Forest UCM PnCP LinAirRes

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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]


[math]b=[/math]coefficient of air resistance
[math]v_t = \frac{mg}{b} =[/math] Terminal speed
[math] v_t -v = \frac{m}{b} \frac{dv}{dt}[/math]
[math] \frac{b}{m} dt= \frac{dv}{v_t -v} [/math]
[math] -\frac{b}{m} dt= \frac{dv}{v -v_t} [/math]
[math] -\int_0^t \frac{b}{m} dt= \int_{v_0}^v \frac{dv}{v -v_t} [/math]
[math] -\frac{b}{m}t = \ln{\left( v -v_t \right)} - \ln{\left ( v_0-v_t \right )}[/math]
[math] -\frac{b}{m}t = \ln \left(\frac{ v -v_t }{v_0-v_t}\right )[/math]
[math] e^{-\frac{b}{m}t} = \left(\frac{ v -v_t }{v_0-v_t}\right )[/math]
[math] v -v_t = \left ( v_0-v_t\right )e^{-\frac{b}{m}t}[/math]
[math] v = v_0e^{-\frac{b}{m}t} + v_t \left (1 -e^{-\frac{b}{m}t}\right )[/math]

The posiiton as a function of time may be determined by directly integrating the above equation

[math] \frac{dy}{dt} = v_0e^{-bt} + v_t \left (1 -e^{-\frac{b}{m}t}\right )[/math]
[math] \int_0^y = \int_0^t \left ( v_0e^{-\frac{b}{m}t} + v_t \left (1 -e^{-\frac{b}{m}t}\right ) \right ) dt[/math]
[math]y = \int_0^t v_0e^{-\frac{b}{m}t}dt + \int_0^t v_t \left (1 -e^{-b\frac{b}{m}t}\right ) dt[/math]
[math]= \frac{v_0}{-\frac{b}{m}}\left ( e^{-\frac{b}{m}t}-e^{-b0} \right ) + v_t t + \frac{mv_t}{b}\left ( e^{-\frac{b}{m}t} - e^{-\frac{b}{m}0}\right ) [/math]
[math]= \frac{v_0}{\frac{b}{m}}\left ( 1- e^{-\frac{b}{m}t} \right ) + v_t t + \frac{v_t}{b}\left ( e^{-bt} - 1\right ) [/math]
[math]= v_t t + \frac{m}{b}\left ( v_0 - v_t) \right ) \left ( 1- e^{-\frac{b}{m}t} \right ) [/math]