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{vdv}{\left ( mg -bv^2 \right ) }[/math]
- [math]y = \int_{v_i}^{v_f} \frac{vdv}{\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 \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