Motion in Two Dimensions and Projectile Motion: Study Notes for PHYSICS 371

Motion in Two Dimensions

Independence of Motion in x and y Directions

In two-dimensional motion, the movement along the x-axis is completely independent from the movement along the y-axis. This principle allows us to analyze each direction separately when solving problems.

• Key Point 1: The equations of motion for the x and y directions are solved independently.

• Key Point 2: Each direction may have different initial velocities and accelerations.

• Example: A projectile launched horizontally will have constant velocity in the x-direction and constant acceleration in the y-direction due to gravity.

Kinematic Variables and Equations

To analyze motion, we define variables for each direction and use kinematic equations.

x-direction	y-direction
$x_i$ = initial position	$y_i$ = initial position
$x_f$ = final position	$y_f$ = final position
$v_{ix}$ = initial velocity	$v_{iy}$ = initial velocity
$v_{fx}$ = final velocity	$v_{fy}$ = final velocity
$a_x$ = acceleration	$a_y$ = acceleration
$t$ = time	$t$ = time

• Kinematic Equations:

For constant acceleration:

• $v_{fx} = v_{ix} + a_x t$

• $x_f = x_i + v_{ix} t + rac{1}{2} a_x t^2$

• $v_{fy} = v_{iy} + a_y t$

• $y_f = y_i + v_{iy} t + rac{1}{2} a_y t^2$

Projectile Motion

Characteristics of Projectile Motion

Projectile motion refers to the motion of an object that is projected into the air and is subject only to gravity (neglecting air resistance).

• Key Point 1: Vertical acceleration is always $9.8 ext{ m/s}^2$ downward for the entire trip.

• Key Point 2: Horizontal acceleration is zero ($a_x = 0$).

• Key Point 3: If an object rolls off a horizontal surface, its initial vertical velocity is zero ($v_{iy} = 0$).

• Key Point 4: If an object is dropped, $v_{iy} = 0$; if thrown, $v_{iy} eq 0$.

• Example: A ball rolling off a table has $v_{ix} eq 0$, $v_{iy} = 0$, $a_x = 0$, $a_y = -9.8 ext{ m/s}^2$.

Sample Problem: Ball Rolling Off a Table

A small ball rolls horizontally off the edge of a tabletop that is 1.20 m high. It strikes the floor at a point 1.47 m horizontally away from the edge of the table. How fast was the ball moving when it left the table? Neglect air resistance.

$x_i$	$y_i$
0	1.20 m
$x_f$	$y_f$
1.47 m	0
$v_{ix}$	$v_{iy}$
?	0
$v_{fx}$	$v_{fy}$
$v_{ix}$	$v_{fy}$
$a_x$	$a_y$
0	-9.8 m/s2
$t$	$t$
?	?

• Solution Steps:

  1. Find time to fall using $y$-direction: $y_f = y_i + v_{iy} t + rac{1}{2} a_y t^2$

  2. Use $x$-direction to find $v_{ix}$: $x_f = x_i + v_{ix} t$

Trigonometry and Vectors

Trigonometric Relationships

Trigonometry is essential for resolving vectors into components.

• Key Point 1: $ext{Sin} heta = rac{ ext{opp}}{ ext{hyp}}$

• Key Point 2: $ext{Cos} heta = rac{ ext{adj}}{ ext{hyp}}$

• Key Point 3: $ext{Tan} heta = rac{ ext{opp}}{ ext{adj}}$

• Example: For a velocity vector at angle $heta$, $v_x = v ext{Cos} heta$, $v_y = v ext{Sin} heta$.

Vector Components and Magnitude

Vectors can be decomposed into components and recombined using the Pythagorean theorem.

• Key Point 1: $v^2 = v_x^2 + v_y^2 + v_z^2$

• Key Point 2: $|v| = ext{sqrt}(v_x^2 + v_y^2 + v_z^2)$

• Key Point 3: $heta = ext{Sin}^{-1} igg( rac{v_y}{v} igg )$

Projectile at an Angle

Solving for Range and Final Speed

When a projectile is launched at an angle, its initial velocity must be resolved into horizontal and vertical components.

• Key Point 1: $v_{ix} = v_i ext{Cos} heta$

• Key Point 2: $v_{iy} = v_i ext{Sin} heta$

• Key Point 3: The range is found using $x_f = x_i + v_{ix} t$ where $t$ is the total time of flight.

