Introduction: Moving Beyond One Dimension
Welcome to one of the most exciting parts of Further Mechanics! So far in your maths journey, you’ve mostly looked at objects moving in straight lines. But the real world happens in 3D. Whether it’s a drone weaving through trees, a planet orbiting a star, or a football curving through the air, we need vectors to describe this motion accurately. In this chapter, we’ll learn how to use calculus with vectors to model forces that change over time. Don't worry if it sounds complex—we'll take it one step at a time!
1. The Language of Kinematics in 2D and 3D
To talk about motion in space, we use three main vector quantities. These are usually written in terms of unit vectors i, j, and k.
- Position Vector \((\mathbf{r})\): This tells us where an object is relative to a fixed origin.
- Velocity Vector \((\mathbf{v})\): This tells us how fast the object is moving and in which direction. The magnitude of this vector is the speed.
- Acceleration Vector \((\mathbf{a})\): This tells us how the velocity is changing. The magnitude of this vector is simply the magnitude of acceleration.
Prerequisite Check: Displacement vs. Distance
Remember: Displacement is the vector pointing from your start point to your end point. Distance travelled is the total length of the path you actually took. Think of a bee flying around a flower; its displacement might be small if it ends up near where it started, but its distance travelled could be huge!
Quick Review: The Calculus Connection
Just like in 1D mechanics, we use calculus to move between these quantities:
1. To go "down" (Position \(\rightarrow\) Velocity \(\rightarrow\) Acceleration), we differentiate with respect to time \(t\):
\(\mathbf{v} = \frac{d\mathbf{r}}{dt} = \mathbf{\dot{r}}\)
\(\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2} = \mathbf{\dot{v}} = \mathbf{\ddot{r}}\)
2. To go "up" (Acceleration \(\rightarrow\) Velocity \(\rightarrow\) Position), we integrate with respect to time \(t\):
\(\mathbf{v} = \int \mathbf{a} \, dt\)
\(\mathbf{r} = \int \mathbf{v} \, dt\)
Key Takeaway: Differentiation and integration are performed on each component (\(\mathbf{i, j, k}\)) separately. It’s like doing three 1D problems at the same time!
2. Working with Variable Forces
In the "real world," forces aren't always constant. A gust of wind might push harder at some times than others. We use Newton's Second Law in vector form to handle this:
\(\mathbf{F} = m\mathbf{a}\)
If the force \(\mathbf{F}\) is a function of time, then the acceleration \(\mathbf{a}\) will also change over time. To find the velocity or position, we follow these steps:
Step-by-Step: Finding Position from a Variable Force
- Use \(\mathbf{a} = \frac{\mathbf{F}}{m}\) to find the acceleration vector.
- Integrate \(\mathbf{a}\) with respect to \(t\) to find \(\mathbf{v}\). Crucial: Don't forget the constant vector \(\mathbf{c}_1\)! You find this using "initial conditions" (e.g., "at \(t=0, \mathbf{v} = 3\mathbf{i}\)").
- Integrate \(\mathbf{v}\) with respect to \(t\) to find \(\mathbf{r}\). Again, add a constant vector \(\mathbf{c}_2\) and solve using the initial position.
Example: If \(\mathbf{a} = (6t)\mathbf{i} + \sin(t)\mathbf{j}\), then integrating once gives \(\mathbf{v} = (3t^2)\mathbf{i} - \cos(t)\mathbf{j} + \mathbf{c}\).
Common Mistake: Forgetting that the constant of integration is a vector (like \(c_x\mathbf{i} + c_y\mathbf{j}\)), not just a single number!
Key Takeaway: Variable forces lead to variable acceleration. Use integration to "climb the ladder" from force to position, finding your constant vectors at each step.
3. The Equation of the Path (Trajectory)
Sometimes we don't care when a particle is at a certain spot; we just want to know the shape of the path it takes. This is called the Cartesian equation of the path (usually an equation involving just \(x\) and \(y\)).
