Difference between revisions of "Forest UCM PnCP LinAirRes"

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; Horizontal motion  
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Linear Air Resistance=
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=Horizontal motion =
  
 
If <math>n</math> is unity then the velocity is exponentially approaching zero.
 
If <math>n</math> is unity then the velocity is exponentially approaching zero.
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:: <math>=  v_i \left ( \frac{m}{b} e^{-\frac{b}{m}0} -\frac{m}{b} e^{-\frac{b}{m}t} \right ) </math>
 
:: <math>=  v_i \left ( \frac{m}{b} e^{-\frac{b}{m}0} -\frac{m}{b} e^{-\frac{b}{m}t} \right ) </math>
 
:: <math>=  \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )</math>
 
:: <math>=  \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )</math>
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=Example: falling object with linear air friction=
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Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity
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:<math>\sum \vec{F}_{ext} = mg -bv = m \frac{dv}{dt}</math>
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let
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:<math>v_t = \frac{mg}{b}</math>
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:<math> v_t -v = \frac{1}{b} \frac{dv}{dt}</math>
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:<math>  b dt= \frac{dv}{v_t -v} </math>
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:<math>  -b dt= \frac{dv}{v -v_t} </math>
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:<math>  -\int_0^t b dt= \int_{v_0}^v \frac{dv}{v -v_t} </math>
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:<math>  -bt = \ln{\left( v -v_t \right)} - \ln{\left ( v_0-v_t \right )}</math>
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:<math>  -bt = \ln \left(\frac{ v -v_t  }{v_0-v_t}\right )</math>
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:<math>  e^{-bt} =  \left(\frac{ v -v_t  }{v_0-v_t}\right )</math>
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:<math>  v -v_t = \left ( v_0-v_t\right )e^{-bt}</math>
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:<math>  v  = v_0e^{-bt} + v_t \left (1 -e^{-bt}\right )</math>
  
  
 
[[Forest_UCM_PnCP#Linear_Air_Resistance]]
 
[[Forest_UCM_PnCP#Linear_Air_Resistance]]

Revision as of 13:54, 31 August 2014

Linear Air Resistance=

Horizontal motion

If [math]n[/math] is unity then the velocity is exponentially approaching zero.

[math]F(v) = -bv[/math]: negative sign indicates a retarding force and [math]b[/math] is a proportionality constant
[math]\sum \vec {F}_{ext} = -bv = m \frac{dv}{dt}[/math]
[math]\Rightarrow \int_{v_i}^{v_f} \frac{dv}{v} = \int_{t_i}^{t_f} \frac{-b}{m}dt[/math]
[math]\ln\frac{v_f}{v_i} = \frac{-b}{m}t[/math]; [math]t_i \equiv 0[/math]
[math]v_f = v_i e^{-\frac{b}{m}t}[/math]

The displacement is given by

[math]x = \int_0^t v_i e^{-\frac{b}{m}t} dt[/math]
[math]= \left . v_i \left ( \frac {e^{-\frac{b}{m}t}}{-\frac{b}{m}} \right ) \right |_0^t[/math]
[math]= \left . v_i \left ( -\frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_0^t[/math]
[math]= \left . v_i \left ( \frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_t^0[/math]
[math]= v_i \left ( \frac{m}{b} e^{-\frac{b}{m}0} -\frac{m}{b} e^{-\frac{b}{m}t} \right ) [/math]
[math]= \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )[/math]


Example: falling object with linear air friction

Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity

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

let

[math]v_t = \frac{mg}{b}[/math]
[math] v_t -v = \frac{1}{b} \frac{dv}{dt}[/math]
[math] b dt= \frac{dv}{v_t -v} [/math]
[math] -b dt= \frac{dv}{v -v_t} [/math]
[math] -\int_0^t b dt= \int_{v_0}^v \frac{dv}{v -v_t} [/math]
[math] -bt = \ln{\left( v -v_t \right)} - \ln{\left ( v_0-v_t \right )}[/math]
[math] -bt = \ln \left(\frac{ v -v_t }{v_0-v_t}\right )[/math]
[math] e^{-bt} = \left(\frac{ v -v_t }{v_0-v_t}\right )[/math]
[math] v -v_t = \left ( v_0-v_t\right )e^{-bt}[/math]
[math] v = v_0e^{-bt} + v_t \left (1 -e^{-bt}\right )[/math]


Forest_UCM_PnCP#Linear_Air_Resistance