🚀 Further Mechanics (9665) Study Notes: Application of Differential Equations
Hello Future Mathematician! This chapter is where the abstract world of calculus meets the concrete reality of physics. We are going to use the powerful tool of Differential Equations (DEs) to describe exactly how things move. If you've enjoyed Mechanics (FM1) and learned about separation of variables in Pure Mathematics, you're perfectly equipped for this!
In this unit, we analyze motion in a straight line, particularly when the forces acting on an object (like air resistance) depend on its speed or position. We turn Newton's Second Law into an equation we can solve!
Key Takeaway from the Introduction
We use calculus to solve real-world problems involving force, acceleration, velocity, and displacement, primarily focusing on motion governed by Newton's Second Law.
1. The Foundation: Newton's Second Law as a DE
The core of every problem in this chapter is Newton's Second Law:
\(F = ma\)
Where \(F\) is the net force acting on the object, \(m\) is its mass, and \(a\) is the acceleration.
In Further Mechanics, we stop treating acceleration \(a\) as just a number and start treating it as a derivative. This is what turns the physics equation into a differential equation.
Understanding Acceleration (a)
Since acceleration is the rate of change of velocity (\(v\)) with respect to time (\(t\)), and velocity is the rate of change of displacement (\(x\)) with respect to time, we have three key expressions for \(a\):
- Acceleration in terms of Time:
\(a = \frac{dv}{dt}\)
- Acceleration in terms of Displacement:
\(a = \frac{d^2x}{dt^2}\)
- Acceleration in terms of Velocity and Displacement:
\(a = v \frac{dv}{dx}\)
Substituting these into \(F=ma\) gives us the DEs we need to solve.
🧠 Memory Aid: When to Use Which 'a'?
The choice of which form of \(a\) to use depends entirely on what the Force \(F\) depends on, and what you are asked to find.
-
If \(F\) is a function of \(t\) (time) only, OR if you need to find \(v\) in terms of \(t\):
Use \(m \frac{dv}{dt} = F(t)\)
-
If \(F\) is a function of \(v\) (velocity) only, OR if you need to find \(v\) in terms of \(t\):
Use \(m \frac{dv}{dt} = F(v)\)
-
If \(F\) is a function of \(x\) (displacement) only, OR if you need to find \(v\) in terms of \(x\):
Use \(m v \frac{dv}{dx} = F(x)\)
Don't worry if this seems tricky at first. Practice makes perfect! The question usually hints at the variables you need to link.
Key Takeaway on Setup
The first and most important step is setting up the equation \(F = m \times (\text{correct form of } a)\). Always ensure the net force \(F\) accounts for all forces acting on the particle (gravity, tension, and especially resistance).
2. Dealing with Resistance Forces (R)
In realistic mechanics problems, objects often face resistance (like air resistance or drag), and this resistance usually depends on the object's speed \(v\).
Syllabus Constraint: Permitted Resistance Forms
The syllabus restricts the resistance force \(R\) at speed \(v\) to one of two forms:
- Linear Resistance: \(R = a + bv\)
- Quadratic Resistance: \(R = a + bv^2\)
(Where \(a\) and \(b\) are constants, often simplified to \(R=kv\) or \(R=kv^2\) if the constant term is zero.)
Important Rule for Force F: Direction Matters!
Resistance always opposes the direction of motion. When setting up \(F=ma\), if you define the positive direction as the direction of motion, then resistance must be included as a negative term.
Example: A particle of mass \(m\) falling downwards under gravity (\(mg\)), experiencing resistance \(R=kv\).
If we define downwards as positive:
Net Force \(F = (\text{Force Down}) - (\text{Force Up})\)
\(F = mg - R\)
The resulting DE is: \(m \frac{dv}{dt} = mg - kv\)
Quick Review: Modelling the Force
- Identify all forces (Gravity \(mg\), Tension, Resistance \(R\)).
- Choose a positive direction (usually the direction of initial motion).
- Set \(F = m \times a\). Forces in the positive direction are positive; forces opposing motion (like resistance) are negative.
3. Solving Differential Equations in Mechanics
All the DEs derived from \(F=ma\) in this chapter are separable. This means we can rearrange the equation so that all terms involving one variable (like \(v\)) are on one side with its differential (\(dv\)), and all terms involving the other variable (like \(t\) or \(x\)) are on the other side with its differential.
Analogy: Sorting Laundry
Think of separating variables like sorting laundry: you put all the 'velocity socks' on the left side with \(dv\), and all the 'time shirts' on the right side with \(dt\). Once separated, you integrate both sides.
We will explore the three main types of separable DEs that arise from mechanics problems.
Case A: Force Depends on Time, \(F = F(t)\)
This is the simplest case, often involving forces that are explicitly changing over time.
