{1}{3} = \frac{\tan(\theta)}{4} \implies \tan(\theta) = \frac{4}{3} \implies \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ\n * 2. Ball kicked into air at angle \theta; at highest point:\n * Option (c) is true: Its velocity is horizontal while acceleration is vertically downward, making velocity perpendicular to acceleration.\n * 3. Horizontal throw vs vertical drop from equal height:\n * Both hit the ground simultaneously because vertical motion is completely independent of horizontal motion.\n\n# Inclined Projection Quantitative Examples\n\n* **Example 2.3 (Football Kicked at Angle)**:\n * *Problem*: A football player kicks a ball at an angle of 37^\circ$with initial speed$40\,\text{m/s}$. Using$g = 10\,\text{m/s}^2, find (a) maximum height, (b) horizontal range.\n * *Given*: v_0 = 40\,\text{m/s}$,$\theta = 37^\circ$,$\sin(37^\circ) = 0.6$,$\sin(74^\circ) = 0.9613$,$g = 10\,\text{m/s}^2$.
  • Solution (a):         $H = \frac{v_0^2 \sin^2(\theta)}{2g} = \frac{(40\,\text{m/s})^2 \sin(37^\circ) \sin(37^\circ)}{2 \times 10\,\text{m/s}^2} = \frac{1600 \times 0.6 \times 0.6}{20} = 28.8\,\text{m}$
  • Solution (b):         $R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{(40\,\text{m/s})^2 \sin(74^\circ)}{10\,\text{m/s}^2} = \frac{1600 \times 0.9613}{10} = 153.8\,\text{m}$
• Example 2.4 (Ball Kicked Toward Wall):
  • Problem: A ball is kicked from ground level at $25\,\text{m/s}$ at an angle of $53^\circ$ toward a wall $24\,\text{m}$ away. Take $g = 10\,\text{m/s}^2$, $\cos(53^\circ) = 0.6$, \sin(53^\circ) = 0.8$.\n * *(a) Time to reach wall*:\n        \Delta x = v_0 \cos(\theta) t\n        24\,\text{m} = (25\,\text{m/s}) \times 0.6 \times t = 15 t \implies t = \frac{24}{15} = 1.6\,\text{s}\n * *(b) Height above ground level where ball strikes wall*:\n        \Delta y = v_0 \sin(\theta) t + \frac{1}{2} g t^2\n        \Delta y = (25 \times 0.8 \times 1.6) + \frac{1}{2} (-10) (1.6)^2 = 32 - 12.8 = 19.2\,\text{m}\n * *(c) Horizontal and vertical components of velocity at impact*:\n        v_x = v_0 \cos(\theta) = 25 \times 0.6 = 15\,\text{m/s}\n        v_y = v_0 \sin(\theta) + g t = (25 \times 0.8) + (-10 \times 1.6) = 20 - 16 = 4\,\text{m/s}\n * *(d) Resultant velocity at impact*:\n        v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 4^2} = \sqrt{225 + 16} = \sqrt{241} \approx 15.5\,\text{m/s}\n* **Activity 2.4 (V-Shaped Track Conservation of Energy Launcher)**:\n * *Procedure*: Release ball from height h$on long incline track. Speed leaving short incline at height$h_{\text{table}}$at angle$\theta$is determined via conservation of energy ($m g h = \frac{1}{2} m v^2 \implies v = \sqrt{2gh}$). Calculate flight time$t = \frac{2v_0 \sin(\theta)}{g}$and range$R = \frac{v_0^2 \sin(2\theta)}{g}$, and catch ball in a cup at predicted range$R.\n* **Discussion Question 2.5**:\n * 1. Along parabolic trajectory:\n * (a) Velocity and acceleration vectors are perpendicular *only* at maximum height (velocity is horizontal, acceleration is vertically downward).\n * (b) Velocity and acceleration vectors are *never* parallel to each other.\n * 2. Projectile motion statements (ignoring air resistance):\n * (e) All of the above statements are true (horizontal/vertical independent; force constant; acceleration constant; path depends on velocity, not mass).\n * 3. Projectile on Earth vs Moon (g_{\text{Moon}} = 1.6\,\text{m/s}^2):\n * The Moon projectile has both greater range and greater maximum height because gravitational acceleration g$appears in the denominator of both formulas ($R \propto \frac{1}{g}$,$H \propto \frac{1}{g}).\n* **Exercise 2.1 (Practice Problems)**:\n * 1. *Minimum speed position*: **B. at maximum height** (where vertical speed is zero).\n * 2. *Gun aiming*: Muzzle speed 500\,\text{m/s}$, target$50\,\text{m}$away. Time$t = \frac{50}{500} = 0.1\,\text{s}$. Drop$\Delta y = \frac{1}{2} g t^2 = \frac{1}{2} (10) (0.1)^2 = 0.05\,\text{m} = 5\,\text{cm}. Gun must be aimed **C. 5 cm high above the target**.\n * 3. *Horizontal throw*: v_{0x} = 20\,\text{m/s}$, height$90\,\text{m}$,$g = 10\,\text{m/s}^2$.
    • a) Time: $-90 = -5 t^2 \implies t = \sqrt{18} \approx 4.24\,\text{s}$.
    • b) Range: $\Delta x = 20 \times 4.24 = 84.8\,\text{m}$.
    • c) Velocity: $v_x = 20\,\text{m/s}$, $v_y = -10 \times 4.24 = -42.4\,\text{m/s}$. Resultant $v = \sqrt{20^2 + (-42.4)^2} = \sqrt{400 + 1797.76} = 46.88\,\text{m/s}$.
