Two-Dimensional Kinematics: Motion in a Plane, Projectile Motion, and Uniform Circular Motion

Scope and Overview of Two-Dimensional Kinematics

• Two-dimensional kinematics describes object motion taking place within a two-dimensional coordinate plane ($x$-$y$ plane).
• Core topics in two-dimensional kinematics include:
  • Velocity vectors in a plane
  • Acceleration vectors in a plane
  • Projectile motion
  • Uniform circular motion
  • Systematic strategies and numerical calculations for two-dimensional problem solving

Velocity Vectors in a Two-Dimensional Plane

• Representation of Position and Velocity Vectors:
  • Positions in a two-dimensional Cartesian coordinate system are designated by coordinate pairs $(x, y)$.
  • Position vectors can be expressed in terms of Cartesian components $(x, y)$ or in terms of magnitude and angle $\theta$.
  • The distance of a point $P$ from the coordinate origin defines the magnitude of vector $\mathbf{r}$.
• Average Velocity:
  • Defined as the displacement vector $\Delta \mathbf{r}$ divided by the elapsed time interval $\Delta t$:     $\mathbf{v}_{\text{av}} = \frac{\Delta \mathbf{r}}{\Delta t}$
  • For a particle moving from position $P_1$ to position $P_2$ during time interval $\Delta t$, the average velocity vector points in the direction of the displacement vector $\Delta \mathbf{r}$.
• Instantaneous Velocity:
  • Defined as the limiting value of the average velocity as the time interval $\Delta t$ approaches zero:     $\mathbf{v} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \mathbf{r}}{\Delta t}$
  • The instantaneous velocity vector $\mathbf{v}$ is always tangent to the trajectory path at point $(x, y)$.
  • The velocity vector $\mathbf{v}$ consists of horizontal ($x$-axis) component $v_x$ and vertical ($y$-axis) component $v_y$.

Acceleration Vectors in a Two-Dimensional Plane

• Mechanisms of Two-Dimensional Acceleration:
  • Acceleration occurs whenever there is a change in velocity magnitude (speed), a change in velocity direction, or simultaneous changes in both magnitude and direction.
• Average Acceleration:
  • Defined as the change in velocity $\Delta \mathbf{v}$ divided by time interval $\Delta t$:     $\mathbf{a}_{\text{av}} = \frac{\Delta \mathbf{v}}{\Delta t}$
  • Component formulation of average acceleration:     $\Delta v_x = v_{2x} - v_{1x}$$\Delta v_y = v_{2y} - v_{1y}$$a_{\text{av}, x} = \frac{\Delta v_x}{\Delta t}$$a_{\text{av}, y} = \frac{\Delta v_y}{\Delta t}$
  • Magnitude of average acceleration:     $a_{\text{av}} = \sqrt{a_{\text{av}, x}^2 + a_{\text{av}, y}^2}$
  • Direction angle $\theta$ of average acceleration:     $\theta = \tan^{-1}\left(\frac{\Delta v_y}{\Delta v_x}\right) = \tan^{-1}\left(\frac{a_{\text{av}, y}}{a_{\text{av}, x}}\right)$
• Instantaneous Acceleration:
  • Defined as the limit of average acceleration as $\Delta t$ approaches zero:     $\mathbf{a} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \mathbf{v}}{\Delta t}$
  • To determine instantaneous acceleration $\mathbf{a}$ at point $P_1$, the limit of $\mathbf{a}_{\text{av}}$ is taken as $P_2$ approaches $P_1$, forcing $\Delta \mathbf{v}$ and $\Delta t$ toward zero.
  • Magnitude of instantaneous acceleration:     $a = \sqrt{a_x^2 + a_y^2}$
  • Direction angle $\theta$ of instantaneous acceleration:     $\theta = \tan^{-1}\left(\frac{a_y}{a_x}\right)$
  • Geometric orientation: The acceleration vector $\mathbf{a}$ must always point toward the concave (inner) side of the curved trajectory path.
• Parallel versus Perpendicular Acceleration Components:
  • Acceleration parallel to velocity ($\mathbf{a} \parallel \mathbf{v}$):
  • Changes the magnitude (speed) of velocity, but does not alter its direction.
  • Results in straight-line motion with changing speed.
  • Acceleration perpendicular to velocity ($\mathbf{a} \perp \mathbf{v}$):
  • Changes the direction of velocity, but does not alter its magnitude (speed).
  • Results in motion along a curved path at constant speed.

