User:Egm6321.f11.team1.colsonfe/HW1

= [R1.1] Second Total Time Derivative =

Given
$$\displaystyle \dot Y :=\frac{dY^1(t)}{dt}$$

$$\displaystyle f_{,S}(Y^1,t):=\frac{\partial{f(Y^1,t)}}{\partial{S}}$$

$$\displaystyle f_{,S t}(Y^1,t):=\frac{\partial^2{f(Y^1,t)}}{\partial{S}\partial{t}}$$

Find
Show that the following equality is true.

$$\displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1+f_{,SS}(Y^1,t)(\dot Y^1)^2+2f_{,S t}(Y^1,t)+f_{tt}(Y^1,t)$$

Basic Equations
First Total Derivative:


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$$ $$
 * $$\displaystyle \frac{d}{dt}f(Y^1(t),t)=\frac{\partial{f(Y^1(t),t)}}{\partial{S}}\dot{Y}^1+\frac{\partial{f(Y^1(t),t)}}{\partial{t}}
 * $$\displaystyle \frac{d}{dt}f(Y^1(t),t)=\frac{\partial{f(Y^1(t),t)}}{\partial{S}}\dot{Y}^1+\frac{\partial{f(Y^1(t),t)}}{\partial{t}}
 * $$\displaystyle (Equation\;1.1.1)
 * }
 * }

Chain Rule: 

If $$y=f(u),u=g(x)$$, and the derivatives $$\frac{dy}{du}$$ and $$\frac{du}{dx}$$ both exist, then the composite function defined by $$y=f(g(x))$$ has a derivative given by

$$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=f'(u)g'(x)=f'(g(x))g'(x) $$

Solution
The following steps make use of algebra of derivatives and the chain rule.

$$ \displaystyle \frac{d^2f}{dt^2}=\frac{d}{dt}\left [ \frac{df}{dt} \right ] $$

$$ \displaystyle \frac{d^2f}{dt^2}=\frac{d}{dt}\left [ \frac{\partial{f(Y^1(t),t)}}{\partial{S}}\dot{Y}^1+\frac{\partial{f(Y^1(t),t)}}{\partial{t}} \right ] $$

$$ \displaystyle \frac{d^2f}{dt^2}=\frac{d}{dt}\left[\frac{\partial}{\partial S}f(Y^1(t),t)\dot Y^1 \right ]+\frac{d}{dt}\left[\frac{\partial}{\partial t}f(Y^1(t),t) \right ] $$

$$ \displaystyle \frac{d^2f}{dt^2}=\left [\frac{\partial^2}{\partial S\partial t}f(Y^1(t),t)\dot Y^1+\frac{\partial}{\partial S}\frac{\partial f(Y^1(t),t)}{\partial t}\dot Y^1+\frac{\partial}{\partial S}f(Y^1(t),t)\ddot Y^1 \right]+\left [\frac{\partial^2}{\partial t^2}f(Y^1(t),t)+\frac{\partial}{\partial t}\frac{\partial f(Y^1(t),t)}{\partial t}\right] $$

NOTE: $$ \displaystyle \frac{\partial}{\partial t}\frac{\partial f(Y^1(t),t)}{\partial t}=\frac{\partial}{\partial S}\frac{\partial S}{\partial t}\left [\frac{\partial f}{\partial S}\frac{\partial S}{\partial t}\right]=\frac{\partial}{\partial S}\dot Y^1\left[\frac{\partial f}{\partial S}\dot Y^1\right]=\frac{\partial^2 f(Y^1(t),t)}{\partial^2 S}(\dot Y^1)^2 $$

$$ \displaystyle \frac{d^2f}{dt^2}=2\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\dot Y^1+\frac{\partial f(Y^1(t),t)}{\partial S}\ddot Y^1+\frac{\partial^2 f(Y^1(t),t)}{\partial S^2}(\dot Y^1)^2 +\frac{\partial^2 f(Y^1(t),t)}{\partial t^2} $$

$$ \displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1+f_{SS}(Y^1,t)(\dot Y^1)^2+2f_{S t}(Y^1,t)\dot Y^1+f_{tt}(Y^1,t) $$

 Solution by Egm6321.f11.team1.colsonfe 21:15, 7 September 2011 (UTC)

= [R1.2] Coriolis Force Derivation =

Find
Derive the following two equations:

$$\displaystyle \frac{d}{dt}f(Y^1(t),t)=\frac{\partial{f(Y^1(t),t)}}{\partial{S}}\dot{Y}^1+\frac{\partial{f(Y^1(t),t)}}{\partial{t}}$$

$$\displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1+f_{,SS}(Y^1,t)(\dot Y^1)^2+2f_{,S t}(Y^1,t)+f_{tt}(Y^1,t)$$

Show similarity between the above derivations and derivation of Coriolis Force.

