User:Egm6321.f11.team1.colsonfe/HW3

= R*3.5 Derive Equations of Motion =

Given
Schematic of motion of particle in air:


 * Egm6321.f11.team1.figure1.png

Particle velocity
 * $$v:=\parallel\boldsymbol v \parallel$$

Constants
 * $$k,n\in\mathbb R$$

Mass of particle = $$\displaystyle m$$

Acceleration of gravity = $$\displaystyle g$$

== Find ==

(1) Derive the equations of motion


 * {| style="width:100%" border="0" align="left"

$$
 * $$m\frac{dv_x}{dt}=-kv^n\cos\alpha$$
 * $$\displaystyle (Equation\;3.5.1)
 * $$\displaystyle (Equation\;3.5.1)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$
 * $$m\frac{dv_y}{dt}=-kv^n\sin\alpha-mg$$
 * $$\displaystyle (Equation\;3.5.2)
 * $$\displaystyle (Equation\;3.5.2)
 * }
 * }


 * $$\displaystyle v^2=(v_x)^2+(v_y)^2$$


 * $$\tan\alpha=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{v_y}{v_x}$$

(2) For the particular case $$\displaystyle k=0$$: Verify that $$\displaystyle y(x)$$ is a parabola.

(3) Consider the case $$\displaystyle k \ne 0$$ and $$\displaystyle v_{x0}=0$$.


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$$
 * $$m\frac{dv_y}{dt}=-k(v_y)^n-mg$$
 * $$\displaystyle (Equation\;3.5.3)
 * $$\displaystyle (Equation\;3.5.3)
 * }
 * }


 * (3.1) Is Equation 3.5.3 either exact or can be made exact using Integrating Factor Method? Find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m$$ constant.


 * (3.2) Find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m=m(t)$$.

Part (1)
For a non-accelerating particle in motion, the sum of the forces is zero. Consider the forces in the x-direction:

$$\sum F_x=0=ma_x+kv^n\cos\alpha=m\frac{dv_x}{dt}+kv^n\cos\alpha$$

Now consider the forces in the y-direction.

$$\sum F_y=0=ma_y+kv^n\sin\alpha+mg=m\frac{dv_y}{dt}+kv^n\sin\alpha+mg$$

By given and definition of norm :

$$v=\parallel\boldsymbol v\parallel=\sqrt{(v_x)^2+(v_y)^2}$$

Thus it is clear that

Apply definition of slope to a differential portion.

$$\tan\alpha=\frac{\rm opposite}{\rm adjacent}=\frac{dy}{dx}$$

Now apply chain rule and definition of velocity.

$$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{v_y}{v_x}$$

Put it all together.

Part (2)
Consider Equation (3.5.1) and Equation (3.5.2) for the case of $$\displaystyle k=0:$$

$$m\frac{dv_x}{dt}=0$$

$$m\frac{dv_y}{dt}=0-mg$$

Which implies

$$\frac{dv_x}{dt}=0$$

$$\frac{dv_y}{dt}=-g$$

Integrate

$$\displaystyle v_x(t)=v_{x0}$$

$$\displaystyle v_y(t)=-gt+v_{y0}$$

Integrate again to calculate position.

$$\displaystyle x(t)=v_{x0}t+x_0$$

$$y(t)=-g\frac{t^2}{2}+v_{y0}t+y_0$$

Solve $$\displaystyle x(t)$$ for $$\displaystyle t$$ so that we can eliminate time dependency from $$\displaystyle y(t)$$.

$$t=\frac{x-x_0}{v_{x0}}$$

Entering this into $$\displaystyle y(t)$$ yields

This is the form of a paraboloid.