  • 4. Long jumper: Angle $20.0^\circ$, speed $11.0\,\text{m/s}$, g = 10\,\text{m/s}^2$.\n * a) Range: R = \frac{11^2 \sin(40^\circ)}{10} = \frac{121 \times 0.6428}{10} = 7.78\,\text{m}.\n * b) Height: H = \frac{11^2 \sin^2(20^\circ)}{20} = \frac{121 \times (0.3420)^2}{20} = 0.708\,\text{m}.\n * 5. *Projected object*: v_0 = 30\,\text{m/s}$,$t_{\text{peak}} = 1.5\,\text{s}$,$g = 10\,\text{m/s}^2$.
    • t_{\text{peak}} = \frac{v_0 \sin(\theta)}{g} \implies 1.5 = \frac{30 \sin(\theta)}{10} \implies \sin(\theta) = 0.5 \implies \theta = 30^\circ$.\n * Range: R = \frac{30^2 \sin(60^\circ)}{10} = \frac{900 \times 0.866}{10} = 77.94\,\text{m}.\n\n# Rotational Kinematics\n\n* **Definition**: Rotational motion is the movement of an object in a circular path around a fixed axis of rotation.\n* **Rigid Body**: An object with a perfectly defined and unchanging shape. Regardless of applied forces, the distance between any two particles within a rigid body remains strictly constant.\n* **Directions of Rotation**: Rotation around a fixed axis occurs in either a clockwise or anticlockwise (counterclockwise) direction.\n* **Angular Displacement (\Delta \theta)**:\n * When a rigid object rotates, every particle in the object rotates through the exact same angle \Delta \theta.\n * Formula:\n        \Delta \theta = \theta_f - \theta_0\n * *Units*: Measured in radians (rad) or degrees (^\circ).\n * *Conversion Standard*:\n        1\,\text{revolution} = 2\pi\,\text{rad} = 360^\circ\n* **Angular Velocity (\omega)**:\n * The rate of change of angular displacement with respect to time.\n * Average angular velocity equation:\n        \omega_{\text{av}} = \frac{\theta_f - \theta_0}{t_f - t_0} = \frac{\Delta \theta}{\Delta t}\n * *Units*: Radians per second (rad/s).\n* **Angular Acceleration (\alpha)**:\n * The rate of change of angular velocity with respect to time.\n * Average angular acceleration equation:\n        \alpha = \frac{\omega_f - \omega_0}{t_f - t_0} = \frac{\Delta \omega}{\Delta t}\n * *Units*: Radians per second squared (rad/s^2).\n* **Vector Directions of Angular Quantities**:\n * Angular velocity \vec{\omega}$and angular acceleration$\vec{\alpha} are vector quantities directed along the axis of rotation.\n * **Right-Hand Rule (RHR)**: Wrap the four fingers of your right hand around the axis in the direction of rotation; your extended right thumb points in the direction of the angular velocity vector \vec{\omega}.\n * **Direction of \vec{\alpha}$**:$\vec{\alpha}$points in the same direction as$\vec{\omega}$if angular speed is increasing, and points antiparallel (opposite direction) to$\vec{\omega} if angular speed is decreasing.\n\n# Equations of Motion for Constant Angular Acceleration\n\n* Derived analogously to linear kinematic equations:\n 1. \omega_f = \omega_0 + \alpha \Delta t\n 2. \omega_{\text{av}} = \frac{\omega_0 + \omega_f}{2}\n 3. \Delta \theta = \omega_0 \Delta t + \frac{1}{2} \alpha \Delta t^2\n 4. \omega_f^2 = \omega_0^2 + 2 \alpha \Delta \theta\n\n* **Kinematic Analogy Table (Linear vs. Rotational)**:\n\n| Linear Motion (Constant a$) | Rotational Motion (Constant$\alpha) |\n| :--- | :--- |\n| v_f = v_0 + a \Delta t$|$\omega_f = \omega_0 + \alpha \Delta t |\n| v_{\text{av}} = \frac{v_f + v_0}{2}$|$\omega_{\text{av}} = \frac{\omega_f + \omega_0}{2} |\n| \Delta s = \left(\frac{v_f + v_0}{2}\right) \Delta t$|$\Delta \theta = \left(\frac{\omega_f + \omega_0}{2}\right) \Delta t |\n| \Delta s = v_0 t + \frac{1}{2} a \Delta t^2$|$\Delta \theta = \omega_0 \Delta t + \frac{1}{2} \alpha \Delta t^2 |\n| v_f^2 = v_0^2 + 2 a \Delta s$|$\omega_f^2 = \omega_0^2 + 2 \alpha \Delta \theta |\n\n# Relationships Between Rotational and Translational Quantities\n\n* For a point $P$ located at radius r from the axis of rotation on a rigid rotating body:\n* **Arc Length / Linear Displacement (s)**:\n    s = r \theta\n    *(Note: \theta must be expressed strictly in radians)*.\n* **Tangential Velocity (v)**:\n    v = \frac{\Delta s}{\Delta t} = r \frac{\Delta \theta}{\Delta t} = \omega r\n * Every point on a rigid body has the same angular speed \omega$, but tangential speed$v$increases linearly with distance$r from the axis.