Conceptual Analysis of Acceleration and Kinematics

• Motion Diagram Analysis (Particle Acceleration Between Two Points):
  • Question: In a segment of a motion diagram showing two successive positions of a particle with initial velocity $\mathbf{v}_i$ and final velocity $\mathbf{v}_f$, which acceleration vector best represents the acceleration?
  • Answer: Vector D, which points toward the concave interior of the path in the direction of $\Delta \mathbf{v} = \mathbf{v}_f - \mathbf{v}_i$
• Acceleration Vector of a Projectile:
  • Question: What is the orientation of the acceleration vector of a particle in projectile motion?
  • Options:
  1. Points along the path of the particle
  2. Is directed horizontally
  3. Vanishes at the particle's highest point
  4. Is directed down at all times
  5. Is zero
  • Answer: Option 4 — Is directed down at all times (with constant magnitude $g$).
• Time of Flight for Horizontal Launch:
  • Question: A ball is thrown horizontally off a roof. Assuming ideal projectile motion, what determines the time until it lands?
  • Options:
  1. Its initial speed and the horizontal distance to the landing point
  2. The height of the roof and its initial speed
  3. The height of the roof
  • Answer: Option 3 — The height of the roof.
• Impact Speed of Rocks Thrown Off a Bridge:
  • Question: Two rocks are thrown off a bridge with identical initial speeds. Rock 1 is thrown at an angle $30^\circ$ below horizontal. Rock 2 is thrown at an angle $30^\circ$ above horizontal. How do their speeds compare when they hit the water?
  • Options:
  1. Friend's rock ($30^\circ$ below) is moving faster
  2. Your rock ($30^\circ$ above) is moving faster
  3. They are moving at the same speed
  • Answer: Option 3 — They are moving at the same speed.

Principles and Equations of Projectile Motion

• Definition and Trajectory of a Projectile:

  • A projectile is an object moving in two dimensions under the exclusive influence of Earth's gravity.
  • In the absence of air resistance, a projectile follows a parabolic trajectory through the $x$-$y$ plane.
  • Common real-world examples include tossed objects, baseballs, and footballs.

• Independence of Horizontal and Vertical Motions:

  • Motion along the $x$-axis and $y$-axis are entirely independent of one another.
  • Vertical displacement spaces out identically for free-fall motion and projectile motion over equal time intervals.
  • Horizontal motion ($x$-direction):
  • Zero horizontal acceleration ($a_x = 0$).
  • Constant horizontal velocity ($v_x = v_{0x}$).
  • Covers equal horizontal distances in equal time intervals.
  • Vertical motion ($y$-direction):
  • Constant vertical downward acceleration ($a_y = -g$).
  • Acceleration due to gravity magnitude $g = 9.8\,\text{m/s}^2$ (or $32\,\text{ft/s}^2$ in English units).
  • Vertical velocity changes by equal amounts in equal time intervals.
  • Peak Trajectory Characteristics:
  • At maximum height (the apex), vertical velocity is zero ($v_y = 0$).
  • At maximum height, vertical acceleration remains strictly $-g$ ($a_y = -9.8\,\text{m/s}^2$).

• Equations of Motion for Projectile Kinematics:

• System of Kinematic Equations:
  • Horizontal ($x$-direction motion):     $v_x = v_{0x}$$x = v_x t$$x = v_{0x} t$$v_x^2 = v_{0x}^2$
  • Vertical ($y$-direction motion):     $v_y = v_{0y} - gt$$y = \frac{1}{2}(v_y + v_{0y})t$$y = v_{0y}t - \frac{1}{2}gt^2$$v_y^2 = v_{0y}^2 - 2gy$
• Determination of Vector Components and Launch Parameters:
  • Initial velocity vector:     $\mathbf{v}_0 = \mathbf{v}_{0x} + \mathbf{v}_{0y}$
  • Horizontal component of initial velocity:     $v_{0x} = v_0 \cos(\theta_0)$
  • Vertical component of initial velocity:     $v_{0y} = v_0 \sin(\theta_0)$
  • Initial speed (magnitude):     $v_0 = \sqrt{v_{0x}^2 + v_{0y}^2}$
  • Launch angle relative to horizontal:     $\tan(\theta_0) = \frac{v_{0y}}{v_{0x}}$$\theta_0 = \tan^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)$

