University Physics: Motion in Two Dimensions (Kinematics, Projectiles, Circular Motion, and Relative Velocity)

Kinematics in Two Dimensions

• The study of kinematics in two dimensions expands upon one-dimensional motion by considering the vector nature of position, velocity, and acceleration.

• Key focus areas include projectile motion and uniform circular motion, which are treated as special cases of two-dimensional motion.

• The chapter also introduces the concept of relative motion between different frames of reference.

Position, Velocity, and Acceleration Vectors

• Position and Displacement: Displacement in two dimensions is defined as the change in the position vector:   $\Delta \vec{r} \equiv \vec{r}_f - \vec{r}_i$

• Average Velocity: The average velocity of a particle is the ratio of its displacement to the time interval over which that displacement occurs:   $\vec{v}_{avg} \equiv \frac{\Delta \vec{r}}{\Delta t} = \frac{\vec{r}_f - \vec{r}_i}{t_f - t_i}$

• Instantaneous Velocity: This is the limit of the average velocity as the time interval approaches zero. It is equal to the derivative of the position vector with respect to time:   $\vec{v} \equiv \lim_{\Delta t\rightarrow 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}$

• Average Acceleration: The average acceleration of a particle is defined as the change in its instantaneous velocity vector divided by the time interval:   $\vec{a}_{avg} \equiv \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_f - \vec{v}_i}{t_f - t_i}$

• Instantaneous Acceleration: This is the limit of the average acceleration as the time interval approaches zero. It is equal to the derivative of the velocity vector with respect to time:   $\vec{a} \equiv \lim_{\Delta t\rightarrow 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt}$

Kinematic Equations for Two-Dimensional Motion

• When motion in two dimensions occurs with a constant acceleration ($\vec{a}$), the motion can be broken down into $x$ and $y$ components.

• Position Vector: $\vec{r} = x\hat{i} + y\hat{j}$

• Velocity Vector: $\vec{v} = \frac{d\vec{r}}{dt} = v_x\hat{i} + v_y\hat{j}$

• Velocity at Time $t$: $\vec{v}_f = \vec{v}_i + \vec{a}t$

• Position at Time $t$: $\vec{r}_f = \vec{r}_i + \vec{v}_i t + \frac{1}{2} \vec{a} t^2$

  • This can be decomposed into:

  • $x_f = x_i + v_{xi} t + \frac{1}{2} a_x t^2$

  • $y_f = y_i + v_{yi} t + \frac{1}{2} a_y t^2$

Projectile Motion Principles

• Projectile motion is a specific form of 2D motion where an object is launched into the air and moves in both $x$ and $y$ directions simultaneously.

• Fundamental Assumptions:

  • The free-fall acceleration ($\vec{g}$) is constant throughout the motion and directed downward.

  • This assumption implies a "flat Earth" model, which is accurate for ranges small compared to the Earth's radius.

  • Air friction is considered negligible.

• Characteristics of the Path:

  • The path of a projectile is a parabola, referred to as its trajectory.

  • The $x$-component of velocity ($v_x$) remains constant because there is no acceleration in the horizontal direction ($a_x = 0$).

  • The $y$-component of velocity is zero at the peak of the path.

Range and Maximum Height of a Projectile

For symmetric projectile motion (where the launch and landing heights are the same):

• Maximum Height ($h$): The highest vertical distance reached by the projectile:   $h = \frac{v_i^2 \sin^2(\theta_i)}{2g}$

• Horizontal Range ($R$): The total horizontal distance traveled:   $R = \frac{v_i^2 \sin(2\theta_i)}{g}$

• Trajectory Specifics:

  • These equations are valid only for symmetric trajectories.

  • A projectile reaches its maximum range at a launch angle of $45^\circ$.

  • Complementary values of the initial angle (e.g., $30^\circ$ and $60^\circ$, or $15^\circ$ and $75^\circ$) result in the same horizontal range, though they reach different maximum heights.

Uniform Circular Motion

• Definition: Occurs when an object moves in a circular path with a constant speed ($v$).