How to Eliminate the Parameter (Time)
If you are given a position vector \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\), where \(x\) and \(y\) are functions of \(t\):
- Write out the separate equations: \(x = f(t)\) and \(y = g(t)\).
- Rearrange the simplest equation (usually the \(x\) one) to make \(t\) the subject.
- Substitute this expression for \(t\) into the other equation.
- Simplify to get an equation in the form \(y = f(x)\).
Analogy: Imagine a pirate map. The vector \(\mathbf{r}(t)\) tells the pirate exactly where to be at every second. The Cartesian equation is just the line drawn on the map showing the trail in the sand.
Did you know? For standard projectile motion, this process always results in a quadratic equation, which is why projectiles follow a parabolic path!
Key Takeaway: To find the path, "get rid of \(t\)" by substituting it out of your component equations.
4. Projectiles on a Uniform Slope
In this section, we look at what happens when you launch a ball up or down a hill. This is a classic "Mechanics Major" topic.
Key Concepts
- The Range: This is the distance from the launch point to where the projectile hits the slope.
- Coordinates: You can solve these by either using standard horizontal/vertical axes or by tilting your axes so one is parallel to the slope. Tilting the axes is often easier!
Standard Modelling Assumptions
Don't worry if these seem a bit simplified—they are the "rules of the game" for your exam:
- No air resistance (the most common simplification).
- The projectile is a particle (we ignore its size and rotation).
- Constant gravity acting vertically downwards.
- The horizontal distance is small enough that we ignore the earth's curvature.
Key Takeaway: Projectile problems on slopes are just standard projectile problems where the "landing" height is defined by the equation of the slope (e.g., \(y = x \tan(\alpha)\)).
5. Differential Equations and Variable Acceleration
Sometimes acceleration isn't given as a function of time \(t\), but as a function of velocity \(v\) or displacement \(s\). In these cases, we need to choose the right form of acceleration to set up a differential equation.
Choosing the Right "Acceleration":
- If you have \(v\) and \(t\), use: \(a = \frac{dv}{dt}\)
- If you have \(v\) and \(s\), use: \(a = v \frac{dv}{ds}\)
Simple Harmonic Motion (SHM) Recognition
The syllabus requires you to recognize SHM in "non-standard" forms. The standard form is \(\mathbf{\ddot{x}} = -\omega^2 x\).
Memory Trick: If you see an equation where the double-dot (acceleration) is equal to negative something times the displacement, it's SHM!
Examples of SHM to watch out for:
- \(\ddot{x} + cx = 0\) (Here, \(\omega^2 = c\))
- \(\ddot{x} = -\omega^2(x + k)\) (This is just SHM centered at \(x = -k\) instead of the origin).
Quick Review: SHM Formulas
For \(\ddot{x} = -\omega^2 x\):
1. Period \(T = \frac{2\pi}{\omega}\)
2. Amplitude \(|A| = \sqrt{p^2 + q^2}\) (if solution is \(p \cos(\omega t) + q \sin(\omega t)\))
3. Velocity-Displacement: \(v^2 = \omega^2(A^2 - x^2)\)
Key Takeaway: Use \(a = v \frac{dv}{ds}\) when time isn't in the equation. Always look for the \(\ddot{x} = -\omega^2 x\) pattern to identify SHM.
Summary: Your "Vectors and Variable Forces" Toolkit
- Calculus is King: Differentiate to go from position to acceleration; integrate to go back.
- Don't Forget \(\mathbf{c}\): Constants of integration are vectors in these problems.
- Eliminate \(t\): To find the shape of a path, substitute \(t\) out of your \(x\) and \(y\) functions.
- Recognize Patterns: Acceleration proportional to negative displacement is always Simple Harmonic Motion.
Keep practicing! Mechanics is all about seeing the patterns in how things move. You've got this!