\n* **Tangential Acceleration (a_t)**:\n    a_t = \frac{\Delta v}{\Delta t} = r \frac{\Delta \omega}{\Delta t} = \alpha r\n\n# Rotational Kinematics Quantitative Examples\n\n* **Example 2.5 (Average Angular Velocity)**:\n * *Problem*: A wheel changes angular speed from 30\,\text{rad/s}$to$50\,\text{rad/s}$in$2\,\text{s}. Calculate average angular acceleration.\n * *Solution*: \alpha_{\text{av}} = \frac{50\,\text{rad/s} - 30\,\text{rad/s}}{2\,\text{s}} = 10\,\text{rad/s}^2\n* **Example 2.6 (Wheel Angular Acceleration & Distance)**:\n * *Problem*: Wheel has \omega_0 = 10\,\text{rad/s}$and$\alpha = 2.5\,\text{rad/s}^2$. (a) How many revolutions in$30\,\text{s}$? (b) Angular speed at$t = 20\,\text{s}?\n * *Solution (a)*:\n        \Delta \theta = (10 \times 30) + \frac{1}{2} (2.5) (30)^2 = 300 + 1125 = 1425\,\text{rad}\n        \Delta \theta = 1425\,\text{rad} \times \frac{1\,\text{rev}}{2\pi\,\text{rad}} = 226.9\,\text{rev}\n * *Solution (b)*:\n        \omega_f = 10 + (2.5 \times 20) = 60\,\text{rad/s}\n* **Example 2.7 (Car Wheel Speed After Revolutions)**:\n * *Problem*: Car wheel initial \omega_0 = 6\,\text{rad/s}$,$\alpha = 3\,\text{rad/s}^2$. Calculate$\omega_f$after$100\,\text{rev}.\n * *Solution*:\n        \Delta \theta = 100\,\text{rev} \times 2\pi = 628\,\text{rad}\n        \omega_f^2 = 6^2 + 2 (3) (628) = 36 + 3768 = 3804\n        \omega_f = \sqrt{3804} = 61.68\,\text{rad/s}\n* **Example 2.8 (Tangential Acceleration of Pulley Tip)**:\n * *Problem*: Wheel radius r = 20\,\text{cm} = 0.2\,\text{m}$, accelerates from rest to$15\,\text{rev/s}$in$30\,\text{s}. Find tangential acceleration.\n * *Solution*:\n        \alpha = \frac{15 - 0}{30} = 0.5\,\text{rev/s}^2 = 0.5 \times 2\pi = 3.14\,\text{rad/s}^2\n        a_t = \alpha r = 3.14\,\text{rad/s}^2 \times 0.2\,\text{m} = 0.6\,\text{m/s}^2\n* **Example 2.9 (Accelerating Car Wheel)**:\n * *Problem*: Car accelerates from 20\,\text{m/s}$to$24\,\text{m/s}$in$5\,\text{s}$. Radius$r = 40\,\text{cm} = 0.4\,\text{m}$. Calculate$\alpha$.
  • Solution:         $a_t = \frac{24 - 20}{5} = 0.8\,\text{m/s}^2$$\alpha = \frac{a_t}{r} = \frac{0.8\,\text{m/s}^2}{0.4\,\text{m}} = 2\,\text{rad/s}^2$
• Example 2.10 (Bicycle Ride Parameters):
  • Problem: Bicycle ridden for $5\,\text{min}$ ($300\,\text{s}$). Wheel radius $r = 30\,\text{cm} = 0.3\,\text{m}$ completes $2000\,\text{rev}$. Find (a) average angular velocity, (b) distance traveled.
  • Solution (a):         $\omega_{\text{av}} = \frac{2000\,\text{rev}}{300\,\text{s}} = 6.67\,\text{rev/s} = 6.67 \times 2\pi = 41.9\,\text{rad/s}$
  • Solution (b):         $\Delta \theta = 2000 \times 2\pi = 12560\,\text{rad}$$\Delta s = r \Delta \theta = 0.3\,\text{m} \times 12560\,\text{rad} = 3768\,\text{m}$
• Example 2.11 (Pulley Winding):
  • Problem: Pulley radius $50\,\text{cm} = 0.5\,\text{m}$ lifts bucket $20\,\text{m}$. Find revolutions required.
  • Solution:         $\Delta \theta = \frac{\Delta s}{r} = \frac{20\,\text{m}}{0.5\,\text{m}} = 40\,\text{rad}$$\Delta \theta = \frac{40}{2\pi} = 6.34\,\text{rev}$
• Example 2.12 (Bicycle Speed from Angular Velocity):
  • Problem: Wheel $\omega = 18\,\text{rad/s}$, $r = 40\,\text{cm} = 0.4\,\text{m}$. Find linear speed.
  • Solution: $v = \omega r = 18\,\text{rad/s} \times 0.4\,\text{m} = 7.2\,\text{m/s}$
• Example 2.13 (Two Particles at Different Radii):
  • Problem: Flat disk speeds from rest to $20\,\text{rad/s}$ in $4\,\text{s}$. Particle A at $r_A = 20\,\text{cm} = 0.2\,\text{m}$; Particle B at $r_B = 40\,\text{cm} = 0.4\,\text{m}$. Calculate $\alpha_{\text{av}}$ and linear acceleration $a$ for both.
  • Solution (a - Particle B):         $\alpha_{\text{av}} = \frac{20 - 0}{4} = 5\,\text{rad/s}^2$$a_B = \alpha r_B = 5 \times 0.4 = 2\,\text{m/s}^2$
  • Solution (b - Particle A):         $\alpha_{\text{av}} = 5\,\text{rad/s}^2\,\text{(same for all points)}$$a_A = \alpha r_A = 5 \times 0.2 = 1\,\text{m/s}^2$
• Discussion Question 2.7:
  • 1. Second hand of clock: Period $T = 60\,\text{s}$. Angular speed $\omega = \frac{2\pi}{60} = \frac{\pi}{30} \approx 0.105\,\text{rad/s}$. Direction of $\vec{\omega}$ for clock on vertical wall: pointing directly into the wall away from observer.
  • 2. Counterclockwise wheel in xy plane: $\vec{\omega}$ points along positive z-axis (out of page). If angular velocity decreases, $\vec{\alpha}$ points antiparallel, along negative z-axis (into page).
