Projectile Motion and Normal-Tangential Coordinates

Hello! Welcome back to our course on mechanical engineering fundamentals.

In our last lesson, we introduced the kinematics of particles, covering motion in a straight line (rectilinear) and along a curve (curvilinear) using rectangular (x-y) coordinates. We ended with a brief look at projectile motion as a classic application of this method.

Today, we'll build directly on that foundation to meet our learning outcome: Analyze projectile motion and motion in normal-tangential coordinates.

This lesson is split into two parts:

1. A deeper analysis of projectile motion: We'll move beyond the basic equations to develop a robust problem-solving strategy for finding key parameters like range, maximum height, and time of flight.
2. Introduction to Normal-Tangential (n-t) coordinates: We'll learn a new, powerful coordinate system that's especially useful for analyzing motion along a curved path, providing direct insight into how an object's speed and direction are changing. This is fundamental for analyzing aircraft maneuvers and orbital mechanics.

Let's get started.

1. Projectile Motion: A Problem-Solving Approach

Recall from last lesson that projectile motion (neglecting air resistance) is governed by two simple, independent kinematic rules:

• Horizontal motion: Constant velocity ($a_x=0$).
• Vertical motion: Constant acceleration due to gravity ($a_y=-g$, where $g\approx 9.81{\text{m/s}}^2$ or $32.2{\text{ft/s}}^2$).

This allows us to use the constant acceleration equations we've already learned, applied separately to the x and y directions. The key to solving any projectile motion problem is to systematically break it down.

The video below provides an excellent step-by-step guide to tackling these problems.

A General Strategy for Analysis

As demonstrated in the video, a reliable strategy is:

1. Establish a Coordinate System: Usually, it's best to place the origin at the point of launch, with the y-axis vertical (positive upwards) and the x-axis horizontal.
2. Resolve Initial Velocity: Break the initial velocity $v_0$ into its horizontal $(v_0{)}_x=v_0\cos \theta$ and vertical $(v_0{)}_y=v_0\sin \theta$ components.
3. List Knowns and Unknowns: Create separate lists for the horizontal and vertical motions. Remember that time, $t$, is the common variable linking both.
   • Horizontal: $x_0,x,(v_0{)}_x,a_x=0,t$
   • Vertical: $y_0,y,(v_0{)}_y,(v_f{)}_y,a_y=-g,t$
4. Apply Kinematic Equations: Select the appropriate constant acceleration equations for each direction to solve for your unknowns. You often end up with a system of two equations and two unknowns (e.g., $v_0$ and $t$).

Let's watch another example that applies this strategy to a common scenario.

Key Formulas for Projectile Motion

For the special case where a projectile is launched from and lands on a flat horizontal surface ($y_{final}=y_{initial}$), we can derive some very useful formulas for maximum height, time of flight, and range.

These formulas are direct results of applying the problem-solving strategy. For example, the maximum height $H$ occurs when the vertical velocity $v_y$ is momentarily zero.

Test your understanding!

An early catapult for launching aircraft from ships could accelerate a plane to a speed of 30 m/s. If the plane is launched at an angle of 15° above the horizontal from a height of 5 m above the water, how far does it travel horizontally before hitting the water? (Use $g=9.81{\text{m/s}}^2$).

Show answer

1. Coordinate System: Origin at the launch point. $x_0=0$, $y_0=0$. The water is at $y=-5$ m.

2. Initial Velocity Components:

   • $(v_0{)}_x=30\cos (15^{\circ })\approx 28.98$ m/s
   • $(v_0{)}_y=30\sin (15^{\circ })\approx 7.76$ m/s

3. Find Time of Flight (t): We use the vertical motion equation, as this is where we know the displacement ($-5$ m).
   $y=y_0+(v_0{)}_{y}t+\frac{1}{2}{a}_y{t}^2$
   $-5=0+(7.76)t+\frac{1}{2}(-9.81)t^2$
   Rearranging into a quadratic equation:
   $4.905t^2-7.76t-5=0$
   Using the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:
   $t=\frac{7.76\pm \sqrt{(-7.76{)}^2-4(4.905)(-5)}}{2(4.905)}$
   $t=\frac{7.76\pm \sqrt{60.22+98.1}}{9.81}=\frac{7.76\pm 12.58}{9.81}$
   We take the positive root for time: $t\approx 2.07$ s.

4. Find Horizontal Distance (Range): Now use the time in the horizontal motion equation.
   $x=x_0+(v_0{)}_{x}t$
   $x=0+(28.98)(2.07)\approx 59.99\text{m}$

Answer: The plane travels approximately 60 meters horizontally before hitting the water.

2. Normal and Tangential (n-t) Coordinates

While rectangular coordinates are useful, they don't always provide the most intuitive description of motion. When an airplane pulls out of a dive, the pilot feels a force pushing them into their seat. This sensation is related to the change in direction of their velocity, not just their speed. The n-t coordinate system is designed to separate these effects.

