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Hi I'm stuck on the following problem:

Given a projectile with position $\mathbf{r}(t)=[r_x(t),r_y(t)]$, velocity $\mathbf{v}(t)=[v_x(t),v_y(t)]$ and acceleration $\mathbf{a}(t)$, we can approximate the drag force $\mathbf{F}_d$ acting in the opposite direction to the projectile's trajectory using

$$\mathbf{F}_d = \frac{1}{2} \rho c_d A |\mathbf{v}|\mathbf{v},$$

where $\rho$ is the air density, $c_d$ is the drag coefficiant, $A$ is the frontal area and $\mathbf{v}$ is the velocity defined above.

Using Newton's second law, we can get

$$\mathbf{F}=m\mathbf{a} \Rightarrow \mathbf{a}(t) = -\frac{1}{m}(-\rho c_d A|\mathbf{v}|\mathbf{v} - m\mathbf{g}),$$

Where m is the mass of the projectile and $\mathbf{g} = (0,g)$ is the gravitational acceleration.

We can use the fact that

$$\mathbf{a}(t)=\frac{\delta \mathbf{v}}{\delta t} = \mathbf{\dot{v}}(t)$$

and

$$\mathbf{v}(t)=\frac{\delta \mathbf{r}}{\delta t} = \mathbf{\dot{r}}(t)$$

to give us the non-linear, second order Ordinary Differential Equation

$$\mathbf{\ddot{r}}(t)=\frac{1}{m}(-\frac{1}{2}\rho c_d A |\mathbf{\dot{r}}|\mathbf{\dot{r}}-m\mathbf{g}). \;\;\; (1)$$

We can also consider the initial conditions at time $t=0$. The projectile is initially fired from height $h$ at an angle of $\alpha$ degrees with initial speed $v_0$, which gives:

$$\mathbf{r}(0) = (0, h)$$ and $$\mathbf{\dot{r}}(0) = \mathbf{v}(0)=[v_0 cos(\frac{\pi\alpha}{180}), v_0 sin(\frac{\pi\alpha}{180})].$$

At this point I need to transform equation (1) as a system of first-order equations for the two components of velocity and two components of position. Mathematically this will be expressed as

$$\mathbf{\dot{y}} = \mathbf{F}(t, \mathbf{y}(t)),$$

where $\mathbf{y}(t) = [v_x(t), v_y(t), r_x(t), r_y(t)]$.

I'm familiar with a method to transfrom a second order ODE to a system of linear first-order ODEs, but the method I'm thinking of runs into a problem due to the fact we have $|\mathbf{\dot{r}}|\mathbf{\dot{r}}$ in the equation and I'm not sure how to handle this. Could anyone show me how to do this?

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    $\begingroup$ Since your differential equation doesn't depend on $\mathbf{r}$ directly, then just set $\mathbf{v} = \mathbf{\dot r}$ to get a first order system. $\endgroup$ – Jacky Chong Nov 21 '18 at 7:07
  • $\begingroup$ @JackyChong So substituting $\mathbf{v} = \mathbf{\dot{r}}$ into the equation tells gives use the first order equations for the velocity components, but what about the position components? We simply have that at time $t$, $\mathbf{\dot{r}}(t) = \mathbf{v}$, yes? $\endgroup$ – Thomas M Nov 21 '18 at 7:37
  • $\begingroup$ Correct. Once you find the velocity, you just integrate to get the position. $\endgroup$ – Andrei Nov 21 '18 at 19:36

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