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AIM

Derive the equations of motion (hence of velocity and acceleration too) in each coordinate for the following problem.

SETTING

First of all, this question is very similar (in a sense) to this one Projectile Motion with Air Resistance and Wind. However my problem is way simpler than that. First of all the $z$ axis points upwards, the $x$ axis is horizontal, positive to the right and therefore $y$

I was thinking of a projectile of mass $m$, launched at a $\textit{speed}$ $V$ and at an angle $\theta$ from the horizontal. Now I want to take into account the force of gravity $\underline{F}_g = -mg\underline{k}$ (where $\underline{k}$ is the standard unit vector in the $z$ direction).

The drag proportional to the velocity (not the square!) $\underline{F}_d = -k\underline{v}_{pa}$, where I have written $\underline{v}_{pa}$ to emphasize that it is the velocity of the projectile with respect to the air, and $k$ is the drag coefficient.

Finally there is a wind blowing in the $\underline{j}$ direction with speed equals to 2, so the velocity of the wind is $V_w = 2\underline{j}$

MY TRIAL

x-coordinate (horizontal): The only force is the drag due to the initial velocity in the $x$ direction so: $\underline{F}_x = m\ddot{x} = -k\dot{x}$, which is the standard equation in the projectile motion with air resistance and hence I get: \begin{equation} \dot{x} = V\cos{\theta}e^{\frac{-kt}{m}} \,\,\,\, \text{and} \,\,\,\,\, x = \frac{mV\cos{\theta}}{k}(1-e^{\frac{-kt}{m}}) \end{equation}

z-coordinate (vertical): Here I have the force of gravity and the drag, hence $\underline{F}_z = m\ddot{z} = -mg\underline{k} -k\dot{z}$ and therefore I get the following: \begin{equation} \dot{z} = -\frac{mg}{k} + (V\sin{\theta} + \frac{mg}{k})e^{-\frac{kt}{m}} \end{equation} And also: \begin{equation} z = -\frac{mg}{k}t + \frac{m}{k}(V\sin{\theta} + \frac{mg}{k})(1 - e^{-\frac{kt}{m}}) \end{equation}

ISSUE HERE IS WHERE I NEED HELP I THINK

now for the $y$ coordinate I don't understand how to find the equations. Indeed I launch the projectile in the x-z plane, so the initial velocity in the $y$ direction should be $0$ right? I don't understand how to use the wind blowing in $2\underline{j}$. Should I say that since there is this wind, then the initial velocity of the projectile will be the same as the one of the wind? Or how should I do it?

I understand that the only force that will be acting n that direction will be the drag, opposing to the velocity given by the wind. Do you have any idea of how I should solve it?

Thank you

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  • $\begingroup$ No, I mean the velocity $2\underline{j}$ is the velocity of the air (wind) relative to the ground. If I understand your comment $\endgroup$ – Euler_Salter Nov 9 '16 at 21:41
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The system can be specified on the orthogonal axes easily if you give it the initial conditions in cartesian axis: $\vec{v}=v_{x_0}\vec{i}+v_{0_y}\vec{j}+v_{0_z}\vec{k}$. If the velocity of the wind is $\vec{w}=w_y\vec{j}$ we can express the equations in

$\dot{\vec{v}} = -\vec{g} - k(\vec{v}-\vec{w})$

For any axis and using the given wind:

$\vec{i}: \dot{v_x} = -kv_x$

$\vec{j}: \dot{v_y} = -k(v_y-w_y)$

$\vec{k}: \dot{v_z} = -g -kv_z $

These are three independent first order differential equations that you can solve for, the first two are fairly simple and the last one you can find for a solution in this post.Then you can integrate them again to obtain the position.

The y axis is actually the simplest to solve. The wind is going to drag slowly the the projectile out the $y=0$ path and eventually it will converge to the wind velocity.

$v_y=(v_{0_y}-w_y)e^{-kt}+w_y$

You can find the whole problem solution here.

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  • $\begingroup$ Sorry, I don't understand what $v_y$ should be in $\vec{v}$. Should it be $0$ then? And why is it $-k(v_y-w_y)$ and not $-k(v_y+w_y)$? $\endgroup$ – Euler_Salter Nov 9 '16 at 22:30
  • $\begingroup$ In your statement you were saying it was $v_y=0$, but more generally, you could choose a firing heading angle the same way you choose a fire tilt angle. $\endgroup$ – ebabio Nov 10 '16 at 0:48
  • $\begingroup$ Also, it is the difference between the projectile and wind speed that will decay with time (it is a drag force). So this way the force opposes that difference and it becomes zero when the relative speed is zero, i.e. when $v_y - w-y = 0$. If there is a sum there would be a force even if there is no relative movement. $\endgroup$ – ebabio Nov 10 '16 at 0:53
  • $\begingroup$ thank you. Sorry for asking so much, but also I do not understand how it can be dimensionally correct. Normally we take the drag force $D = -k\underline{v}$, so $-k\underline{v}$ is in Newton. But in your equations it has the same dimension as the acceleration $\endgroup$ – Euler_Salter Nov 10 '16 at 8:35
  • $\begingroup$ From what I remember from my physics classes, the Force slowing down the movement in presence of drag = cw * A * V * V where A = the projected area of the object and V the speed of the object and cw is a coefficient accounting for the shape of the projectile. Not sure how "drag" is defined when you think it is proportional to the speed instead of the square of the speed. Maybe your units in your equation are worth checking. $\endgroup$ – BitTickler Apr 6 '18 at 22:36

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