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rg2004
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Homework Statement
Consider a projectile subject to air resistance. The drag force is [itex]F=-\hat{v} C v^{a}[/itex] where C and a are constants and v=[itex]\dot{r}[/itex]. Restrict the problem to 2-D and take y in the vertical direction. Take the magnitude of initial velocity to be [itex]v_{0}[/itex] and the initial position to be the origin. Introduce dimensionless coordinates X,Y,Z by relations
[itex]x=\frac{v_{0}^{2}}{g}X[/itex]
[itex]y=\frac{v_{0}^{2}}{g}Y[/itex]
[itex]t=\frac{v_{0}}{g}T[/itex]
Given the above definitions show that
[itex]\frac{d^2Y}{dT^2}=-1-k V^{a-1} \frac{dY}{dT}[/itex]
[itex]\frac{d^2X}{dT^2}=-k V^{a-1} \frac{dX}{dT}[/itex]
where [itex]k=\frac{C v_{0}^a}{m g}[/itex]
[itex]V=\sqrt{(\frac{dY}{dT})^2+(\frac{dX}{dT})^2}[/itex]
The Attempt at a Solution
[itex]x=\frac{v_{0}^{2}}{g}X[/itex][itex]dx=\frac{v_{0}^{2}}{g}dX[/itex]
[itex]y=\frac{v_{0}^{2}}{g}Y[/itex]
[itex]dy=\frac{v_{0}^{2}}{g}dY[/itex]
[itex]t=\frac{v_{0}}{g}T[/itex]
[itex]dt=\frac{v_{0}}{g}dT[/itex]
finding d?/dT equations
[itex]\frac{dx}{dt}=\frac{\frac{v_{0}^{2}}{g}dX}{\frac{v_{0}}{g}dT}
=v_0\frac{dX}{dT}
[/itex]
[itex]\frac{dy}{dt}=\frac{\frac{v_{0}^{2}}{g}dY}{\frac{v_{0}}{g}dT}
=v_0\frac{dY}{dT}
[/itex]
finding [itex]\frac{d^2?}{dT^2}[/itex]
[itex]\frac{d^2x}{dt^2}=\frac{dx}{dt}\frac{dx}{dt}=v_0^2\frac{d^2X}{dT^2}[/itex]
[itex]\frac{d^2y}{dt^2}=\frac{dy}{dt}\frac{dy}{dt}=v_0^2\frac{d^2Y}{dT^2}[/itex] From force equation:
[itex]F_{d,x}=m \ddot{x}=-\hat{v} C v^{a}=\frac{-\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} C (\sqrt{\dot{x}^2+\dot{y}^2})^{a}[/itex]
[itex]=-\dot{x} C \sqrt{\dot{x}^2+\dot{y}^2}^{a-1}[/itex][itex]m v_0^2\frac{d^2X}{dT^2}=-(v_0\frac{dX}{dT}) C \sqrt{(v_0\frac{dX}{dT})^2+(v_0\frac{dY}{dT})^2}^{a-1}[/itex]
[itex]m v_0^2\frac{d^2X}{dT^2}=-v_0^a \frac{dX}{dT} C \sqrt{(\frac{dX}{dT})^2+(\frac{dY}{dT})^2}^{a-1}[/itex]
[itex]\frac{d^2X}{dT^2}=-\frac{v_0^a C}{m v_0^2} V^{a-1} \frac{dX}{dT} [/itex]
[itex]\frac{d^2X}{dT^2}=-k \frac{g}{v_0^2} V^{a-1} \frac{dX}{dT} [/itex]
Which is not what I'm supposed to get. I must be messing up a rule or two somewhere.
Thank you.
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