Forest UCM PnCP QuadAirRes - Revision history

Foretony at 11:54, 10 September 2014

2014-09-10T11:54:44Z

Foretony at 11:53, 10 September 2014

2014-09-10T11:53:59Z

Foretony at 11:53, 10 September 2014

2014-09-10T11:53:46Z

Foretony: Created page with " Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared : $\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}<…"$

2014-08-31T14:06:10Z

Created page with " Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared :<math>\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}<…"

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

:<math>\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}</math>

Find the fall distance

Here is a trick to convert the integral over time to one over distance so you don't need to integrate twice as inthe previous example

:<math>\frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v\frac{dv}{dy}</math>

The integral becomes

:<math>mg -bv^2 = m v\frac{dv}{dy}</math>
:<math>\int_{y_i}^{y_f} dy = \int_{v_i}^{v_f} m \frac{dv}{\left ( mg -bv^2 \right ) }</math>
:<math>y = \int_{v_i}^{v_f} \frac{dv}{\left ( g -\frac{b}{m}v^2 \right ) }</math>

let <math>u = g -\frac{b}{m}v^2</math>

then <math>du = -2\frac{b}{m}v dv</math>

:<math>y =\int_{v_i}^{v_f} \frac{-m}{2b} \frac{du}{u } = \frac{b}{m} \int_{v_f}^{v_i} \ln {g -\frac{b}{m}v^2} =</math>
:<math>y =\frac{m}{2b} \ln \left ( \frac {g -\frac{b}{m}v_i^2}{g -\frac{b}{m}v_f^2} \right ) </math>

[[Forest_UCM_PnCP#quadratic_friction]]

@@ Line 13: / Line 13: @@
 :<math>mg -bv^2 = m v\frac{dv}{dy}</math>
 :<math>\int_{y_i}^{y_f} dy  = \int_{v_i}^{v_f} m \frac{vdv}{\left ( mg -bv^2 \right ) }</math>
-:<math>y  = \int_{v_i}^{v_f} \frac{dv}{\left ( g -\frac{b}{m}v^2 \right ) }</math>
+:<math>y  = \int_{v_i}^{v_f} \frac{vdv}{\left ( g -\frac{b}{m}v^2 \right ) }</math>

@@ Line 20: / Line 20: @@
 then <math>du = -2\frac{b}{m}v dv</math>
-:<math>y  =\int_{v_i}^{v_f} \frac{-m}{2b} \frac{du}{u } = \frac{b}{m} \int_{v_f}^{v_i} \ln {g -\frac{b}{m}v^2} =</math>
+:<math>y  =\int_{v_i}^{v_f} \frac{-m}{2b} \frac{du}{u } = \frac{b}{m} \int_{v_f}^{v_i} \ln \left ( g -\frac{b}{m}v^2 \right ) =</math>
 :<math>y  =\frac{m}{2b}  \ln \left ( \frac {g -\frac{b}{m}v_i^2}{g -\frac{b}{m}v_f^2} \right ) </math>
 [[Forest_UCM_PnCP#quadratic_friction]]

@@ Line 1: / Line 1: @@
 Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared
@@ Line 13: / Line 12: @@
 :<math>mg -bv^2 = m v\frac{dv}{dy}</math>
-:<math>\int_{y_i}^{y_f} dy  = \int_{v_i}^{v_f} m \frac{dv}{\left ( mg -bv^2 \right ) }</math>
+:<math>\int_{y_i}^{y_f} dy  = \int_{v_i}^{v_f} m \frac{vdv}{\left ( mg -bv^2 \right ) }</math>
 :<math>y  = \int_{v_i}^{v_f} \frac{dv}{\left ( g -\frac{b}{m}v^2 \right ) }</math>

Forest UCM PnCP QuadAirRes - Revision history

Foretony at 11:54, 10 September 2014

Foretony at 11:53, 10 September 2014

Foretony at 11:53, 10 September 2014

Foretony: Created page with " Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared :\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}<…"

Foretony: Created page with " Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared : $\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}<…"$