Two-Dimensional Analysis: Projectile and Uniform Circular Motion Study Guide

The study of two-dimensional analysis focuses on two primary types of movement: Projectile Motion and Circular Motion.
The central objective is to analyze projectile motion by relating horizontal and vertical components. This analysis allows for the determination of critical quantities, including:
- Range
- Maximum height
- Time of flight
- Velocity

Projectile motion is defined as the curved path an object follows when it is thrown into the air, specifically moving under the influence of the force of gravity.
Key parameters of a projectile's flight include:
- Range ( $R$ ): This is the horizontal distance covered by the object during its entire flight.
- Height ( $H$ ): This is the vertical distance measured from the ground to the object's highest point.
- Highest Point Characteristics: As the projectile follows its curved path, it reaches an apex where its vertical velocity ( $V_y$ ) is exactly equal to zero ( $0$ ).
- Launch Angle ( $\theta$ ): The specific angle at which the projectile is launched significantly affects its trajectory.

To analyze the motion, the initial velocity ( $V_i$ ) must be resolved into its horizontal ( $x$ ) and vertical ( $y$ ) components:
- For the x-component: $V_x = V_i \times \text{cos}(\theta)$
- For the y-component: $V_y = V_i \times \text{sin}(\theta)$
The magnitude of the resultant velocity vector is given by: $v = \text{sqr}(V_x^2 + V_y^2)$ [Note: Represented as $v = \text{sqr}(v1^2 + v2^2)$ in some notations].
The angle of the trajectory can be found using the inverse tangent function: $\theta = \text{tan}^{-1}(\frac{V_y}{V_x})$

Scenario: A sipa is kicked at a $30^\circ$ angle with an initial velocity of $6\,m/s$ .
Given Values:
- Initial Velocity ( $V_i$ ) = $6\,m/s$
- Angle ( $\theta$ ) = $30^\circ$
Calculations for x-component ( $V_x$ ):
- $V_x = V_i \times \text{cos}(\theta)$
- $V_x = 6\,m/s \times \text{cos}(30^\circ)$
- $V_x = 6\,m/s \times 0.8660$
- $V_x = 5.2\,m/s$
Calculations for y-component ( $V_y$ ):
- $V_y = V_i \times \text{sin}(\theta)$
- $V_y = 6\,m/s \times \text{sin}(30^\circ)$
- $V_y = 6\,m/s \times 0.5$
- $V_y = 3\,m/s$

Specific formulas are utilized to determine the physical properties of a projectile's path:
- Total Time ( $T$ ): The duration of the flight is calculated as: $T = \frac{2V_i \times \text{sin}(\theta)}{g}$
- Maximum Height ( $H$ ): The peak vertical distance is calculated as: $H = \frac{(V_i \times \text{sin}(\theta))^2}{2g}$
- Range ( $R$ ): The total horizontal distance is calculated as: $R = \frac{V_i^2 \times \text{sin}(2\theta)}{g}$

Scenario: A takraw ball has an initial velocity of $5\,m/s$ and is projected at an angle of $15^\circ$ . The goal is to find the maximum height, total time, and range as it lands.
Given Values:
- $V_i = 5\,m/s$
- $\theta = 15^\circ$
- $g = 9.8\,m/s^2$
Calculation of Maximum Height ( $H$ ):
- $H = \frac{(5\,m/s \times \text{sin}(15^\circ))^2}{2 \times 9.8\,m/s^2}$
- $H = 0.09\,m$
Calculation of Range ( $R$ ):
- $R = \frac{(5\,m/s)^2 \times \text{sin}(2 \times 15^\circ)}{9.8\,m/s^2}$
- $R = 1.28\,m$
Calculation of Total Time ( $T$ ):
- $T = \frac{2 \times (5\,m/s) \times \text{sin}(15^\circ)}{9.8\,m/s^2}$
- $T = 0.26\,s$

The exercise involves finding the maximum height, total time, and range for a sipa kicked at an initial velocity of $5\,m/s$ .
Different angles are assigned based on the student's category:
- For Male Students: $30^\circ$ and $50^\circ$
- For Female Students: $40^\circ$ and $60^\circ$

Uniform Circular Motion is a type of motion that allows an object to follow a circular path while maintaining a constant speed.
Common Examples of UCM:
- Carousel
- Centrifuge
- Microwave
- Washing machine
- Machine dryer

Changing Direction: As an object moves in a circle, it is constantly changing its direction. Even though the speed remains constant, the velocity changes because velocity is a vector quantity (having both magnitude and direction).
Tangential Velocity: At all instances, the object is moving tangent to the circle. The velocity vector is directed tangent to the circle at every point.
Perpendicular Relationship: The directions of acceleration and velocity are perpendicular to each other while the speed remains constant.
Related Terms:
- Period ( $T$ ): The amount of time needed to complete one full rotation.
- Frequency ( $f$ ): The number of rotations completed per unit of time.

Centripetal Force ( $F_c$ ): This is the net force acting upon an object in circular motion, directed toward the center of the circle. It is what allows the object to accelerate in a circular path.
Centripetal Acceleration ( $a_c$ ): This is the acceleration of a body traversing a circular path. It is caused by the constant change in direction of the velocity vector.
- Direction: Directed radially toward the center of the circle.
- Magnitude formula: $a_c = \frac{v^2}{r}$
- Units: Measured in meters per second squared ( $m/s^2$ ).
Centripetal Force Formula: Based on Newton's Second Law ( $F = ma$ ), the centripetal force is calculated as:
- $F_c = m \times a_c$
- Units:
  - mass ( $m$ ) is measured in kilograms ( $kg$ ).
  - force ( $F_c$ ) is measured in Newtons ( $N$ ).
  - acceleration ( $a_c$ ) is measured in $m/s^2$ .

Scenario: Shiela is riding in a carousel moving at a constant speed of $7.0\,m/s$ with a radius of $1.3\,m$ .
Part A: Centripetal Acceleration:
- Given:
  - $v = 7.0\,m/s$
  - $r = 1.3\,m$
- Formula: $a_c = \frac{v^2}{r}$
- Solution:
  - $a_c = \frac{(7.0\,m/s)^2}{1.3\,m}$
  - $a_c = \frac{49.0\,m^2/s^2}{1.3\,m}$
  - $a_c = 37.69\,m/s^2$
Part B: Centripetal Force:
- Requirement: Find the force if the mass of the carousel (or system component) is $26\,kgs$ .
- Given:
  - $m = 26\,kgs$
  - $a_c = 37.69\,m/s^2$
- Formula: $F_c = m \times a_c$
- Solution:
  - $F_c = 26\,kg \times 37.69\,m/s^2$
  - $F_c = 979.94\,kg \cdot m/s^2$ or $979.94\,N$
Calculation Note: Always round off results to the nearest hundredths.