  • 3. Wheel rotation: (a) All points have identical angular speed $\omega$. (b) Points at different radii have different linear speeds $v = \omega r$.
• Exercise 2.2:
  • 1. True: All points on a rigid rotating body share identical angular speed.
  • 2. Invalid unit for angular displacement: D. rpm (revolutions per minute is a unit of angular velocity, not displacement).
  • 3. Pulley $r = 20\,\text{cm} = 0.2\,\text{m}$, lifts bucket $10\,\text{m}$ in $5\,\text{s}$. $\Delta \theta = \frac{10}{0.2} = 50\,\text{rad}$. $\omega_{\text{av}} = \frac{50}{5} = 10\,\text{rad/s}$.
  • 4. Circle $r = 1.50\,\text{m}$, arc $s = 2.50\,\text{m}$. $\theta = \frac{2.50}{1.50} = 1.67\,\text{rad}$. Degree value: 1.67 \times \frac{180^\circ}{\pi} = 95.5^\circ$.\n * 5. Deceleration \alpha = -5\,\text{rad/s}^2$,$\omega_0 = 3\,\text{rad/s}$,$\omega_f = 0$.$0 = 3^2 + 2(-5)\Delta \theta \implies 10\Delta \theta = 9 \implies \Delta \theta = 0.9\,\text{rad}.\n * 6. Speed increases from 200\,\text{rpm}$to$600\,\text{rpm}$in$8\,\text{s}.\n * \omega_0 = \frac{200 \times 2\pi}{60} = 20.94\,\text{rad/s}$,$\omega_f = \frac{600 \times 2\pi}{60} = 62.83\,\text{rad/s}.\n * a) \alpha = \frac{62.83 - 20.94}{8} = 5.24\,\text{rad/s}^2\n * b) \text{Revolutions} = \omega_{\text{av}} \times t = \left(\frac{200 + 600}{2}\right) \text{rpm} \times \frac{8}{60}\,\text{min} = 400 \times 0.1333 = 53.33\,\text{rev}.\n\n# Rotational Dynamics & Moment of Inertia\n\n* **Torque (\tau)**:\n * Torque is the rotational quantitative analog of force. It represents the rotational effectiveness of an applied force and causes angular acceleration.\n * Formula:\n        \tau = r F \sin(\theta)\n        Where r$is distance from axis of rotation to force application point,$F$is applied force, and$\theta$is angle between position vector$\vec{r}$and force vector$\vec{F}.\n * *Properties*: Vector quantity, SI unit is Newton-meter (N\cdotm). Direction is determined by right-hand rule (curl fingers along direction of force-induced rotation, thumb points along torque vector).\n* **Moment of Inertia ($I$)**:\n * Quantitative measure of rotational inertia (resistance of a rigid body to changes in its angular velocity), analogous to mass m in linear motion.\n * Depends on total mass and distribution of mass relative to axis of rotation.\n * Single point mass formula:\n        I = m r^2\n        *Units*: kg\cdot$m$^2. Scalar quantity.\n * System of multiple point masses formula:\n        I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + \dots = \sum m_i r_i^2\n* **Newton's Second Law for Rotation**:\n * The net torque applied to a rotating object is directly proportional to its angular acceleration:\n        \tau_{\text{net}} = I \alpha\n\n# Rotational Dynamics Quantitative Examples\n\n* **Example 2.14 (Net Torque on Pivoted Object)**:\n * *Problem*: Object pivoted at O subject to three forces: F_1 = 10\,\text{N}$at$r_1 = 3.0\,\text{m}$($\theta_1 = 120^\circ$);$F_2 = 16\,\text{N}$at$r_2 = 4.0\,\text{m}$($\theta_2 = 150^\circ$);$F_3 = 19\,\text{N}$at$r_3 = 8.0\,\text{m}$($\theta_3 = 45^\circ). Find net torque.\n * *Calculation*:\n        \tau_1 = -(3.0 \times 10 \times \sin(120^\circ)) = -25.9\,\text{N}\cdot\text{m}\,\text{(clockwise)}\n        \tau_2 = 4.0 \times 16 \times \sin(150^\circ) = 32.0\,\text{N}\cdot\text{m}\,\text{(counterclockwise)}\n        \tau_3 = 8.0 \times 19 \times \sin(45^\circ) = 107.4\,\text{N}\cdot\text{m}\,\text{(counterclockwise)}\n        \tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = -25.9 + 32.0 + 107.4 = 113.5\,\text{N}\cdot\text{m}\,\text{(counterclockwise)}\n* **Example 2.15 (Moment of Inertia of Point Masses)**:\n * *Problem*: Three masses along y-axis rotating around x-axis: m_1 = 4\,\text{kg}$at$y_1 = 3\,\text{m}$,$m_2 = 2\,\text{kg}$at$y_2 = 2\,\text{m}$,$m_3 = 3\,\text{kg}$at$y_3 = 4\,\text{m}$. System rotates at$\omega = 2\,\text{rad/s}. Find moment of inertia about x-axis.\n * *Solution*:\n        I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 = 4(3)^2 + 2(2)^2 + 3(4)^2 = 36 + 8 + 48 = 92\,\text{kg}\cdot\text{m}^2\n        *(Note: Text lists tabular sum evaluation as 164\,\text{kg}\cdot\text{m}^2)*.\n* **Example 2.16 (Torque and Rotational Inertia)**:\n * *Problem*: Applied torque \tau = 36\,\text{N}\cdot\text{m}$produces$\alpha = 24\,\text{rad/s}^2$. Find$I$.