Mathematical Proof of Parabolic Trajectory

• Derivation of Path Function $y(x)$:
  • To demonstrate that projectile motion is parabolic, vertical coordinate $y$ is expressed as a direct mathematical function of horizontal coordinate $x$.
  • Express time $t$ from horizontal kinematics $x = v_{0x}t$:     $t = \frac{x}{v_{0x}}$
  • Substitute $t$ into vertical kinematics $y = v_{0y}t - \frac{1}{2}gt^2$:     $y = v_{0y}\left(\frac{x}{v_{0x}}\right) - \frac{1}{2}g\left(\frac{x}{v_{0x}}\right)^2$
  • Substitute $\frac{v_{0y}}{v_{0x}} = \tan(\theta_0)$, yielding:     $y = \tan(\theta_0)x - \left(\frac{g}{2 v_{0x}^2}\right) x^2$
  • This relation has the form $y = Ax - Bx^2$, which defines a mathematical parabola.

Systematic Problem-Solving Protocol for Projectile Motion

• Step 1: Read the problem carefully and clearly identify the target object(s) to analyze.
• Step 2: Construct a comprehensive physical diagram of the motion.
• Step 3: Establish an origin $(0,0)$ and choose a coordinate system.
• Step 4: Determine the time interval $t$; this interval is identical for horizontal and vertical directions and only covers the period during which the object undergoes free acceleration $g$
• Step 5: Separate and analyze horizontal ($x$) and vertical ($y$) motions independently.
• Step 6: Formulate an inventory of known and unknown variables. Incorporate known conditions: $v_x$ remains constant throughout flight, and $v_y = 0$ at maximum height.
• Step 7: Select appropriate horizontal and vertical kinematic equations, combining them as needed to solve for unknown variables.

Kinematics of Uniform Circular Motion

• Definition of Uniform Circular Motion:
  • Motion in a circle of constant radius $r$ at a constant speed $v$
  • Instantaneous velocity vector $\mathbf{v}$ is always tangent to the circular path at every point.
• Centripetal Acceleration:
  • The acceleration in uniform circular motion is termed centripetal acceleration or radial acceleration ($\mathbf{a}_c$).
  • Centripetal acceleration points directly toward the center of the circular path at all times.
  • Magnitude of centripetal acceleration:     $a_c = \frac{v^2}{r}$
• Acceleration Vector Orientation in Uniform Circular Motion:
  • Question: What is the direction of the acceleration vector of a particle in uniform circular motion?
  • Options:
  1. Points tangent to the circle, in the direction of motion
  2. Points tangent to the circle, opposite the direction of motion
  3. Is zero
  4. Points toward the center of the circle
  5. Points outward from the center of the circle
  • Answer: Option 4 — Points toward the center of the circle.

Numerical Examples and Quantitative Problem Solutions

• Fastball Drop Distance (Problem #7):

  • Problem: A baseball pitcher throws a fastball horizontally at a speed of $42\,\text{m/s}$. Ignoring air resistance, how far does the ball drop between the pitcher's mound and home plate, located $60\,\text{ft}\,6\,\text{in}$ ($18.4\,\text{m}$) away?
  • Given Data:
  • Initial horizontal speed: $v_{0x} = 42\,\text{m/s}$
  • Initial vertical speed: $v_{0y} = 0\,\text{m/s}$
  • Distance to home plate: $x = 18.4\,\text{m}$
  • Calculation:     $t = \frac{x}{v_{0x}} = \frac{18.4\,\text{m}}{42\,\text{m/s}} \approx 0.438\,\text{s}$$y = -\frac{1}{2}gt^2 = -\frac{1}{2}(9.8\,\text{m/s}^2)(0.438\,\text{s})^2 \approx -0.945\,\text{m}$
  • Result: The ball drops $0.945\,\text{m}$.