Form: \(m \frac{dv}{dt} = F(t)\)
Step-by-Step Solution:
1. **Substitute and Separate:**
\(\frac{dv}{dt} = \frac{1}{m} F(t)\)
\(dv = \frac{1}{m} F(t) \, dt\) (All \(v\) terms on the left, all \(t\) terms on the right).
2. **Integrate:**
\(\int dv = \int \frac{1}{m} F(t) \, dt\)
3. **Apply Initial/Boundary Conditions:** Use the given starting conditions (e.g., \(v=0\) when \(t=0\)) to find the constant of integration, \(C\).
Example: If \(F=mt^2\), then \(\int dv = \int t^2 dt\), giving \(v = \frac{1}{3}t^3 + C\).
Case B: Force Depends on Velocity, \(F = F(v)\)
This is the most common case, usually involving resistance proportional to \(v\) or \(v^2\).
Form: \(m \frac{dv}{dt} = F(v)\)
Step-by-Step Solution:
1. **Separate:** Move the \(F(v)\) term to the side with \(dv\), and move \(dt\) to the other side.
\(\frac{m}{F(v)} \, dv = dt\)
2. **Integrate:**
\(\int \frac{m}{F(v)} \, dv = \int 1 \, dt\)
3. **Solve:** The left side often requires standard integration techniques (like substitution or partial fractions). The right side is simply \(t + C\).
🌟 Terminal Velocity (Important Concept)
When solving problems of this type (e.g., falling objects with resistance), the velocity often approaches a maximum constant speed called the Terminal Velocity.
This happens when the net force is zero (\(F=0\)), meaning the acceleration is zero (\(a=0\)).
To find Terminal Velocity (\(V_T\)):
Set the net force equation \(F=0\) and solve for \(v\).
If \(m \frac{dv}{dt} = mg - kv\), then Terminal Velocity occurs when \(mg - kv = 0\), so \(V_T = \frac{mg}{k}\).
Case C: Force Depends on Displacement, \(F = F(x)\)
This case arises when force is linked to position (e.g., elastic strings/springs, or gravitational force varying with distance). Since the force depends on \(x\), we need to find \(v\) in terms of \(x\).
Crucial Identity: We must use the chain rule identity for acceleration:
$$
a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}
$$
Form: \(m v \frac{dv}{dx} = F(x)\)
Step-by-Step Solution:
1. **Separate:**
\(m v \, dv = F(x) \, dx\) (All \(v\) terms on the left, all \(x\) terms on the right).
2. **Integrate:**
\(\int m v \, dv = \int F(x) \, dx\)
3. **Solve:** The left side integrates easily to \(\frac{1}{2} m v^2\). The right side depends on \(F(x)\). You usually end up with an equation relating \(v^2\) and \(x\).
Did you know? (Kinetic Energy Connection)
The left side, \(\int m v \, dv = \frac{1}{2} m v^2 + C\), looks exactly like the formula for Kinetic Energy. This is no accident! Integrating \(F\) with respect to displacement \(x\) is calculating the Work Done, relating this DE application directly to the Work-Energy Principle in physics.
4. Common Pitfalls and Tips for Success
⚠️ Common Mistake 1: Forgetting the Minus Sign
The most frequent error is incorrectly setting up the net force \(F\). Remember: Resistance is always negative relative to the direction of motion.
If a particle is moving right (positive direction), and resistance \(R\) is acting left, then \(F_{net} = \text{Driving Force} - R\).
⚠️ Common Mistake 2: Using the Wrong Form of 'a'
If the force depends on position \(x\), you MUST use \(v \frac{dv}{dx}\). If you try to use \(\frac{dv}{dt}\), you will have three variables (\(v, t, x\)) and an insoluble, non-separable equation!
Tip 1: Always Use Definite Integrals if Limits are Known
It is often cleaner and safer to use definite integrals rather than indefinite integrals (\( + C \)), especially when boundary conditions (initial speed, initial time) are given:
Instead of: \(\int \frac{1}{mg-kv} dv = \int dt\), then finding \(C\)...
Use: \(\int_{v_0}^{v} \frac{1}{mg-kv} dv = \int_{t_0}^{t} dt\)
Tip 2: Logarithm Care
When integrating functions like \(\frac{1}{a-bv}\), remember the chain rule reversal (or substitution):
\(\int \frac{1}{a-bv} dv = -\frac{1}{b} \ln |a-bv| + C\)
Do not forget the factor of \(-1/b\) from the derivative of the denominator!
Quick Revision Box: Essential Steps
Follow these four steps for every problem in this chapter:
- Define Direction: Establish a positive direction and draw a simple force diagram.
- Formulate \(F\): Write the net force \(F\), ensuring resistance is correctly subtracted.
- Select \(a\): Choose the correct form of acceleration (\(\frac{dv}{dt}\) or \(v \frac{dv}{dx}\)) based on the variables in \(F\).
- Separate and Integrate: Rearrange the resulting DE so variables are separated and integrate both sides, using boundary conditions to find the particular solution.