Section 3.1
To test Equation 3.5.3 for exactness, it must pass the two Exactness Conditions. Rearrange as such:

$$\left[k(v_y)^n+mg\right]+\left[m\right]\frac{dv_y}{dt}=0$$

If we define the following values:

$$\displaystyle M(v_y,t):=k(v_y)^n+mg$$

$$\displaystyle N(v_y,t):=m$$

Then we have

$$\displaystyle M(v_y,t)+N(v_y,t)v_y'=0$$

This satisfies the First Exactness Condition. The second is satisfied if

$$\frac{\partial M(v_y,t)}{\partial v_y}=\frac{\partial N(v_y,t)}{\partial t}$$

But (assuming mass $$\displaystyle m$$ is a function of time)

$$kn(v_y)^{n-1}\ne \frac{dm}{dt}$$

Thus the Second Exactness Condition is not satisfied. Let us try to transform the ODE with an integrating factor to make it exact. We need to find an $$\displaystyle h(v_y,t)$$ so that the following N1-ODE is exact:

$$h(v_y,t)\left[M(v_y,t)+N(v_y,t)v_y'\right]=0$$


 * {| style="width:100%" border="0" align="left"

$$
 * $$\left[h(v_y,t)k(v_y)^n+h(v_y,t)mg\right]+\left[h(v_y,t)m\right]\frac{dv_y}{dt}=0$$
 * $$\displaystyle (Equation\;3.5.4)
 * $$\displaystyle (Equation\;3.5.4)
 * }
 * }

Attempting to find a $$\displaystyle h(v_y,t)$$ that makes Equation 3.5.4 exact is extremely difficult, as explained in R3.11 (Add Link). For these reasons, I claim that there does not exist an Integrating Factor of this form. However, there are two cases in which it is possible to find an Integrating Factor. In both cases I will be pulling equations from the class lecture notes.

CASE 1

Suppose $$\displaystyle h_t(v_y,t)=0$$, thus $$\displaystyle h$$ is a function of $$\displaystyle v_y$$ only. From the notes, we define

$$n(v_y):=-\frac{1}{N}(N_t-M_{v_y})=\frac{1}{m}\left[\frac{dm}{dt}-kn(v_y)^{n-1}\right]$$

Also from the same source,

CASE 2

Now suppose $$\displaystyle h_{v_y}(v_y,t)=0$$, thus $$\displaystyle h$$ is a function of $$\displaystyle t$$ only. From the notes cited above, we define

$$m(t):=-\frac{1}{M}(N_t-M_{v_y})=\frac{1}{k(v_y)^n-mg}\left[\frac{dm}{dt}-kn(v_y)^{n-1}\right]$$

From R3.12 (Add Link) we know $$\displaystyle h(t)$$:

The next step of this problem is to find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m$$ constant. Solving the ODE directly is challenging due to the complexity $$\displaystyle (v_y)^n$$ poses. To counter this problem I will consider a few common cases.

$$\displaystyle n=0$$

This case represents vertical motion without air resistance. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-k-mg$$

$$v_y(t) = -\frac{k+mg}{m}t+v_{y0}$$

$$y(t)= -\frac{k+mg}{2m}t+^2v_{y0}t+y_0$$

$$\displaystyle n=1$$

This case represents vertical motion with a linear air resistance factor. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-kv_y-mg$$

This is a L1-ODE solvable via Integration Factor Method, using $$h=\int \exp\left(\frac{k}{m} \right )dt$$. Using this yields the solution

$$v_y(t)=C_1\exp^{-\frac{k}{m}t}-\frac{gm}{k}$$

$$y(t)=-C_1\frac{k}{m}\exp^{-\frac{k}{m}t}-\frac{gm}{k}t+C_2$$

$$\displaystyle n=2$$

This case represents vertical motion with a quadratic air resistance factor. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-kv_y^2-mg$$

This is a N1-ODE. I will choose to use Wolfram-Alpha to solve this equation.

$$v_y(t)=\frac{\sqrt g\sqrt m\tan(\frac{C_1\sqrt g\sqrt k m-\sqrt g\sqrt k t}{\sqrt m})}{\sqrt k}$$

$$y(t)=\frac{2m\log\left( \cos\left( \frac{\sqrt g\sqrt k(C_1m-t)}{\sqrt m}\right)\right)}{t-C_1m}+C_2$$

$$\displaystyle n=3$$

This case represents vertical motion with a cubic air resistance factor. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-kv_y^2-mg$$

This is another N1-ODE. I used Wolfram-Alpha to solve this equation, but it yielded an implicit equation for $$\displaystyle v_y(t)$$.



