User:Egm6321.f12.team07.liu/report3

=R*3.9=

Problem 9: IFM application in the motion of a particle in the air
Report problem 9 from.

Given: the motion of a particle in the air
The schematic of the motion of a particle in the air:

Find: Verify the solution to the equation
1.  Derive the equations of motion:

2.  Particular case $$k=0$$:Verify that $$y(x)$$ is parabola

3.  Consider the case $$k\neq 0$$ and $$v_{x0}=0$$: ($$) turns into

(1). Is ($$) for $$n=0,1,2$$ either exact or can be made exact using IFM? Find $$v_y(t)$$ and $$y(t)$$ for $$m$$ constant.

(2). Find $$v_y(t)$$ and $$y(t)$$ for $$m=m(t)$$ varying with $$t$$ as following:

Solution

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On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
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1. Derive the equations of motion

According to ($$):

First, we divide $$v$$ into $$v_x$$ and $$v_y$$ components, and we have: $$v^2=(v_x)^2+(v_y)^2$$ and $$tan\alpha=\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}=\frac{v_y}{v_x}$$.

Then, we could use Newton's Second Law: $$F=ma$$, respectively, in X direction and Y direction:


 * X direction: $$-kv^n*cos\alpha=m\frac{dv_x}{dt}$$
 * Y direction: $$-kv^n*sin\alpha-mg=m\frac{dv_y}{dt}$$

Thus, ($$), ($$), ($$) and ($$) are derived.

2. Particular case $$k=0:$$ Verify that $$y(x)$$ is parabola.

When $$k=0$$, ($$) is turned into

Thus, through integration, we obtain:

3.  Consider the case $$k\neq 0$$ and $$v_{x0}=0$$:

When $$k\neq 0$$, ($$) is turned into

(1). Is ($$) for $$n=0,1,2$$ either exact or can be made exact using IFM? Find $$v_y(t)$$ and $$y(t)$$ for $$m$$ constant.

Rewrite ($$):

FIRST, suppose $$\frac{dM}{v_y}=\frac{dN}{dt}$$, under such condition, ($$) is exact:

we get, $$n=0$$. thus, when $$n=0$$, ($$) is exact.

THEN, suppose ($$) is not exact but could be made exact. We introduce Euler Integrating Factor $$h(t)$$ and dot it with ($$):

then, we have:

moreover,

we find $$n=1$$ satisfies ($$), but $$n=2$$ doesn't.

THUS: When $$n=0$$ ($$) is exact.

When $$n=1$$ ($$) is not exact but could be made exact.

When $$n=2$$ ($$) is neither exact nor could be made exact.

NEXT, we will find $$v_y(t)$$ and $$y(t)$$ for $$m$$ constant:

When $$n=0$$,($$) is reduced to:

through integration, we obtain:

When $$n\neq 0$$, we start from ($$):

through integration, we obtain:

where $$C_1$$ is the integration constant.

institue ($$) into ($$), we obtain:

through integration:

(2). Find $$v_y(t)$$ and $$y(t)$$ for $$m=m(t)$$ varying with $$t$$ as ($$):

Under such condition:

we can verify that when $$n=0,1,2$$, $$\frac{dM'}{dv_y}\neq \frac{dN'}{dt}$$ i.e. ($$) is not exact.

Thus we need to introduce a Euler Integrating Factor $$H(t)$$ to make it exact:

dot ($$) with $$H(t)$$:

then we have:

make it explicit:

According to the graphic ($$): {{NumBlk|:|$$m=\left\{\begin{matrix} K_0t+m_0\;(0\leqslant t\leqslant t_1)\\m_1\;\;\;\;(t_1< t) \end{matrix}\right.$$|$$}} where: $$K_0=-\frac{m_0-m_1}{t_1}$$.

When $$n=0$$, we have:

thus:

institute ($$) into ($$) and take integration, we obtain:

{{NumBlk|:|$$v_y=\left\{\begin{matrix} -gt-\frac{k}{K_0}\log(\frac{K_0}{m_0}t+1)+v_{y_0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\\-gt-\frac{k}{K_0}\log(\frac{K_0}{m_0}t_1+1)+v_{y_0}+\frac{k}{m_1}(t-t_1)\;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

{{NumBlk|:|$$y(t)=\left\{\begin{matrix}-\frac{1}{2}gt^2-\frac{k}{K_0}\int_{0}^{t}\log (\frac{K_0}{m_0}t+1)dt+v_{y_0}t\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\\-\frac{1}{2}(\frac{k}{m_1}-g)^2-(\frac{kt_1}{m_1}+\frac{k}{K_0}\log \frac{m_1}{m_0}-v_{y_0})t\;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

When $$n=1$$, we have:

thus: {{NumBlk|:|$$H=\left\{\begin{matrix} m^{\frac{k}{K_0}}\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\frac{e^{\frac{kt}{m_1}}}{m}\;\;\;\;(t\geqslant t_1)\end{matrix}\right.$$|$$}}

institute ($$) into ($$) and take integration, we obtain:

{{NumBlk|:|$$v_y=\left\{\begin{matrix}   -\frac{mg}{k+K_0}+\frac{m_0^{k/k_0} m^{(K_0-k)/K_0}g}{k+K_0}+v_{y_0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\\  \frac{mg(m_0^{k/K_0}-m_1^{k/K_0})}{e^{kt/m_1}(k+K_0)}+\frac{m_1(e^{kt/m_1}-e^{kt_1/m_1})}{k}+v_{y_0}\;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

{{NumBlk|:|$$y(t)=\left\{\begin{matrix}  -\frac{(\frac{1}{2}K_0t^2+m_0t)g}{k+K_0}+\frac{K_0m_0^{k/K_0}m^{2-k/K_0}g}{(2K_0-k)(k+K_0)}+v_{y_0t}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\frac{(\frac{1}{2}K_0t^2+m_0t)g(m_0^{k/K_0}-m_1^{k/K_0})}{e^{kt/m_1}(k+K_0)}+(m_1/k)^2e^{kt/m_1}-\frac{m_1t}{k}e^{kt_1/m_1}+v_{y_0}t   \;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

When $$n=2$$, we have:

Obviously, ($$) could not be satisfied. Thus as for $$n=2$$, we still can't make it exact.