It is a moving coordinate system that travels with the particle.

• The tangential axis (t) is tangent to the path and points in the direction of motion.
• The normal axis (n) is perpendicular to the t-axis and points toward the center of curvature of the path.

Velocity and Acceleration in n-t Coordinates

• Velocity: Since velocity is always tangent to the path, the expression is simple:
  $\vec{v}=v{\vec{u}}_t$
  where $v=||\vec{v}||$ is the particle's speed.

• Acceleration: This is where the power of the n-t system becomes clear. The acceleration vector is broken into two physically meaningful components:
  $\vec{a}=a_t{\vec{u}}_t+a_n{\vec{u}}_n$

  • Tangential Acceleration ($a_t$): This component acts along the t-axis and represents the rate of change of speed.
    $a_t=\dot{v}=\frac{dv}{dt}$
    If you're speeding up, $a_t$ is positive. If you're slowing down, $a_t$ is negative. If your speed is constant, $a_t=0$.
  • Normal (Centripetal) Acceleration ($a_n$): This component acts along the n-axis and represents the rate of change of direction. It is always directed towards the center of curvature.
    $a_n=\frac{v^2}{\rho }$
    Here, $\rho$ (rho) is the instantaneous radius of curvature of the path. If you are moving along a curve, even at a constant speed, you must have a normal acceleration.

Think of driving a car:

• Pressing the accelerator or brake produces tangential acceleration, $a_t$.
• Turning the steering wheel produces normal acceleration, $a_n$.
• Doing both simultaneously creates an acceleration vector with both components.

Calculating n-t Components

As summarized in the reading, we have two primary ways to calculate the components $a_t$ and $a_n$:

1. If you can describe the motion with vectors $\vec{v}$ and $\vec{a}$ in rectangular (x-y) coordinates first:

   • Tangential component: $a_t=\frac{\vec{v}\cdot \vec{a}}{||\vec{v}||}$ (The projection of $\vec{a}$ onto the direction of $\vec{v}$)
   • Normal component: $a_n=\frac{||\vec{v}\times \vec{a}||}{||\vec{v}||}$

2. Once you have one component, you can find the other using the magnitude of the total acceleration, $||\vec{a}|{|}^2=a_t^2+a_n^2$:

   • $a_n=\sqrt{||\vec{a}|{|}^2-a_t^2}$

Your aerospace goal makes this coordinate system particularly important. The "G-force" an astronaut or pilot experiences is directly related to the normal component of their acceleration. For an aircraft in a level, coordinated turn at constant speed, $a_t=0$, but the normal acceleration $a_n=v^2/\rho$ can be significant, leading to high G-loads.

Test your understanding!

A race car enters a circular turn of radius $\rho =200$ m at a speed of 180 km/h. At that instant, the driver is braking, causing a tangential deceleration of $a_t=-5{\text{m/s}}^2$. What is the magnitude of the car's total acceleration at this moment?

Show answer

1. Convert Units: First, convert the speed to m/s.
   $v=180\frac{\text{km}}{\text{h}}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\text{s}}=50\text{m/s}$

2. Calculate Normal Acceleration ($a_n$): This component is due to the car turning.
   $a_n=\frac{v^2}{\rho }=\frac{(50\text{m/s}{)}^2}{200\text{m}}=\frac{2500}{200}=12.5{\text{m/s}}^2$

3. Calculate Total Acceleration Magnitude: The total acceleration is the vector sum of the tangential and normal components. Since they are perpendicular, we use the Pythagorean theorem.
   $||\vec{a}||=\sqrt{a_t^2+a_n^2}=\sqrt{(-5{)}^2+(12.5{)}^2}$
   $||\vec{a}||=\sqrt{25+156.25}=\sqrt{181.25}\approx 13.46{\text{m/s}}^2$

Answer: The magnitude of the car's total acceleration is $13.46{\text{m/s}}^2$.

Conclusion

In this lesson, we mastered two essential methods for analyzing motion along a curved path.

Key Takeaways:

• Projectile motion problems are solved by separating them into two independent rectilinear motion problems: constant velocity horizontally and constant acceleration vertically.
• Normal-Tangential (n-t) coordinates provide a powerful way to describe curvilinear motion by separating acceleration into two components:
  • $a_t=\dot{v}$, which measures the change in speed.
  • $a_n=v^2/\rho$, which measures the change in direction.
• Total acceleration is $\vec{a}=a_t{\vec{u}}_t+a_n{\vec{u}}_n$, with magnitude $||\vec{a}||=\sqrt{a_t^2+a_n^2}$.

Next Lesson Preview:

So far, we have only described how particles move (kinematics). We haven't yet addressed why they move. In our next lesson, we will make that crucial connection by introducing kinetics. We will apply Newton's Second Law ($\vec{F}=m\vec{a}$) to relate the forces acting on a particle to the acceleration we've learned how to describe. This will allow us to start solving for the forces required to produce a given motion, or the motion that results from a given set of forces.