TSSFL TECHNOLOGY STACK

Posted: **Tue May 12, 2015 6:11 pm**

In this topic, we formulate and solve a number of differential equations that originate from real-world problems.

Example 1

A ball of mass $m$ is thrown straight up from the ground with an initial velocity of $v_{0}$ . Ignoring air resistance, find the maximum height and the time the ball takes to attain the maximum height.

Solution

Let the height of the ball from the ground at any time $t$ be $y(t)$ .

Since the force acting on the ball is $-mg$ , Newton's second law of motion, $F = ma$ , gives the relation $F = -mg.$

Thus

$m\dfrac{d^{2}y}{dt^{2}} + mg = 0$

$\dfrac{d^{2}y}{dt^{2}} + g = 0$

Integrating both sides,

\begin{equation}\tag{eq1}
m\dfrac{dy}{dt} + gt + c = 0
\end{equation}

Velocity $\dfrac{dy}{dt} = v_{0}$ at $t = 0$ , so (eq1) yields

$v_{0} + c = 0$ , this $\implies \ c = -v_{0}$

Now, we have

\begin{equation}\tag{eq2} \dfrac{dy}{dt} + gt - v_{0} = 0 \end{equation}

Integrating (eq2),

$y(t) + \frac{1}{2}gt^{2} - v_{0}t + c_{1} = 0$

Since $y(0) = 0, \ c_{1} = 0$

The equation of position at any time $t$ is

\begin{equation}\tag{eq3} y(t) = -\frac{1}{2}gt^{2} + v_{0}t \end{equation}

To find time ( $t_{max}$ ) at the maximum height, we set the first derivative of (eq3) equals to $0$ (set velocity = $0$ ) to have

$-gt + v_{0} = 0$ , $\implies$ $t_{max} = \frac{v_{0}}{g}$ .

Check that the second derivative at this point is negative, i.e., $y''\left(\frac{v_{0}}{g}\right) = -g$ .

Maximum height possible the ball can attain is

$y(t)_{max} = -\frac{1}{2}g\left(\dfrac{v_{0}}{g} \right)^{2} + v_{0}\left(\frac{v_{0}}{g}\right) = -\frac{1}{2}\frac{v_{0}^{2}}{g} + \frac{v_{0}^{2}}{g} = \frac{v_{0}^{2}}{2g}$ .

Example 2

A ball weighing $\frac{3}{4} lb$ is thrown vertically upwards from a point $6 ft$ above the surface of the earth with an initial velocity of $20 ft/sec$ . As it rises it is acted upon by air resistance that is numerically equal to $\frac{1 }{64}v$ (in pounds), where $v$ is the velocity in feet per second. How high will the ball rise?

Solution

Let the height of the ball at at any time $t$ from the ground be $y(t)$ , and assume that the upward direction is positive. From Newton's second law of motion $F = ma$ , where $m$ is the mass of an object and $a$ is acceleration due to gravity,

$m\dfrac{dv}{dt} = -mg - kv$ , $k = \frac{1}{64}$

$\dfrac{dv}{dt} = -g - \frac{\frac{1}{64}}{m}v$

$\dfrac{dv}{dt} = -g - \frac{2}{3}v$

$3\dfrac{dv}{dt} = -96 - 2v$

Separating variables,

$-\frac{3}{2}\int\frac{dv}{v+48} = \int dt$

Integrating,

\begin{equation}\tag{1}-\frac{3}{2}ln(v+48) = t + c_{1} \end{equation}

Applying $v(0) = 20$ ,

$c_{1} = -\frac{3}{2}ln(68)$

From $ln(v+48) = -\frac{2t + 2c_{1}}{3}$

\end{equation}\tag{2} v = e^-\left({\dfrac{2t + c_{2}}{3}}\right) - 48[/tex], where $[tex]$ c_2 = -3ln(68) \end{equation}

At the maximum height, $y_{max}$ , $v = 0$

So $0 = e^-\left({\dfrac{2t_{max} + c_{2}}{3}}\right) - 48$ , where $t_{max}$ is the time the ball takes to reach maximum height.

$48 = e^-\left({\dfrac{2t_{max} + c_{2}}{3}}\right)$

$2t_{max} + c_{2} = -3ln(48)$

$\implies t_{max} = \dfrac{3ln(68) - 3ln(48)}{2}$

Hence \begin{equation}\tag{3} y(t)_{max} = \int_{0}^{t_{max}}\left( e^-\left({\dfrac{2t + c_{2}}{3}}\right) - 48 \right)dt \ + 6 \end{equation}

$y(t)_{max} = -\frac{3}{2}\left[e^-\left({\dfrac{2t + c_{2}}{3}} \right)- 48t \right]_{0}^{t_{max}} + 6$

$y(t)_{max} = -\frac{3}{2}e^-\left( \dfrac{3(ln(68) - ln(48)) - 3ln(68)}{3} \right ) - 72\left(ln(68) - ln(48) \right) - \left( -\frac{3}{2}e^-{\left(-ln(68) \right)}\right) + 6$

$y(t)_{max} = -\frac{3}{2}e^{ln(48)} - 72\left(ln(68) - ln(48) \right) + \frac{3}{2}e^{ln(68)} + 6$

$y(t)_{max} = -72 + 102 - 72\left(ln(68) - ln(48) \right) + 6 = 10.92191801 ft.$

To be continued.

Posted: **Sat Oct 01, 2016 1:05 am**

I tried to simulate Example 1:

Simulation helps to visualize what's real happen when the ball is thrown upward.

I modified the question but mathematical concepts are the same as highlighted by gaDgeT above. I will have three patterns regarding the ball

a) The ball thrown upward and bounces back on the surface on the ground, assuming it losses some energy upon every impact
b)The ball thrown upward and bounces back and no loss of energy upon impact
c) The ball thrown upward and doesn't bounce upon reaching the surface.

$F=-mg$

$ma=-mg$

$ma+mg=0$

$\frac{md^{2}x}{dt^{2}} + mg = 0$

$\frac{d^{2}x}{dt^{2}} + g = 0$ This is second order differential equation that's why we have double integration in the model to get position 'x'

The simulation will describe position 'x' of the ball at time t, and Velocity of the ball at time t. Maximum height reached and time can be just seen on the graph with initial conditions given.

Assumptions:

The ball is at initial height $10m$
An initial velocity is $10m/s$
Coefficient of restitution 0.8
Gravitation acceleration $9.8m/s^2$

The model

a)The ball thrown upward and bounces back on the surface on the ground, assuming it losses some energy upon every impact

b)The ball thrown upward and bounces back and no loss of energy upon impact

c)The ball thrown upward and doesn't bounce upon reaching the surface.

Posted: **Sat Oct 01, 2016 11:19 am**

Great stuff! That's why mathematics makes a lot of sense when visualized! Technology in teaching and learning is increasingly becoming inevitable. @Simulink

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