  • Solution: $I = \frac{\tau}{\alpha} = \frac{36}{24} = 1.5\,\text{kg}\cdot\text{m}^2$
• Example 2.17 (Flywheel Motor Acceleration):
  • Problem: Motor produces constant torque $\tau = 100\,\text{N}\cdot\text{m}$, max speed $\omega_{\text{max}} = 150\,\text{rad/s}$, attached to flywheel $I = 0.1\,\text{kg}\cdot\text{m}^2$. Find (a) $\alpha$, (b) time from rest to max speed.
  • Solution (a): $\alpha = \frac{\tau}{I} = \frac{100}{0.1} = 1000\,\text{rad/s}^2$
  • Solution (b): $t = \frac{\omega_f - \omega_0}{\alpha} = \frac{150 - 0}{1000} = 0.15\,\text{s}$
• Exercise 2.3:
  • 1. Force $F = 400\,\text{N}$ applied at $r = 5\,\text{m}$ perpendicularly ($\theta = 90^\circ$). $\tau = 5 \times 400 = 2000\,\text{N}\cdot\text{m}$.
  • 2. Three masses $m$ at corners of equilateral triangle side $L$. Axis through one corner perpendicular to plane. Distance for corner mass is 0, for other two is $L$. Moment of inertia I = m(0)^2 + m(L)^2 + m(L)^2 = 2 m L^2$.\n * 3. Disc I = 2\,\text{kg}\cdot\text{m}^2$,$\omega_0 = 3\,\text{rad/s}$,$\omega_f = 8\,\text{rad/s}$,$\tau = 50\,\text{N}\cdot\text{m}$.$\alpha = \frac{50}{2} = 25\,\text{rad/s}^2$. Time$t = \frac{8 - 3}{25} = \frac{5}{25} = 0.2\,\text{s}.\n\n# Planetary Motion and Kepler's Laws\n\n* **Historical Evolution of Models**:\n * **Geocentric Model**: Proposed/formalized by Claudius Ptolemy (c. 100–c. 170 CE) in the 2nd century. Placed Earth static at the center of the Universe. Accepted for ~1400 years.\n * **Heliocentric Model**: Proposed by Nicolaus Copernicus (1473–1543) in 1543. Asserted Earth and planets revolve in circular orbits around the Sun.\n * **Observational Data**: Tycho Brahe (1546–1601) made high-precision planetary measurements using a large sextant and compass (pre-telescope era).\n * **Mathematical Model**: Johannes Kepler acquired Brahe's data and spent 16 years deriving planetary motion laws based on Mars' orbit.\n* **Kepler's First Law (Law of Ellipses)**:\n * The orbit of a planet around the Sun is an ellipse with the Sun located at one focus.\n * *Ellipse Properties*: Closed curve such that the sum of distances from any point on the curve to two fixed foci (f_1$and$f_2$) is constant ($r_1 + r_2 = \text{constant}).\n * Because orbits are elliptical, planet-to-Sun distance varies continuously.\n* **Kepler's Second Law (Law of Equal Areas)**:\n * An imaginary line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal intervals of time (A_1 = A_2 = A_3).\n * *Variable Orbital Speed*: Planets do not move at constant speed.\n * **Perihelion**: Point of closest approach to the Sun. Orbital speed is **fastest**.\n * **Aphelion**: Point of greatest separation from the Sun. Orbital speed is **slowest**.\n* **Kepler's Third Law (Law of Harmonies / Periods)**:\n * The ratio of the square of a planet's orbital period (T$) to the cube of its average distance from the Sun ($R) is constant for all planets orbiting the Sun:\n        \frac{T^2}{R^3} = K\n * The constant K is independent of planet mass.\n * *Orbital Period Examples*: Mercury (88\,\text{days}$), Earth ($365\,\text{days}$), Saturn ($10\,759\,\text{days}).\n * *Earth vs. Mars Kepler Data*:\n * Earth: T = 3.156 \times 10^7\,\text{s}$,$R = 1.4957 \times 10^{11}\,\text{m}$,$\frac{T^2}{R^3} = 2.977 \times 10^{-19}\,\text{s}^2/\text{m}^3\n * Mars: T = 5.93 \times 10^7\,\text{s}$,$R = 2.278 \times 10^{11}\,\text{m}$,$\frac{T^2}{R^3} = 2.975 \times 10^{-19}\,\text{s}^2/\text{m}^3\n* **Example 2.18 (Pluto Orbital Period Calculation)**:\n * *Given*: Earth T_E = 365\,\text{days}$,$R_E = 1.495 \times 10^8\,\text{km}$. Pluto$R_P = 5.896 \times 10^9\,\text{km}.\n * *Solution*:\n        \frac{T_E^2}{R_E^3} = \frac{T_P^2}{R_P^3} \implies \frac{365^2}{(1.495 \times 10^8)^3} = \frac{T_P^2}{(5.896 \times 10^9)^3}\n        T_P = 9.0 \times 10^{14}\,\text{days}\,\text{(as specified in text calculations)}\n* **Example 2.19 (Saturn Orbital Period Calculation)**:\n * *Given*: Saturn distance R_S = 9 R_E$, Earth period$T_E = 1\,\text{year}.\n * *Solution*:\n        \frac{T_E^2}{R_E^3} = \frac{T_S^2}{(9 R_E)^3} \implies \frac{1}{1} = \frac{T_S^2}{729} \implies T_S^2 = 729 \implies T_S = 27\,\text{years}\n* **Exercise 2.4**:\n * 1. True: Satellite increases speed near Sun (perihelion) and decreases away (aphelion).\n * 2. Moon: T_M = 27.3\,\text{days}$,$R_M = 3.84 \times 10^8\,\text{m}$. Satellite altitude$1500\,\text{km}$, Earth radius$6380\,\text{km}$, total orbital radius$R_s = 6380 + 1500 = 7880\,\text{km} = 7.88 \times 10^6\,\text{m}.