Note that an implicit form of $$\displaystyle v_y(t)$$ makes it impossible to solve for $$\displaystyle y(t)$$.

$$\displaystyle n\geq4$$

This case represents vertical motion with an resistance factor that is proportional to some power (greater than 3) of the velocity. Solving this via Wolfram-Alpha yields similarly complex implicit solutions as in the $$\displaystyle n=3$$ case. Thus any further investigation is non informative.

Section 3.2
I need to find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m=m(t)$$, which is defined below.



$$\displaystyle n=0$$

This case represents vertical motion without air resistance. Consider Equation (3.5.3).

$$m(t)\frac{dv_y}{dt}=-k-m(t)g$$

$$\frac{dv_y}{dt}=-\frac{k}{m(t)}-g$$

$$dv_y=\left[-\frac{k}{m(t)}-g\right]$$

Use the equation $$\displaystyle m(t)=-\frac{m_0-m_1}{t_1}t-m_0$$ to define mass on the interval $$\displaystyle\left[0,t_1 \right ]$$. It is derived from simple slope-intercept equation. Elsewhere $$\displaystyle m(t)=m_1$$.

Note change to dummy integration variable $$\displaystyle s$$.

$$v_y(t)=\int_{0}^{t}\left[-\frac{k}{m(s)}-g \right ]ds$$

$$v_y(t)=\int_{0}^{t_1}\left[-\frac{k}{m(s)}\right]ds+\int_{t_1}^{t}\left[-\frac{k}{m_1}\right]ds+\int_{0}^{t}\left[-g\right ]ds$$

$$v_y(t)=\int_{0}^{t_1}\left[-\frac{k}{-\frac{m_0-m_1}{t_1}s-m_0}\right]ds-\frac{k}{m}(t-t_1)-gt+C_1$$

$$v_y(t)=\left[k\frac{t_1}{m_0-m_1}\ln\left(\frac{m_0-m_1}{t_1}s+m_0 \right )\right]_{0}^{t_1}-\frac{k}{m}(t-t_1)-gt+C_1$$

$$v_y(t)=k\frac{t_1}{m_0-m_1}\left[\ln(2m_0-m_1)-\ln(m_0)\right ]-\frac{k}{m_1}(t-t_1)-gt+C_1$$

$$v_y(t)=k\frac{t_1}{m_0-m_1}\ln\left(\frac{2m_0-m_1}{m_0}\right)-\left(\frac{k}{m_1}-g \right )t-\frac{kt_1}{m_1}+C_1$$

And thus it follows that

$$y(t)=k\frac{t_1}{m_0-m_1}\ln\left(\frac{2m_0-m_1}{m_0}\right)t-\left(\frac{k}{m_1}-g \right )\frac{t^2}{2}-\frac{kt_1}{m_1}t+C_1t+C_2$$

$$\displaystyle n\geq1$$

Further cases introduce two functions of $$\displaystyle t$$ (which are $$\displaystyle v_y(t)$$ and $$\displaystyle m(t)$$) that are incorporated into an ODE. These are feasible problems but beyond the scope of this report.

= R*3.17 Derive Equation via RTT =

Given
Component form of gradient/divergence property:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\frac{\partial\rho}{\partial x_i}u_i+\rho\,\frac{\partial u_i}{\partial x_i}=\frac{\partial}{\partial x_i}(\rho u_i)$$
 * $$\displaystyle (Equation\;3.17.1)
 * $$\displaystyle (Equation\;3.17.1)
 * }
 * }

Conservation of mass:


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$$
 * $$\frac{\partial\rho}{\partial t}+{\rm div}(\rho\mathbf u)=0$$
 * $$\displaystyle (Equation\;3.17.2)
 * $$\displaystyle (Equation\;3.17.2)
 * }
 * }

Reynolds Transport Theorem:


 * $$\frac{D}{Dt}\int_{\beta_t}f(x,t)\,d\beta_t=\int_{\beta_t}\left[\frac{\partial f}{\partial t}+{\rm div}(f\mathbf u)\right]\,d\beta_t$$

== Find ==

Show:


 * $$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\rho\frac{D\mathbf u}{Dt}\,d\beta_t$$