Basic Equations
Note the following notation of using $$S_0$$ for inertial (non-moving) and $$S$$ for rotating frames.

Note that I define the position vector $$\mathbf r =x \hat{\mathbf x}+y \hat{\mathbf y}+z \hat{\mathbf z}$$.

Newton's Second Law:

$$m\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\mathbf{F}$$

Time Derivative in a Rotating Frame: 


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\left(\frac{d\mathbf{Q}}{dt}\right)_{S_0}=\left(\frac{d\mathbf{Q}}{dt}\right)_S+\mathbf\Omega\times\mathbf Q $$ $$ Note this identity relates the derivative of any one vector $$\mathbf Q$$ as measured in the inertial frame $$S_0$$ to the corresponding derivative in the rotating frame $$S$$.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Equation\;1.2.1)
 * }
 * }

Note that $$\mathbf\Omega$$ is the angular velocity of the rotating frame.

Solution
Derive Equation 1.1.1:

Via application of chain rule we can begin with

$$\frac{df}{dt}=\frac{\partial f}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial t} $$

$$ \frac{df}{dt}=\frac{\partial f}{\partial S}\dot Y^1+\frac{\partial f}{\partial t}\cdot 1 $$

$$ \frac{df(Y^1(t),t)}{dt}=\frac{\partial f(Y^1(t),t)}{\partial S}\dot Y^1+\frac{\partial f(Y^1(t),t)}{\partial t} $$

$$f$$ Derivations Summary

Coriolis Derivation

Let's use Equation 1.2.1 to help us express $$\mathbf r$$ in terms of derivatives.

$$\left(\frac{d\mathbf{r}}{dt}\right)_{S_0}=\left(\frac{d\mathbf{r}}{dt}\right)_S+\mathbf\Omega\times\mathbf r$$

Differentiating a second time we find

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt}\right)_{S_0}\left(\frac{d\mathbf{r}}{dt}\right)_{S_0}$$

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt}\right)_{S_0}\left[\left(\frac{d\mathbf{r}}{dt}\right)_{S}+\mathbf\Omega\times\mathbf r\right]$$

Thus again applying our rotating frame equation we find

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt}\right)_{S}\left[\left(\frac{d\mathbf{r}}{dt}\right)_{S}+\mathbf\Omega\times\mathbf r\right]+\mathbf\Omega\times\left[\left(\frac{d\mathbf{r}}{dt}\right)_{S}+\mathbf\Omega\times\mathbf r\right]$$

Once expanded this messy result can be rewritten using dot notation:

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\mathbf{\ddot r}+2\mathbf\Omega\times\mathbf{\dot r}+\mathbf\Omega\times(\mathbf\Omega\times\mathbf r)$$

Now substitute the result into Newton's Second Law

$$m\mathbf{\ddot r}=\mathbf F +2m\mathbf{\dot r}\times\mathbf{\Omega}+m(\mathbf\Omega\times\mathbf r)\times\mathbf\Omega$$

$$\mathbf F$$ denotes the sum of all the forces as identified in any inertial frame. However, there are two additional terms on the right side of the equation. The final term is the centrifugal force. The middle term is what we are interested in here, this is the Coriolis force.

$$\mathbf F_{cor} = 2m\mathbf{\dot r}\times\mathbf\Omega$$

Coriolis Derivations Summary

Qualitatively we can visualize the parallelism between the first derivatives and second derivatives by noting the difference between the two tables presented above. We can do a term by term comparison of the first derivative of $$f(Y^1(t),t)$$ and velocity and the second derivative of $$f(Y^1(t),t)$$ and acceleration.

 Solution by Egm6321.f11.team1.colsonfe 02:17, 12 September 2011 (UTC)

= References =