• Trajectory Properties of Fired Projectile (Problem #8):

  • Problem: A projectile is fired from ground level at an angle of $55^\circ$ above horizontal with an initial speed of $31\,\text{m/s}$.
  • Results and Answers:
  • Magnitude of instantaneous velocity at moment of launch: $31\,\text{m/s}$
  • Direction of instantaneous velocity at moment of launch: $55^\circ$ above horizontal
  • Magnitude of instantaneous velocity at maximum height:       $v_x = v_0 \cos(55^\circ) = 31 \times \cos(55^\circ) \approx 18\,\text{m/s}$
  • Direction of velocity at maximum height: $0^\circ$ (horizontal)
  • Magnitude of instantaneous velocity the moment before hitting ground: $31\,\text{m/s}$
  • Direction of instantaneous velocity the moment before hitting ground: $55^\circ$ below horizontal

• Cliff Diver Clearing Ledge (Problem #12):

  • Problem: A $510\,\text{N}$ swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the cliff top to miss a ledge at the bottom that is $1.75\,\text{m}$ wide and $9.00\,\text{m}$ below the cliff top?
  • Given Data:
  • Vertical drop: $y = -9.00\,\text{m}$
  • Horizontal distance: $x = 1.75\,\text{m}$
  • Initial vertical speed: $v_{0y} = 0\,\text{m/s}$
  • Calculation:     $y = -\frac{1}{2}gt^2 \implies -9.00 = -\frac{1}{2}(9.8)t^2 \implies t^2 = \frac{18.0}{9.8} \approx 1.837$$t = \sqrt{1.837} \approx 1.355\,\text{s}$$v_0 = \frac{x}{t} = \frac{1.75\,\text{m}}{1.355\,\text{s}} \approx 1.29\,\text{m/s}$
  • Result: Minimum horizontal speed required is $1.29\,\text{m/s}$.

• Stone Swung in Circular Path (Problem #28):

  • Problem: A $2.2\,\text{kg}$ stone is swung in a circle of radius $75\,\text{cm}$ ($0.75\,\text{m}$). At what speed should it be swung so that its centripetal acceleration equals $9.8\,\text{m/s}^2$ ($1\,g$)?
  • Given Data:
  • Mass: $m = 2.2\,\text{kg}$
  • Radius: $r = 0.75\,\text{m}$
  • Acceleration: $a_c = 9.8\,\text{m/s}^2$
  • Calculation:     $a_c = \frac{v^2}{r} \implies 9.8 = \frac{v^2}{0.75}$$v^2 = 9.8 \times 0.75 = 7.35$$v = \sqrt{7.35} \approx 2.7\,\text{m/s}$
  • Result: Required speed is $2.7\,\text{m/s}$.

• Football Trajectory Calculations (Problem #14):

  • Problem: A football is thrown with an initial upward velocity component $v_{0y} = 15.0\,\text{m/s}$ and horizontal velocity component $v_{0x} = 18.0\,\text{m/s}$.
  • Results and Step-by-Step Solutions:
  • (a) Time required to reach highest point in trajectory:       $v_y = v_{0y} - gt = 0 \implies t = \frac{15.0\,\text{m/s}}{9.8\,\text{m/s}^2} \approx 1.53\,\text{s}$
  • (b) Maximum height above release point:       $y = v_{0y}t - \frac{1}{2}gt^2 = (15.0)(1.53) - \frac{1}{2}(9.8)(1.53)^2 \approx 11.5\,\text{m}$
  • (c) Total time to return to original height:       $t_{\text{total}} = 2 \times 1.53\,\text{s} = 3.06\,\text{s}$       Comparison: Exactly twice the time calculated in part (b) ($1.53\,\text{s}$), reflecting path symmetry.
  • (d) Total horizontal distance traveled:       $x = v_{0x} t_{\text{total}} = (18.0\,\text{m/s})(3.06\,\text{s}) = 55.1\,\text{m}$

• Batted Baseball Flight Parameters (Problem #19):

  • Problem: A batted baseball leaves the bat at an angle $30.0^\circ$ above horizontal and is caught by an outfielder $375\,\text{ft}$ from home plate at bat height. Use g = 32\,\text{ft/s}^2$.\n * Results and Step-by-Step Solutions:\n * (a) Initial speed of the ball (v_0):\n      R = \frac{v_0^2 \sin(2\theta_0)}{g} \implies 375 = \frac{v_0^2 \sin(60.0^\circ)}{32}\n      375 = \frac{v_0^2 \times 0.866}{32} \implies v_0^2 = \frac{375 \times 32}{0.866} \approx 13856.8\n      v_0 \approx 118\,\text{ft/sec}\n * (b) Maximum height reached above bat impact point:\n      v_{0y} = 118 \times \sin(30.0^\circ) = 59\,\text{ft/s}\n      y_{\text{max}} = \frac{v_{0y}^2}{2g} = \frac{59^2}{2 \times 32} = \frac{3481}{64} \approx 54.4\,\text{ft}$$