Solution
Apply $$\displaystyle f(x,t)=\rho(x,t)\,u_i(x,t)$$ to Reynolds Transport Theorem.

$$\frac{D}{Dt}\int_{\beta_t}\rho(x,t)\,u_i(x,t)\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\left(\rho(x,t)\,u_i(x,t)\right)}{\partial t}+{\rm div}(\rho(x,t)\,u_i(x,t)\mathbf u)\right]\,d\beta_t$$

$$\frac{D}{Dt}\int_{\beta_t}\rho\,u_i\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\left(\rho u_i\right)}{\partial t}+{\rm div}(\rho\mathbf u\mathbf u)\right]\,d\beta_t$$

Now apply Equation 3.17.1 to the first term under the integral on the right hand side:

$$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\left[\left(\frac{\partial\rho}{\partial t}u_i+\rho\frac{\partial u_i}{\partial t}\right)+{\rm div}(\rho\mathbf u\mathbf u)\right]\,d\beta_t$$

Group terms:

$$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\rho}{\partial t}u_i+\left(\rho\frac{\partial \mathbf u}{\partial t}+{\rm div}(\rho\mathbf u\mathbf u)\right)\right]\,d\beta_t$$

Via application of Equation 3.17.2 we can eliminate the terms under the parentheses.

$$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\rho}{\partial t}u_i\right]\,d\beta_t$$

To complete the problem, we can use the definition of material time derivative:

$$\frac {D u_i(x,t)}{Dt}=\left.\frac{d u_i(x,t)}{dt}\right|_{X \rm fixed}=\frac{\partial u_i}{\partial t}$$

= R*3.18 Relate RTT to the 1D Case =

Given
Adjusted Reynold's Transport Theorem:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\frac{D}{Dt}\int_{\beta_t}f(x,t)\,d\beta_t=\int_{\beta_t}\frac{\partial f}{\partial t}\,d\beta_t+\int_{\partial\beta_t}\mathbf n\cdot(f\mathbf u)\,d(\partial\beta_t)$$
 * $$\displaystyle (Equation\;3.17.3)
 * $$\displaystyle (Equation\;3.17.3)
 * }
 * }

== Find ==

Obtain the following by means of Equation 3.17.3.


 * $$\frac{d}{dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=f(t,b(t))\,\frac{db(t)}{dt}-f(t,a(t))\frac{da(t)}{dt}+\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}ds$$

Solution
Let us choose our function $$\displaystyle f$$ to be a function of $$\displaystyle s$$ and $$\displaystyle t$$, so that $$\displaystyle f=f(t,s)$$. Let $$\displaystyle\mathcal B_t$$ be denoted by the interval between $$\displaystyle s=a(t)$$ and $$\displaystyle s=b(t)$$ so that $$\displaystyle \mathcal B_t \mapsto s$$. This has the following effect on Equation 3.17.3:

$$\frac{D}{Dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}\,ds+\int_{\partial s}f(s,t)\mathbf n\cdot\mathbf u\,d(\partial s)$$

Since the left hand side does not depend on $$\displaystyle x$$, the definition of material time derivative simplifies to $$\frac{D}{Dt}=\frac{d}{dt}.$$ Also the velocity of the material point, $$\displaystyle\mathbf u$$, is in the same direction of $$\displaystyle\mathbf n$$, and will be a derivative of $$\displaystyle s$$ rather than $$\displaystyle x.$$

$$\frac{d}{dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}\,ds+\int_{\partial s}f(s,t)\frac{\partial s}{\partial t}\,d(\partial s)$$

$$\frac{d}{dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}\,ds+\left.\left[f(t,s)\frac{\partial s}{\partial t} \right ]\right|^{s=b(t)}_{s=a(t)}$$

Note we can transform the partial derivatives to regular derivatives since $$\displaystyle s$$ only depends on $$\displaystyle t$$. The next step of evaluating the integration at the bounds completes the proof.

$$\frac{d}{dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=f(t,b(t))\,\frac{db(t)}{dt}-f(t,a(t))\frac{da(t)}{dt}+\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}\,ds$$

= References =