1
$\begingroup$

We have a projectile, which we will consider spherical, of radius $r$ density $\rho_m$ and mass $m$, and we begin by firing it from a point $(0,h_0)$ at an initial velocity $v_0$ at an angle $\theta_0$. From this we can say that: $$x'(0)=v_0\cos(\theta_0),\,y'(0)=v_0\sin(\theta_0)\tag{1}$$ If we neglect all forces acting except those due to gravity then we can say: $$x''=0,\,y''=-g\tag{2.1}$$ Integrating to give: $$x'=C_1,\,y'=-gt+C_2\tag{2.2}$$ $$x=C_1t+C_3,\,y=-\frac{g}{2}t^2+C_2+C_4\tag{2.3}$$ Now by applying initial conditions we get: $$x=v_0\cos(\theta_0)t\tag{2.4.1}$$ $$y=-\frac g2t^2+v_0\sin(\theta_0)t+h_0\tag{2.4.2}$$ However, I would like to do the same thing as this but not neglecting air resistance (and if possible I would like to include the effect of spin). My thoughts to doing this would be: If $F_x,F_y$ denote force in the $x$ and $y$ directions, then we can say: $$\sum F_x=mx''\,\,,\,\,\sum F_y=my''\tag{3.1}$$ I have found that for an object moving through air, drag is equal to: $$F_{drag}=\frac 12CA\rho_{f}v^2\tag{3.2}$$ Since this is the total velocity of the object in question, we would have to say: $$v=\sqrt{(x')^2+(y')^2}\tag{3.3}$$ We now know the total drag force but obviously need the force in each direction. I believe the drag would always act in the opposite direction to motion. Because of this I would say: $$x''=-\frac{CA\rho_f}{2m}\left[(x')^2+(y')^2\right]\cos(\theta)\tag{3.4.1}$$ $$y''=-\frac{CA\rho_f}{2m}\left[(x')^2+(y')^2\right]\sin(\theta)-g\tag{3.4.2}$$ For some reason I can't visualise how I would get this $\cos\theta$ and $\sin\theta$ in terms of $x$ and $y$, maybe it would be clearer if I converted to vectors? I'm not sure how to continue from here to get a differential equation for $x''$ and $y''$. Thanks

EDIT:1

After getting help from users I end up with the differential equation: $$\frac{d^2}{dt^2}\begin{pmatrix}x\\y\end{pmatrix}=-\frac{CA\rho_f}{2m}\sqrt{(x')^2+(y')^2}\frac{d}{dt}\begin{pmatrix}x\\y\end{pmatrix}-\begin{pmatrix}0\\g\end{pmatrix}$$

$\endgroup$
1
$\begingroup$

Notice that the drag force is always opposite to velocity. You can then write $$\cos\theta=\frac{v_x}v=\frac{x'}{\sqrt{x'^2+y'^2}}\\\sin\theta=\frac{v_y}v=\frac{y'}{\sqrt{x'^2+y'^2}}$$

$\endgroup$
  • $\begingroup$ Exactly what I was looking for. Thank you! $\endgroup$ – Henry Lee May 27 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.