\n        \frac{T_s^2}{(7.88 \times 10^6)^3} = \frac{(27.3)^2}{(3.84 \times 10^8)^3}\n        T_s = 27.3 \times \left(\frac{7.88 \times 10^6}{3.84 \times 10^8}\right)^{3/2} = 27.3 \times (0.02052)^{1.5} = 27.3 \times 0.00294 = 0.0802\,\text{days} \approx 1.92\,\text{hours}\n * 3. If orbital radius doubles (R_2 = 2 R_1):\n        T_2^2 \propto (2 R_1)^3 = 8 R_1^3 \implies T_2 = \sqrt{8} T_1 = 2\sqrt{2} T_1\n        Correct Option: **C. The period would increase by a factor of 2\sqrt{2}**.\n\n# Newton's Law of Universal Gravitation\n\n* **Brainstorming Question 2.2**: If the Sun's gravity were suddenly switched off, planets would immediately cease orbiting in curved paths and move off in straight lines tangential to their orbits at constant speed (Newton's first law).\n* **Historical Insight**: Galileo Galilei established that all masses fall at identical acceleration under gravity in the absence of air resistance. Sir Isaac Newton (1666) unified terrestrial drop with celestial orbit, recognizing gravity as universal.\n* **Statement of Universal Gravitation Law**: Every particle of matter in the Universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.\n* **Mathematical Formula**:\n    F_g = \frac{G m_1 m_2}{r^2}\n * m_1, m_2: masses of the interacting bodies\n * r: center-to-center separation distance\n * G: Universal Gravitational Constant\n        G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2\,\text{(or } 6.673 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2\text{)}\n* **Vector Nature**: Always attractive force directed along line joining center of masses. Forms Newton's third law action-reaction pair (F_{12} = -F_{21}).\n* **Example 2.20 (Attraction Between Two Small Masses)**:\n * *Problem*: m_1 = 10\,\text{kg}$,$m_2 = 100\,\text{kg}$, separation$r = 1\,\text{m}. Find gravitational force.\n * *Solution*:\n        F_g = \frac{6.673 \times 10^{-11} \times 10 \times 100}{1^2} = 6.67 \times 10^{-8}\,\text{N}\n* **Example 2.21 (Gravitational Force on Person at Earth's Surface)**:\n * *Problem*: Person mass m = 60.0\,\text{kg}$, Earth mass$M_E = 5.97 \times 10^{24}\,\text{kg}$, radius$r_E = 6.38 \times 10^6\,\text{m}.\n * *Solution via Universal Gravitation*:\n        F_g = \frac{6.673 \times 10^{-11} \times 5.97 \times 10^{24} \times 60}{(6.38 \times 10^6)^2} = 584\,\text{N}\n * *Solution via Weight Equation (F_g = m g)*:\n        F_g = 60 \times 9.8 = 588\,\text{N}\n* **Discussion Question 2.12**:\n * *Part 1*: Present mass, 8 times volume. Volume V \propto r^3 \implies 8 = r_{\text{new}}^3 \implies r_{\text{new}} = 2 r_E$. Weight becomes$\frac{1}{2^2} = \frac{1}{4} of present weight.\n * *Part 2*: Present size, 1/3 mass. Weight becomes \frac{1}{3} of present weight.\n* **Derivation of Surface Gravitational Acceleration (g)**:\n    m g = \frac{G M_E m}{r_E^2} \implies g = \frac{G M_E}{r_E^2}\n    Substituting constants:\n    g = \frac{6.673 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.38 \times 10^6)^2} = 9.8\,\text{m/s}^2\n* **Derivation of Kepler's Third Law from Newton's Gravitation & Centripetal Force**:\n * Centripetal force required for circular orbit:\n        F_c = \frac{m_p v^2}{r}\n * Gravitational force provides centripetal force:\n        \frac{m_p v^2}{r} = \frac{G M_s m_p}{r^2} \implies v^2 = \frac{G M_s}{r}\n * Substitute orbital speed v = \frac{2\pi r}{T}:\n        \frac{(2\pi r)^2}{T^2} = \frac{G M_s}{r} \implies \frac{4\pi^2 r^2}{T^2} = \frac{G M_s}{r}\n * Rearranging terms yields Kepler's Third Law:\n        \frac{T^2}{r^3} = \frac{4\pi^2}{G M_s} \approx 2.97 \times 10^{-19}\,\text{s}^2/\text{m}^3\n * Constant K = \frac{4\pi^2}{G M_s}$. For satellite orbiting Earth, replace Sun mass$M_s$with Earth mass$M_E$($K_E = \frac{4\pi^2}{G M_E}).\n* **Exercise 2.5**:\n * 1. False: By Newton's third law, the force exerted by Earth on the man is equal in magnitude and opposite in direction to the force exerted by the man on Earth.\n * 2. Distance doubled (r \rightarrow 2r$): New force$F_{\text{new}} = \frac{F}{2^2} = \frac{1}{4}F. Correct Option: **B. 1/4F**.\n * 3. Newton's law applies to: **C. All bodies irrespective of their size**.\n * 4. Gravitational force is 30\,\text{N}$. Each mass doubled ($m_1 \rightarrow 2m_1$,$m_2 \rightarrow 2m_2$): New force$F_{\text{new}} = 30 \times (2 \times 2) = 120\,\text{N}.\n * 5. Calculate Mass of Sun (M_s):\n        T = 3.156 \times 10^7\,\text{s}$,$r = 1.496 \times 10^{11}\,\text{m}.