• Example: A ball is shot upward at an angle of $23^ ext{o}$ and $3 ext{ m/s}$. Find how far away it lands and its final speed.

$x_i$	$y_i$
0	0
$x_f$	$y_f$
?	0
$v_{ix}$	$v_{iy}$
$v_i ext{Cos} 23^ ext{o}$	$v_i ext{Sin} 23^ ext{o}$
$v_{fx}$	$v_{fy}$
$v_{ix}$	$v_{fy}$
$a_x$	$a_y$
0	-9.8 m/s2
$t$	$t$
?	?

Acceleration Components

Parallel and Perpendicular Components

Acceleration can be decomposed into components parallel and perpendicular to the direction of motion.

• Key Point 1: The parallel component changes the speed.

• Key Point 2: The perpendicular component changes the direction of motion.

• Example: In circular motion, the perpendicular (radial) component is called centripetal acceleration.

Circular Motion

Uniform Circular Motion and Centripetal Acceleration

When an object moves in a circle at constant speed, it experiences centripetal acceleration directed toward the center of the circle.

• Key Point 1: Centripetal acceleration formula: $a = rac{v^2}{r}$

• Key Point 2: The speed remains constant, but the direction changes continuously.

• Key Point 3: The force causing centripetal acceleration is always perpendicular to the velocity.

Period and Frequency

The period and frequency describe how often an object completes a revolution in circular motion.

• Key Point 1: Period (T): Time for one revolution (seconds).

• Key Point 2: Frequency (f): Number of revolutions per second (hertz).

• Key Point 3: $f = rac{1}{T}$

• Key Point 4: Circumference of a circle: $2 ext{π}r$

Calculus in Kinematics

Velocity and Acceleration as Derivatives

Calculus provides a precise way to define instantaneous velocity and acceleration as derivatives of position and velocity, respectively.

• Key Point 1: Instantaneous velocity: $v = rac{dx}{dt}$

• Key Point 2: Instantaneous acceleration: $a = rac{dv}{dt}$

• Key Point 3: The slope of a position vs. time graph gives velocity; the slope of a velocity vs. time graph gives acceleration.

• Example: If $x(t) = 3t^2 + 2t + 6$, then $v(t) = 6t + 2$, $a(t) = 6$.

Derivatives of Polynomial Functions

For polynomial functions, the derivative follows the power rule.

• Key Point 1: $rac{d}{dt} (t^n) = n t^{n-1}$

• Key Point 2: If $x = t^2$, then $v = 2t$; if $v = t^2$, then $a = 2t$.

• Key Point 3: Constants are preserved in differentiation.

Relative Motion

Relative Position, Velocity, and Acceleration

Relative motion compares the motion of one object to another, often using the ground as a reference.

• Key Point 1: Relative position: $x_{1,2} = x_1 - x_2$

• Key Point 2: Relative velocity: $v_{1,2} = v_1 - v_2$

• Key Point 3: Relative acceleration: $a_{1,2} = a_1 - a_2$

• Example: If two cars move in the same direction, their relative velocity is the difference of their velocities.

Kinematic Equations for Relative Motion

The same kinematic equations apply, but with relative variables.

• Key Point 1: $v_{1,2} = v_{1,2,0} + a_{1,2} t$

• Key Point 2: $x_{1,2} = x_{1,2,0} + v_{1,2,0} t + rac{1}{2} a_{1,2} t^2$

Useful Relationships and Law of Cosines

Law of Cosines

The Law of Cosines is used to relate the sides and angles of a triangle, often in vector addition problems.

• Key Point 1: $c^2 = a^2 + b^2 - 2ab ext{Cos} heta$

• Example: Used to find the magnitude of the resultant vector when two vectors are not perpendicular.

Summary Table: Ball Rolling Off a Table

Variable	x-direction	y-direction
Initial position	$x_i = 0$	$y_i = 1.20$ m
Final position	$x_f = 1.47$ m	$y_f = 0$
Initial velocity	$v_{ix}$ (unknown)	$v_{iy} = 0$
Final velocity	$v_{fx} = v_{ix}$	$v_{fy}$
Acceleration	$a_x = 0$	$a_y = -9.8$ m/s2
Time	$t$	$t$

Additional info: Some context and equations have been expanded for clarity and completeness.