• Centripetal Acceleration ($a_c$):

  • Even if speed is constant, the velocity vector is constantly changing because the direction is changing.

  • The acceleration resulting from this change is directed toward the center of the circle.

  • Formula: $a_c = \frac{v^2}{r}$

  • The SI units are $m/s^2$.

• Period ($T$):

  • The period is the time required for one full revolution around the circle.

  • Since speed is distance divided by time, and distance for one revolution is the circumference ($2\pi r$):     $T = \frac{2\pi r}{v}$

Tangential and Radial Acceleration

• For non-uniform circular motion (where speed is not constant), there are two components of acceleration:

  • Radial Acceleration ($a_r$): directed toward the center of the circle, changing the direction of the velocity.

  • Tangential Acceleration ($a_t$): directed tangent to the path, changing the magnitude of the velocity (speed).

• Total Acceleration:

  • The total acceleration vector is $\vec{a} = \vec{a}_r + \vec{a}_t$

  • The magnitude of the total acceleration is: $a = \sqrt{a_r^2 + a_t^2}$

Relative Velocity and Galilean Transformations

• Measurements of position and velocity depend on the frame of reference of the observer.

• Example: If Observer A is stationary and Observer B is moving, they will record different coordinates for the same point $P$.

• Position Transformation: If frame $S_B$ moves with velocity $\vec{v}_{BA}$ relative to frame $S_A$, the positions are related by:   $\vec{r}_{PA} = \vec{r}_{PB} + \vec{v}_{BA}t$

• Velocity Transformation (Galilean Transformation Equations):

  • $\vec{u}_{PA} = \vec{u}_{PB} + \vec{v}_{BA}$

  • Here, $\vec{u}{PA}$ is the velocity of particle $P$ as measured by observer $A$, and $\vec{u}{PB}$ is the velocity measured by observer $B$.

• Acceleration in Relative Motion:

  • If the relative velocity between frames ($\vec{v}{BA}$) is constant, then its derivative is zero ($d\vec{v}{BA}/dt = 0$).

  • Consequently, $\vec{a}_{PA} = \vec{a}_{PB}$.

  • The acceleration of a particle is measured to be the same by any two observers moving at a constant velocity relative to each other.

Example 4.1: Motion in a Plane

Problem: A particle moves in the $xy$ plane, starting from origin ($t=0$) with $v_{xi} = 20\,m/s$, $v_{yi} = -15\,m/s$, and $a_x = 4.0\,m/s^2$ (with $a_y = 0$).

• (A) Total velocity vector at any time:

  • $\vec{v}_f = (v_{xi} + a_x t)\hat{i} + (v_{yi} + a_y t)\hat{j}$

  • $\vec{v}_f = [20 + (4.0)t]\hat{i} + [-15 + (0)t]\hat{j}$

  • $\vec{v}_f = [(20 + 4.0t)\hat{i} - 15\hat{j}]\,m/s$

• (B) Velocity, speed, and angle at $t = 5.0\,s$:

  • $\vec{v}_f = (40\hat{i} - 15\hat{j})\,m/s$

  • Angle: $\theta = \tan^{-1}\left(\frac{v_{yf}}{v_{xf}}\right) = \tan^{-1}\left(\frac{-15}{40}\right) = -21^\circ$

  • Speed: $v_f = |\vec{v}_f| = \sqrt{40^2 + (-15)^2} = 43\,m/s$

• (C) Position at any time $t$:

  • $x_f = v_{xi} t + \frac{1}{2} a_x t^2 = 20t + 2.0t^2$

  • $y_f = v_{yi} t = -15t$

  • $\vec{r}_f = (20t + 2.0t^2)\hat{i} - 15t\hat{j}$

Example 4.2: The Long Jump

Problem: A jumper leaves the ground at $\theta = 20.0^\circ$ above horizontal at a speed of $11.0\,m/s$.

