A Tale of Returning

September 25, 2023October 4, 2023 / Michael Xue / Leave a comment

A particle is projected vertically upwards under constant gravity $g$ in a medium for which the resistance per unit mass is $kv^n$ , where $v$ is the particle’s speed and $k$ and $n$ are positive constants. If the speed of the projection is $u$ , show that the particle returns to its point of projection with speed $u_1$ such that

if $n=1$ , then $u_1+u=\frac{g}{k}\log(\frac{1+\frac{k}{g}u}{1-\frac{k}{g}u_1});$
if $n=2$ , then $u_1^2 = \frac{u^2}{1+\frac{k}{g}u^2}$ .

Suppose $h$ is a differentiable function such that $h(v)$ gives the displacement from the maximum height when the speed of the downward moving particle is $v$ .

Fig. 1

From the problem statement and the state of the particle at different times (see Fig. 1), we have

$\frac{dv}{dt} = g- kv^n$

and

$h(v+a\Delta t) - h(v) =v\Delta t$ .

Let

$\Delta v = a\Delta t \implies \Delta t = \frac{\Delta v}{a}$ .

Then,

$h(v+\Delta v) - h(v) = v\frac{\Delta v}{a} \implies \frac{h(v+\Delta v) - h(v)}{\Delta v} = \frac{v}{a}.$

Consequently,

$\lim\limits_{\Delta v \rightarrow 0}\frac{h(v+\Delta v) - h(v)}{\Delta v} = \frac{v}{a}.$

Or,

$\frac{dh}{dv} = \frac{v}{a}.$

Integrate from $v = 0$ to $u_1$ gives

$\displaystyle\int\limits_{0}^{u_1}\frac{dh}{dv}\;dv = \int\limits_{0}^{u_1}\frac{v}{a}\;dv.$

Hence, by FTC,

$h(u_1)-h(0) = \displaystyle\int\limits_{0}^{u_1}\frac{v}{g-kv^n}\;dv$ .

Since $h(0) = 0$ , we now have

$h(u_1) = \displaystyle\int\limits_{0}^{u_1}\frac{v}{g-kv^n}\;dv$ .

From “A Journey Skyward“, we see that

$h(u_1) = \displaystyle\int\limits_{0}^{u} \frac{v}{g+kv^2}\;dv$ .

Therefore,

$\displaystyle\int\limits_{0}^{u}\frac{v}{g+kv^n}\;dv =\int\limits_{0}^{u_1}\frac{v}{g-kv^n}\;dv.$

When $n=1$ , it yields (see Fig. 2)

$u_1+u=\frac{g}{k}\log(\frac{1+\frac{k}{g}u}{1-\frac{k}{g}u_1});$

Fig. 2

If $n = 2$ then

$u_1^2 = \frac{u^2}{1+\frac{k}{g}u^2}$ .

Fig. 3

Notice when $n=2, u_1^2 \rightarrow u^2$ as $k \rightarrow 0$ .

Exercise-1 Explain when $n=2, u_1 = -u$ if $k = 0$ .

A Journey Skyward

September 24, 2023October 2, 2023 / Michael Xue / Leave a comment

A ball is projected vertically upwards under gravity with initial velocity $u$ , the resistance of the air producing a retardation $kv^n$ , where $v$ is the speed of the ball and $k$ and $n$ are positive constants. Show that the maximum height reached, $f(u)$ , is given by the integral

$\displaystyle \int\limits_{0}^{u} \frac{v}{g+kv^n}\; dv$ .

Suppose $h$ is a differentiable function such that $h(v)$ gives the greatest height the ball can reach after being projected with speed $v$ .

Fig. 1

From the problem statement and the ball’s states illustrated in Fig. 1 at $t=0, \Delta t$ and $\tau$ when the greatest height is attained, we have

$\frac{dv}{dt} = -g - kv^n = -(g + kv^n)$

and

$h(v) = v\Delta t + h(v-a\Delta t)\implies h(v-a\Delta t) - h(v) = -v\Delta t$

where

$a = g + kv^n.\quad\quad\quad(1)$

Let

$\Delta v = -a\Delta t \implies \Delta t = \frac{-\Delta v}{a}$ .

It means

$h(v+\Delta v) - h(v) = -v\cdot \frac{-\Delta v}{a}=\frac{v\Delta v}{a}$ .

Or,

$\frac{h(v+\Delta v)-h(v)}{\Delta v} = \frac{v}{a}$ .

Consequently,

$\lim\limits_{\Delta v \rightarrow 0}\frac{h(v+\Delta v)-h(v)}{\Delta v} = \lim\limits_{\Delta v \rightarrow 0}\frac{v}{a} = \frac{v}{a}$ .

i.e.,

$\frac{dh}{dv} = \frac{v}{a}$ .

Integrate it from $v = 0$ to $u$ yields

$h(u) - h(0) = \displaystyle\int \limits_{0}^{u}\frac{v}{a}\;dv.$

Since $h(0) = 0$ (a ball projected with zero speed would reach zero height),

$f(u) = h(u) \overset{(1)}{=} \displaystyle\int\limits_{0}^{u} \frac{v}{g+kv^n}\;dv$ .

If $k = 0$ , then $f(u) = \frac{u^2}{2g}$ :

If $n=1$ , then $f(u) =\frac{1}{k}\left(u-\frac{g}{k}\log(1+\frac{ku}{g})\right)$ :

If $n=2$ , then $f(u) = \frac{1}{2k}\log(1+\frac{ku^2}{g})$ :

When $n = 1$ or $2$ , after expanding the function $f$ in a power series in $k$ , we see as $k \rightarrow 0, f \rightarrow \frac{u^2}{2g}$ :

Raw draw2d & plot2d

September 22, 2023September 22, 2023 / Michael Xue / Leave a comment

See also “Higham’s Parametric Curve“.

