Resisted motion is foundational for mechanics component of Maths Extension 2. This blog post will explain the key concepts behind resistive motion and walk through worked examples on how to approach resistive motion mechanics questions.
Resistive motion is an integral part of mechanics, particularly for harder mechanics applications.
Students should already be familiar with motion on a horizontal line, the motion equations and how to manipulate, through integration or differentiating, to achieve velocity and displacement. Also, the use of vectors to represent projectile motion should also be familiar and will be used to decompose force vectors in this subject guide.
Students can refresh their knowledge on projectile motion and calculus applications in the following subject guides:
When in motion, particles often experience resistance in the form of either friction or air resistance. Frictional forces generally are constant whilst air resistance increases as the velocity of the particle increases.
Generally, the resistive forces (denoted by \(R\)), are in a polynomial in terms of \(v\), often \(R=mkv\) or \(R=mkv^2\)
Therefore, when given a question, we would have to account for the addition resistance force applied to the object.
For example, if a falling object experiences air resistance proportional to its velocity squared, our force diagram can be shown as below.
From here, we can equate our forces, taking the upward direction as positive.
\(F_{net}=R-W\)
Since we know that our air resistance is proportional to the velocity squared, this can be written as \(R=mkv^2\). Since \(F_{net}=ma\), we can rewrite the equation above as:
\begin{align*}
ma &= mkv^2 – mg\\
a &= kv^2 -g\\
\end{align*}
In order to find our other motions of equation, we can simply integrate the acceleration equation using the fact that \(\ddot{x} = \frac{dv}{dt}\) or \(\ddot{x} = v \frac{dv}{dx}\).
A mass of \(m\) kilograms is dropped at a height of \(h\) metres above the ground. It experiences air resistance equal to \(mkv^2\), where \(v\) is its speed in metres per second and \(k\) is a positive constant. It experiences a gravitational force of \(mg\). Let \(x\) be the distance, in metres, from its initial position above the ground.
a) Draw a force diagram to show the forces acting on the particle
b) Derive the equation of motion
Taking the downward direction as the positive direction, we can equate our forces on the vertical plane.
\begin{align*}
F_{net}&=ma\\
ma&=mg-mkv^2\\
a&=g-kv^2
\end{align*}
c) Find \(v\) as a function of \(x\). To find the other equations of motion, we would either have to use \(\ddot{x} = \frac{dv}{dt}\) or \(\ddot{x} = v \frac{dv}{dx}\). As the question asks for a function of \(v\) in terms of \(x\), we would use the latter.
\begin{align*}
v \frac{dv}{dx}&=g-kv^2\\
\frac{dv}{dx} &= \frac{g-kv^2}{v}\\
\frac{dx}{dv} &= \frac{v}{g-kv^2}\\
\end{align*}
Now, we can integrate where the initial position and velocity are both 0. (Note where it says “Let \(x\) be the distance, in metres, from its initial position above the ground.”)
\begin{align*}
\int_{0}^{x}dx&= \ – \frac{1}{2k} \int_{0}^{v} \frac{-2kv}{g-kv^2}dv\\
[x] _0^x &= \ – \frac{1}{2k}[ln(g-kv^2)]_0^v\\
x-0 &= \ – \frac{1}{2k}[ln(g-kv^2)-ln(g)]\\
x &= \ – \frac{1}{2k}ln \frac{g-kv^2}{g}\\
\end{align*}
To achieve \(v\) as a function of \(x\), we can simply make \(v\) the subject of our equation to complete the question.
| \begin{align*} -2kx&=ln \frac{g \ – \ kv^2}{g}\\ e^{-2kx}&= \frac{g \ – \ kv^2}{g}\\ ge^{-2kx}&=g-kv^2\\ kv^2&=g-ge^{-2kx}\\ v^2&=\frac{g(1 \ – \ e^{-2kx})}{k}\\ v&= \sqrt{ \frac{g(1 \ – \ e^{-2kx})}{k} } \ \ \text{(because v > 0)} \end{align*} |
Suppose we had a particle of unit mass that is projected from the origin with a speed \(V\) at an angle α to the horizontal. The particle is subject to both gravity and air resistance proportional to its velocity.
To derive the motion equations, we must first understand how the force diagram of the particle can be illustrated.
Air resistance acts in a direction that is directly opposite to the direction of motion.
Therefore, initially we can conclude that the air resistance force is parallel to that of the velocity vector, but in the opposite direction.
As it is also proportional to its velocity, we will then have a resistive force of \(mkv\).
We can illustrate the resistive forces acting on the particle, denoting the upward and rightward directions as both positive.