• (A) Horizontal distance ($R$):

  • $R = \frac{v_i^2 \sin(2\theta_i)}{g} = \frac{(11.0\,m/s)^2 \sin(40.0^\circ)}{9.80\,m/s^2} = 7.94\,m$

• (B) Maximum height ($h$):

  • $h = \frac{v_i^2 \sin^2(\theta_i)}{2g} = \frac{(11.0\,m/s)^2 (\sin(20.0^\circ))^2}{2(9.80\,m/s^2)} = 0.722\,m$

Example 4.4: Stone Thrown from a Building

Problem: Stone thrown from $45.0\,m$ high building at $v_i = 20.0\,m/s$ and $\theta = 30.0^\circ$ above horizontal.

• (A) Time to reach the ground:

  • $v_{xi} = 20.0 \cos(30.0^\circ) = 17.3\,m/s$

  • $v_{yi} = 20.0 \sin(30.0^\circ) = 10.0\,m/s$

  • Vertical position equation: $y_f = y_i + v_{yi} t + \frac{1}{2} a_y t^2$

  • $-45.0\,m = 0 + (10.0\,m/s)t + \frac{1}{2} (-9.80\,m/s^2)t^2$

  • Solving the quadratic for $t$ gives $t = 4.22\,s$.

• (B) Speed just before striking ground:

  • $v_{yf} = v_{yi} + a_y t = 10.0\,m/s + (-9.80\,m/s^2)(4.22\,s) = -31.3\,m/s$

  • $v_{xf} = v_{xi} = 17.3\,m/s$

  • $v_f = \sqrt{(17.3)^2 + (-31.3)^2} = 35.8\,m/s$

Example: Car on a Circular Rise

Problem: A car accelerates at $a_t = 0.300\,m/s^2$ parallel to the road. It passes a circular rise of radius $r = 500\,m$ at speed $v = 6.00\,m/s$.

• Radial Acceleration:

  • $a_r = -\frac{v^2}{r} = -\frac{(6.00\,m/s)^2}{500\,m} = -0.0720\,m/s^2$ (directed downward).

• Magnitude of total acceleration ($a$):

  • $a = \sqrt{a_r^2 + a_t^2} = \sqrt{(-0.0720)^2 + (0.300)^2} = 0.309\,m/s^2$

• Angle ($\phi$) relative to horizontal:

  • $\phi = \tan^{-1}\left(\frac{a_r}{a_t}\right) = \tan^{-1}\left(\frac{-0.0720}{0.300}\right) = -13.5^\circ$

Example 4.8: A Boat Crossing a River

Problem: Boat speed relative to water ($v_{br}$) is $10.0\,km/h$. River speed relative to Earth ($v_{rE}$) is $5.00\,km/h$ east.

• (A) Boat heads due north relative to water:

  • Velocity relative to Earth: $\vec{v}_{bE} = \vec{v}_{br} + \vec{v}_{rE}$

  • Speed: $v_{bE} = \sqrt{(10.0)^2 + (5.00)^2} = 11.2\,km/h$

  • Angle: $\theta = \tan^{-1}\left(\frac{5.00}{10.0}\right) = 26.6^\circ$ east of north.

• (B) Boat must travel due north relative to Earth:

  • The resultant velocity $v_{bE}$ must be vertical (north).

  • The boat must head upstream (west of north).

  • $\vec{v}_{bE}$ is a leg of the triangle, $\vec{v}_{br}$ is the hypotenuse.

  • $v_{bE} = \sqrt{v_{br}^2 - v_{rE}^2} = \sqrt{(10.0)^2 - (5.00)^2} = 8.66\,km/h$

  • Heading direction: $\tan^{-1}\left(\frac{v_{rE}}{v_{bE}}\right) = \tan^{-1}\left(\frac{5.00}{8.66}\right) = 30.0^\circ$ west of north.

Example: Rain and the Bicycle

Problem: Rain falls vertically at $35\,m/s$. A woman rides a bicycle at $12\,m/s$ east to west.

• Relative Velocity:

  • Magnitude: $v_{rel} = \sqrt{(12)^2 + (35)^2} = 37\,m/s$

  • Direction: $\theta = \tan^{-1}\left(\frac{12}{35}\right) = 18.92^\circ$ from the vertical.

  • The woman should hold the umbrella at $18.92^\circ$ to the vertical in the forward direction.