Taking the Plunge

September 20, 2023September 22, 2023 / Michael Xue / Leave a comment

A particle is projected vertically downwards under a constant gravitational force of magnitude $g$ per unit mass. Show that, if the air resistance produces a force of magnitude $kv$ per unit mass opposing the motion (where $k$ is a positive constant) then no matter what initial speed is given to the particle, its speed will eventually tend to $\frac{g}{k}$ (called the terminal speed)

What is the terminal speed if the resistive force is of magnitude $kv^2$ per unit mass?

By Newton’s second law of motion,

$m\frac{dv}{dt} = mg - mkv \implies \frac{dv}{dt}=g-kv.\quad\quad\quad(1)$

Solving $g-kv = 0$ gives a constant solution of (1). Namely,

$v = \frac{g}{k}$ .

This is an asymptotically stable equilibrium (see Exercise-1). Therefore,

$\forall v(0) \ne \frac{g}{k}, v(t) \rightarrow \frac{g}{k}$ as $t \rightarrow \infty$ .

Similarly,

$m\frac{dv}{dt} = mg - mkv^2 \implies \frac{dv}{dt} = g - kv^2.\quad\quad\quad(2)$

Solving $g-kv^2 = 0$ yields the asymptotically stable equilibrium $\sqrt{\frac{g}{k}}$ (see Fig. 1)

Fig. 1

In both cases, we can also obtain the limit after solving an initial-value problem:

Fig. 2 $\frac{dv}{dt} = g-kv, v(0) = u$

Fig. 3 $\frac{dv}{dt} = g-kv^2, v(0) = u$

Exercise-1 Show that $\frac{g}{k}$ is an asymptotically stable equilibrium of (1) (hint: Fig. 1)

A Vertical Blast

September 18, 2023September 21, 2023 / Michael Xue / Leave a comment

A particle of mass $m$ is projected vertically upwards under constant gravity $g$ . The force due to air resistance is $2amv + b^2mv^2$ , where $v$ is the particle’s speed and $a$ and $b$ are constants. Find integral expressions for $h_{max}$ , the greatest height reached, and $\tau$ , the time taken to reach its greatest height, in terms of the speed of projection, $u$ .

Hence deduce that

$b^2h_{max} + a\tau = \frac{1}{2} \log\left(|1+\frac{2au}{g} + \frac{b^2u^2}{g}|\right)$ .

Suppose $h$ is a differentiable function such that $h(v)$ gives the greatest height the particle can reach after being projected with speed $v$ .

Fig. 1

From the depiction of the particle’s states at $t=0, \Delta t$ and $\tau$ (see Fig. 1), we see that by Newton’s second law of motion,

$m\frac{dv}{dt} = -mg-(2amv + b^2mv^2)\implies\frac{dv}{dt} = -(g+2av + b^2v^2)=-\alpha\quad\quad\quad(1)$

and

$h(v) - h(v-\alpha\Delta t) = v\Delta t$

where

$\alpha= g+2av + b^2v^2.\quad\quad\quad(2)$

It follows that

$h(v-\alpha\Delta t) - h(v) = -v\Delta t$ .

Let $\Delta v = -\alpha\Delta t \implies \Delta t = \frac{-\Delta v}{\alpha}$ , we have

$h(v+\Delta v ) - h(v) = -v\cdot\frac{-\Delta v}{\alpha}$

$\frac{h(v+\Delta v) - h(v)}{\Delta v} = \frac{v}{\alpha}$ .

Consequently,

$\lim\limits_{\Delta v \rightarrow 0}\frac{h(v+\Delta v) - h(v)}{\Delta v} = \lim\limits_{\Delta v \rightarrow 0}\frac{v}{\alpha}=\frac{v}{\alpha}$ .

i.e.,

$\frac{dh}{dv} \overset{(2)}{=} \frac{v}{g+2av+b^2v^2}$ .

Integrate from $v=0$ to $u$ gives

$\displaystyle\int\limits_{0}^{u} \frac{dh}{dv}\;dv = \int\limits_{0}^{u} \frac{v}{g+2av+b^2v^2}\;dv$ .

By FTC,

$h(u) - h(0) =\displaystyle \int\limits_{0}^{u} \frac{v}{g+2av+b^2v^2}\;dv$ .

Since $h(u)=h_{max}, h(0) = 0$ (a particle projected with zero speed would reach zero height), we have

$h_{max} = \displaystyle\int\limits_{0}^{u}\frac{v}{g+2av+b^2v^2}\;dv.\quad\quad(*)$

To find $\tau$ , we rewrite (1) as

$\frac{1}{-g-(2av+b^2v^2)}\cdot\frac{dv}{dt} = 1$ .

Integrate it from $t=0$ to $\tau$ :

$\displaystyle\int\limits_{0}^{\tau} \frac{1}{-g-(2av+b^2v^2)}\cdot\frac{dv}{dt} \; dt= \int\limits_{0}^{\tau}1\;dt=\tau$ ,

we obtain

$\displaystyle\int\limits_{u}^{0} \frac{1}{-g-(2av+b^2v^2)} \; dv=\tau$ .

That is,

$\tau = \displaystyle\int\limits_{0}^{u}\frac{1}{g+2av+b^2v^2}\;dv.\quad\quad(**)$

And so,

$b^2h_{max} + a\tau \overset{(*), (**)}{=} \displaystyle\int\limits_{0}^{u}\frac{b^2v}{g+2av+b^2v^2}\;dv +\int\limits_{0}^{u}\frac{a}{g+2av+b^2v^2}\;dv$