\n        M_s = \frac{4\pi^2 r^3}{G T^2} = \frac{4\pi^2 (1.496 \times 10^{11})^3}{(6.673 \times 10^{-11}) (3.156 \times 10^7)^2} = \frac{3.9478 \times 10^{35} \times 3.348 \times 10^{33}}{6.673 \times 10^{-11} \times 9.960 \times 10^{14}} = 1.99 \times 10^{30}\,\text{kg}.\n * 6. Planet mass M_p = 4 M_E$, radius$R_p = 2 R_E.\n        g_p = \frac{G (4 M_E)}{(2 R_E)^2} = \frac{4 G M_E}{4 R_E^2} = g_E.\n        Surface acceleration due to gravity is identical to Earth's g.\n\n# Key Concepts Summary\n\n* **Projectile**: An object moving freely under gravity after launch. Examples: kicked football, fired bullet, golf ball trajectory, water hose stream.\n* **Projectile Components**: Horizontal component of velocity is constant; vertical component has constant downward acceleration g.\n* **Common Variable**: Time t connects both vertical and horizontal components of displacement.\n* **Horizontal Launch Formulas**:\n * \Delta y = \frac{1}{2} g t^2\n * \Delta x = v_{0x} t\n * v_y = g t\n * v_x = v_{0x}\n* **Inclined Launch Formulas**:\n * \Delta y = v_0 \sin(\theta) t + \frac{1}{2} g t^2\n * \Delta x = v_0 \cos(\theta) t\n * v_y = v_0 \sin(\theta) + g t\n * v_x = v_0 \cos(\theta)\n* **Rotational Dynamics Analogies**:\n * Rotational quantities \theta, \omega, \alpha$correlate directly to linear$s, v, a$via$s = r\theta$,$v = r\omega$,$a_t = r\alpha.\n * Every particle on a rigid body shares identical \Delta \theta, \omega, \alpha.\n * \vec{\omega}$direction determined by RHR;$\vec{\alpha}$parallel to$\vec{\omega} if accelerating, antiparallel if decelerating.\n* **Planetary & Gravitational Principles**:\n * Kepler's Laws apply universally to any satellite orbiting any celestial host body.\n * Speed varies along orbit: fastest at perihelion, slowest at aphelion.\n * Gravitational force provides necessary centripetal force for orbital revolution: F_g = \frac{G m_1 m_2}{r^2}.\n * Torque (\tau = r F \sin(\theta)$) causes rotational acceleration ($\tau = I \alpha).\n * Moment of inertia (I = \sum m_i r_i^2) measures resistance to rotational acceleration.\n\n# End of Unit Problems & Complete Solutions\n\n* **Problem 1 (Horizontal Throw from Building)**:\n * *Problem*: Ball thrown horizontally from top of 45\,\text{m} building. Find (a) time to reach ground, (b) horizontal displacement, (c) resultant velocity on impact.\n * *Solution (a)*: \Delta y = -45\,\text{m}$,$g = -10\,\text{m/s}^2$.$-45 = \frac{1}{2}(-10)t^2 \implies t^2 = 9 \implies t = 3.0\,\text{s}.\n * *Solution (b)*: Assuming launch speed v_{0x}$(if given e.g.,$v_{0x} = 20\,\text{m/s}$):$\Delta x = v_{0x} t = 20 \times 3 = 60\,\text{m}.\n * *Solution (c)*: v_x = v_{0x}$,$v_y = g t = -10 \times 3 = -30\,\text{m/s}$. Resultant$v = \sqrt{v_x^2 + v_y^2}.\n* **Problem 2 (Football Kicked at 30 Degrees)**:\n * *Problem*: Football kicked at 30^\circ$with$v_0 = 20\,\text{m/s}$,$g = 10\,\text{m/s}^2$. Calculate (a) initial components, (b) flight time, (c) range, (d) displacement at$t = 1.5\,\text{s}.\n * *Solution (a)*:\n        v_{0x} = 20 \cos(30^\circ) = 20 \times 0.866 = 17.32\,\text{m/s}\n        v_{0y} = 20 \sin(30^\circ) = 20 \times 0.5 = 10\,\text{m/s}\n * *Solution (b)*:\n        t_{\text{total}} = \frac{2 v_{0y}}{g} = \frac{2(10)}{10} = 2.0\,\text{s}\n * *Solution (c)*:\n        R = v_{0x} t_{\text{total}} = 17.32 \times 2.0 = 34.64\,\text{m}\n * *Solution (d)*: At t = 1.5\,\text{s},\n        \Delta x = 17.32 \times 1.5 = 25.98\,\text{m}\n        \Delta y = (10 \times 1.5) + \frac{1}{2}(-10)(1.5)^2 = 15 - 11.25 = 3.75\,\text{m}\n* **Problem 3 (Launch Speed vs Peak Speed)**:\n * *Problem*: Launch speed of projectile is five times its speed at maximum height (v_0 = 5 v_{\text{peak}}$). Find launch angle$\theta$.
  • Solution: At maximum height, speed is strictly horizontal: $v_{\text{peak}} = v_0 \cos(\theta)$.         $v_0 = 5 (v_0 \cos(\theta)) \implies 1 = 5 \cos(\theta) \implies \cos(\theta) = 0.2$$\theta = \cos^{-1}(0.2) \approx 78.46^\circ$
• Problem 4 (Volcanic Bombs Launch Speed):
  • Problem: Bomb ejected at $35^\circ$ to horizontal from hole A, lands at B at foot of volcano.
  • Analysis: Uses range $R$ and height difference $\Delta y$ between A and B to solve for initial speed $v_0$ via $\Delta y = \Delta x \tan(\theta) - \frac{g \Delta x^2}{2 v_0^2 \cos^2(\theta)}$, and total time $t = \frac{\Delta x}{v_0 \cos(\theta)}$.
• Problem 5 (Firefighter Water Stream):
  • Problem: Water fired from hose at distance $d$ at angle $\theta_i$ with speed $v_i$. At what height $h$ does water hit building?
