Difference between revisions of "Forest UCM PnCP ProjMotion"

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:<math>\frac{1}{1+x} = f(x=0) + \left . \frac{d f(x)}{dx} \right |_{x=0} \frac{x-0}{1!} + \left . \frac{d^2 f(x)}{dx^2} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots = \sum_{n=0}^\infty \left . \frac{d^n f(x)}{dx^n} \right |_{x=0} \frac{(x-0)^n}{n!}</math>  
:<math>\frac{1}{1+x} = f(x=0) + \left . \frac{d f(x)}{dx} \right |_{x=0} \frac{x-0}{1!} + \left . \frac{d^2 f(x)}{dx^2} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots = \sum_{n=0}^\infty \left . \frac{d^n f(x)}{dx^n} \right |_{x=0} \frac{(x-0)^n}{n!}</math>  
::<math>=1 - \left . (1+x)^{-2} \right |_{x=0} \frac{(x-0)^1}{1!} + 2 \left . (1+x)^{-3} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots</math>
::<math>=1 - \left . (1+x)^{-2} \right |_{x=0} \frac{(x-0)^1}{1!} + 2 \left . (1+x)^{-3} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots</math>
::<math> = 1 -x + \frac{(x)^2}{2!} - \frac{(x)^3}{3!} + \cdots =</math>  
::<math> = 1 -x + 2\frac{(x)^2}{2!} - 2*3 \frac{(x)^3}{3!} + \cdots </math>
::<math> = 1 -x + x^2 - x^3 + \cdots </math>  

Revision as of 14:05, 1 September 2014

Projectile Motion

Friction depends linearly on velocity

Projectile motion describes the path a mass moving in two dimensions. An example of which is the motion of a projectile shot out of a cannon with an initial velocity [math]v_0[/math] with an angle of inclination [math]\theta[/math].

When the motion in each dimension is independent, the kinematics are separable giving you two equations of motion that depend on the same time.

Using our solutions for the horizontal and vertical motion when friction depends linearly on velocity (Forest_UCM_PnCP_LinAirRes) we can write :

[math]x= \frac{m}{b} v_{ix} \left ( 1-e^{-\frac{b}{m}t} \right )[/math]

where [math]v_{ix}[/math] has replced [math]v_i[/math] so the components are more explicitly identifiable.

in the y-direction however, the directions are changed to represent an object moving upwards instead of falling

Newton's second law for falling

[math]\sum \vec{F}_{ext} = mg -bv = m \frac{dv}{dt}[/math]


[math]\sum \vec{F}_{ext} = -mg +bv = m \frac{dv}{dt}[/math]

for a rising projectile

This changes the signs in front of the [math]v_t[/math] terms such that

[math]y= v_t t + \frac{m}{b}\left ( v_0 - v_t \right ) \left ( 1- e^{-\frac{b}{m}t} \right ) [/math]


[math]y = -v_t t + \frac{m}{b}\left ( v_{iy} + v_t \right ) \left ( 1- e^{-\frac{b}{m}t} \right ) [/math]

where [math]v_{iy}[/math] has replced [math]v_0[/math] so the components are more explicitly identifiable.

We now have a system governed by the following system of two equations


[math]\tau \equiv \frac{m}{b}[/math]

[math]x= \tau v_{ix} \left ( 1-e^{-\frac{t}{\tau}} \right )[/math]
[math]y = \tau\left ( v_{iy} + v_t \right ) \left ( 1- e^{-\frac{t}{\tau}} \right ) -v_t t[/math]

Range equation

To determine how far the projectile will travel in the x-direction (Range) you can solve the above equation for [math]y[/math] in the case that [math]y=0[/math].

since time is the same in both equations you can solve for time in terms of x and substitute for time inthe y-direction equations.

solving for [math]e^{-\frac{t}{\tau}}[/math] using the x-direction equation

[math]x= \tau v_{ix} \left ( 1-e^{-\frac{t}{\tau}} \right )[/math]
[math]\Rightarrow e^{-\frac{t}{\tau}} = 1- \frac{x }{v_{ix} \tau}[/math]

substituting for [math]e^{-\frac{t}{\tau}}[/math]

[math]y = \tau \left ( v_{iy} + v_t \right ) \left ( \frac{x }{v_i \tau} \right ) -v_t t[/math]
[math]= \frac{ v_{iy} + v_t }{v_i} x -v_t t[/math]

now we need to substitute for time [math]t[/math]

[math] e^{-\frac{t}{\tau}} = 1- \frac{x }{v_{ix} \tau}[/math]
[math]\Rightarrow \ln \left ( e^{-\frac{t}{\tau}} \right) = \ln \left ( 1- \frac{x }{v_{ix} \tau}\right)[/math]
[math]-\frac{t}{\tau} = \ln \left ( 1- \frac{x }{v_{ix} \tau}\right)[/math]
[math]t = -\tau\ln \left ( 1- \frac{x }{v_{ix} \tau}\right)[/math]

substituting for time

[math]y =\frac{ v_{iy} + v_t }{v_{ix}} x -v_t t[/math]
[math] =\frac{ v_{iy} + v_t }{v_{ix}} x + v_t \tau\ln \left ( 1- \frac{x }{v_{ix} \tau}\right)[/math]

The Range [math](R)[/math] is defined as the value for [math]x[/math] when [math]y =0[/math]

[math]0 = \frac{ v_{iy} + v_t }{v_{ix}} R + v_t \tau\ln \left ( 1- \frac{R }{v_{ix} \tau}\right)[/math]
[math]v_{ix} = v_0 \cos \theta [/math]
[math]v_{iy} = v_0 \sin \theta[/math]

The above equation does not have an exact analytical solution.

You can try to solve it graphically or by taylor expanding small quantities when they appear as arguments to functions like the [math]\ln[/math] function

Solution by Taylor expansion


ln(1-x) Taylor expansion

Taylor expanding about x = 0 for

[math]\frac{1}{1+x} = f(x=0) + \left . \frac{d f(x)}{dx} \right |_{x=0} \frac{x-0}{1!} + \left . \frac{d^2 f(x)}{dx^2} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots = \sum_{n=0}^\infty \left . \frac{d^n f(x)}{dx^n} \right |_{x=0} \frac{(x-0)^n}{n!}[/math]
[math]=1 - \left . (1+x)^{-2} \right |_{x=0} \frac{(x-0)^1}{1!} + 2 \left . (1+x)^{-3} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots[/math]
[math] = 1 -x + 2\frac{(x)^2}{2!} - 2*3 \frac{(x)^3}{3!} + \cdots [/math]
[math] = 1 -x + x^2 - x^3 + \cdots [/math]