Alternatively, we can write as a vector in terms of its horizontal and vertical component.
\(mkv=mk \binom{ \dot{x}}{ \dot{y}}\)
By adding our resistive forces together, we achieve our motion vector.
\begin{align*}
F_{net} &=ma\\
m \binom{ \ddot{x}}{ \ddot{y}} &= mk\binom{ – \dot{x}}{- \dot{y}} + m \binom{0}{-g}\\
∴ \binom{ \ddot{x}}{ \ddot{y}} &= k\binom{ – \dot{x}}{- \dot{y}} + \binom{0}{-g}\\
\end{align*}
From here, we will be able to decompose our vector into horizontal and vertical motion equations.
\begin{align*}
\ddot{x} &= -k \dot{x}\\
\ddot{y} &= -k \dot{y} \ – \ g\\
\end{align*}
Whilst in this equation, resistive forces were proportional to velocity, there may be scenarios where the force is proportional to velocity squared, has a value or acts only in the horizontal or vertical plane.
However, to approach such problems, draw a force diagram (like the one above) to simplify the problem and make equating the motion equations easier.
A particle is projected from the origin with speed \(V\) at an angle α to the horizontal. The particle is subject to both gravity and an air resistance proportional to its velocity.
a) Use Newton’s second law, explain why the acceleration vector is given by
\(a=\binom{ \ddot{x}}{ \ddot{y}} =\binom{ -k \dot{x}}{ -k \dot{y}-g}\)
Where \(k > 0\) is a constant.
b) By treating \( \ddot{x}= -k \dot{x}\) as a separable differential equation, show that \( \dot{x} = Ve^{-kt} \ cos \ α\)
c) Similarly, solve \( \ddot{y} = -g -k \dot{y}\) to obtain \( \dot{y} = \left( \frac{g}{k} + v \ sin \ α \right) e^{-kt} \ – \frac{g}{k} \)
a) The explanation to this part is identical to the part under the subheading “Resisted Motion in Projectile Motion”
b) Note that the initial velocity should be decomposed into a horizontal and vertical component. Breaking the initial velocity yields \(V \ cos \ α\) in the horizontal component and \(V \ sin \ α\) in the vertical component (used in the next part)
\begin{align*}
a &= -k \dot{x}\\
\frac{d \dot{x}}{dt} &= -k \dot{x}\\
\int_0^t t &= \ – \frac{1}{k} \int_{V \ cos \ α}^{ \dot{x}} \frac{1}{ \dot{x}} d \dot{x}\\
t &= \ – \frac{1}{k} [ln \ \dot{x} \ – \ lnV \cos \ α]\\
t &= \ – \frac{1}{k} ln \frac{ \dot{x}}{V \cos \ α}\\
-kt &= ln \frac{ \dot{x}}{V \cos \ α}\\
e^{-kt} &= \frac{ \dot{x}}{V \cos \ α}\\
\dot{x} &= Ve^{-kt} \ cos \ α\\
\end{align*}
c)
| \begin{align*} a &= -k \dot{y} -g\\ \frac{d \dot{y}}{dt} &= -k \dot{y} -g\\ \int_0^t t &= \ – \frac{1}{k} \int_{V \ cos \ α}^{v} \frac{1}{ k \dot{y} \ + \ g} d \dot{y}\\ t &= \ – \frac{1}{k} [ln(k \dot{y} \ + \ g)] _{V \ cos \ α}^{v}\\ t &= \ – \frac{1}{k} [ln(k \dot{y} \ + \ g) – ln(kV \ cos \ α \ + \ g)]\\ t &= \ – \frac{1}{k} [ln \left( \frac{k \dot{y} \ + \ g} {kV \ cos \ α \ + \ g} \right)]\\ -kt &= \ ln \left( \frac{k \dot{y} \ + \ g} {kV \ cos \ α \ + \ g} \right)\\ e^{-kt} &= \ \frac{k \dot{y} \ + \ g} {kV \ cos \ α \ + \ g}\\ (kV \ cos \ α \ + \ g)e^{-kt} &= \ k \dot{y} \ + \ g\\ ∴ \dot{y} &= \left( \frac{g}{k} +V \ cos \ α \right) e^{-kt} – \frac{g}{k}\\ \end{align*} |
A particle of mass \(3kg\) moves in a straight line against a resistance equal to \(12v^2\), where \(v\) is its velocity in metres per second.
Initially, the particle is at the origin and is travelling with velocity \(10m/s\).
1.
2. \( \ddot{x} = -4v^2 \)
3. Proof
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