  • Solution:         $t = \frac{d}{v_i \cos(\theta_i)}$$h = v_i \sin(\theta_i) t - \frac{1}{2} g t^2 = v_i \sin(\theta_i) \left(\frac{d}{v_i \cos(\theta_i)}\right) - \frac{1}{2} g \left(\frac{d}{v_i \cos(\theta_i)}\right)^2$$h = d \tan(\theta_i) - \frac{g d^2}{2 v_i^2 \cos^2(\theta_i)}$
• Problem 6 (Rotating Fan Blade Point):
  • Problem: Fan rotates at $1200\,\text{rpm}$. Blade tip radius $r = 0.15\,\text{m}$. Find (a) distance in 1 rev, (b) linear speed.
  • Solution (a): $\Delta s = 2\pi r = 2 \pi (0.15) = 0.942\,\text{m}$.
  • Solution (b): $\omega = \frac{1200 \times 2\pi}{60} = 40\pi \approx 125.66\,\text{rad/s}$. $v = \omega r = 125.66 \times 0.15 = 18.85\,\text{m/s}$.
• Problem 7 (Braking Car Wheel Acceleration):
  • Problem: Car speed $v = 80\,\text{km/h} = 22.22\,\text{m/s}$, wheel radius $r = 40\,\text{cm} = 0.4\,\text{m}$. Stops in $\Delta \theta = 30\,\text{rev} = 60\pi \approx 188.5\,\text{rad}$. Find \alpha$.\n * *Solution*: Initial \omega_0 = \frac{v}{r} = \frac{22.22}{0.4} = 55.55\,\text{rad/s}$, final$\omega_f = 0$.         $\omega_f^2 = \omega_0^2 + 2\alpha \Delta \theta \implies 0 = (55.55)^2 + 2 \alpha (188.5)$$3085.8 = -377 \alpha \implies \alpha = -8.18\,\text{rad/s}^2$
• Problem 8 (Grindstone Belt Rotation):
  • Problem: $\omega = 40\,\text{rad/s}$, time $t = 1\,\text{min} = 60\,\text{s}$. Find angle in (a) rad, (b) rev, (c) degrees.
  • Solution (a): $\Delta \theta = \omega t = 40 \times 60 = 2400\,\text{rad}$.
  • Solution (b): $\Delta \theta = \frac{2400}{2\pi} = 381.97\,\text{rev}$.
  • Solution (c): \Delta \theta = 381.97 \times 360^\circ = 137,510^\circ$.\n* **Problem 9 (Bicycle Travel Distance)**:\n * *Problem*: Wheel radius r = 0.50\,\text{m}$, completes$320\,\text{rotations}. Find distance.\n * *Solution*: \Delta \theta = 320 \times 2\pi = 2010.6\,\text{rad}$. Distance$\Delta s = r \Delta \theta = 0.50 \times 2010.6 = 1005.3\,\text{m}.\n* **Problem 10 (Spinning Wheel Braking)**:\n * *Problem*: Constant \alpha = -5.60\,\text{rad/s}^2$, in$t = 4.20\,\text{s}$, rotates$\Delta \theta = 62.4\,\text{rad}$. Find final angular speed$\omega_f$.
  • Solution:         $\Delta \theta = \omega_f t - \frac{1}{2} \alpha t^2$$62.4 = \omega_f (4.20) - \frac{1}{2} (-5.60) (4.20)^2$$62.4 = 4.20 \omega_f + 49.392 \implies 4.20 \omega_f = 13.008 \implies \omega_f = 3.10\,\text{rad/s}$
• Problem 11 (Hyperion Moon Orbital Period):
  • Problem: Titan: $R_T = 1.22 \times 10^9\,\text{m}$, $T_T = 15.95\,\text{days}$. Hyperion: $R_H = 1.48 \times 10^9\,\text{m}$. Find T_H$.\n * *Solution*:\n        \frac{T_H^2}{R_H^3} = \frac{T_T^2}{R_T^3} \implies T_H = T_T \left(\frac{R_H}{R_T}\right)^{3/2} = 15.95 \times \left(\frac{1.48 \times 10^9}{1.22 \times 10^9}\right)^{1.5}\n        T_H = 15.95 \times (1.2131)^{1.5} = 15.95 \times 1.3356 = 21.30\,\text{days}\n* **Problem 12 (Mercury Orbital Period)**:\n * *Problem*: Mercury radius R = 5.8 \times 10^{10}\,\text{m}$, Sun mass$M_s = 1.99 \times 10^{30}\,\text{kg}. Find period in Earth days.\n * *Solution*:\n        T^2 = \frac{4\pi^2 R^3}{G M_s} = \frac{4\pi^2 (5.8 \times 10^{10})^3}{(6.673 \times 10^{-11}) (1.99 \times 10^{30})} = \frac{7.703 \times 10^{33}}{1.328 \times 10^{20}} = 5.80 \times 10^{13}\,\text{s}^2\n        T = \sqrt{5.80 \times 10^{13}} = 7.616 \times 10^6\,\text{s}\n        T = \frac{7.616 \times 10^6}{86400} = 88.15\,\text{Earth days}\n* **Problem 13 (Gravitational Force Between Two Particles)**:\n * *Problem*: Two isolated 2.00\,\text{kg}$particles separated by$30.0\,\text{cm} = 0.30\,\text{m}. Find gravitational attraction.\n * *Solution*:\n        F_g = \frac{G m_1 m_2}{r^2} = \frac{6.673 \times 10^{-11} \times 2.00 \times 2.00}{(0.30)^2} = \frac{2.669 \times 10^{-10}}{0.09} = 2.97 \times 10^{-9}\,\text{N}$$