Calculating the final velocity of a compressed-air launched projectile

In summary, the problem is not really homework-esque but something derived out of curiosity. The homework equations that were given are for a projectile launched from a compressed air chamber. The equation that is used to calculate the kinetic energy of the projectile is the same equation that is used to calculate the kinetic energy of a cannonball. The cannonball has a higher kinetic energy because it has a higher speed. The speed of air molecules will have to be taken into effect when the speed of the projectile inside the launching barrel is of the order of the speed of sound in the gas being used under that pressure.
  • #1
swbluto
12
0
Ehhhhh... this really isn't a "homework" problem but something derived out of curiosity(But I guess it could be considered homework-esque).

Homework Statement



Basically, not accounting for drag or friction, what'd be the final velocity of a projectile launched from a compressed air chamber of volume v_0 connected DIRECTLY to a tube of radius R with launch length L with initial pressure P_0, with the projectile launcher set at angle [tex]\theta[/tex] from the horizontal?

Homework Equations



m is the mass of the projectile
g is the gravitational acceleration.
x will be the length from the beginning of the tube to the given position of the projectile inside the tube.
Area = pi*R^2

P_x = V_0*P_0/(V_0+x*Area) [The pressure where the projectile is down the tube at x-length]
F = pressure*Area
Work = [tex]\int[/tex] F ds
(With the Y-axis perpendicular to the projectile's motion and X-axis parallel to it.)
F_x = Area*P_x -mgsin([tex]\theta[/tex])

The Attempt at a Solution



The kinetic energy of the initial state of the projectile is 0. The net work done along the X-axis, that being along the length of the tube, will be the final kinetic energy. Since the net work is the integral of the Force applied along the X-axis(As there's no net Y_force with my set-up) in respect to distance, then

K_f= Work = [tex]\int[/tex]_L,0 F ds =
[tex]\int[/tex]_L,0 Area*P_x -mgsin([tex]\theta[/tex]) dx =
Area*[tex]\int[/tex]_L,0 P_x dx -mgsin([tex]\theta[/tex])[tex]\int[/tex]_L,0 dx=
Area*[tex]\int[/tex]_L,0 V_0*P_0/(V_0+x*Area) dx -L*m*g*sin([tex]\theta[/tex])= (Area is now A)
V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin([tex]\theta[/tex])

Now that the work ergo final Kinetic energy has been found, relating to the kinetic energy formula 1/2*m*v_f^2= K_f,
v_f = (2K_f/m)^(1/2) =
(2/m)^(1/2)*(V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin([tex]\theta[/tex]))^(1/2)

Is that right? I didn't think the solution would be that simple as I imagine a time-based solution would be much more difficult(even though I haven't even envisaged the solving procedures.). Some of the equations were derived from my own thinking, such as the pressure at length-x equation, so I fear some of my premises may be flawed and so the final solution.
 
Last edited:
Physics news on Phys.org
  • #3
Shooting star said:
There has been a lot of discussion about a few similar concepts at one one time. Perhaps you may benefit from browsing these threads.

https://www.physicsforums.com/showthread.php?p=1471310#post1471310

https://www.physicsforums.com/showthread.php?t=191299

https://www.physicsforums.com/showthread.php?t=192152

Yes, interesting, though most of the content appears to be devoted to addressing oversights that I haven't over-sighted. There are some fairly similar functions, however, that might be useful in relating. Thank you.
 
  • #4
After reading ahead, it appears that my model is a little bit too simplistic. Upon reading about the pressure being directly related to the sum of the force provided by the collision of air-molecules against the sealing membrane, I remembered that an air-launched cannon has a "top speed" since air molecules have a "natural speed limit"(Not an *ultimate one*, but just one under natural physical circumstances) and that's why one team decided to use nitrogen for an ultra-long range projectile. This then shows that the amount of force exerted on the projectile is just not simply due to the "amount of air-molecules inside" and the volume that contains those air molecules(And the area the projectile takes up), but also on the very speed of the projectile(!) as the relative average speed of the air molecules' collisions on the projectile decreases as the projectile's velocity increases, regardless of pressure. So this must be accounted for for an accurate model to be developed(And thus might've been a substantial source of error in others experimentally derived values and expected values that didn't account for this difference.). Any ideas how to mathematically account for it?(Yeah, I know, so does the tube-on-projectile friction and air drag.)
 
  • #5
The speed of air molecules will have to be taken into effect when the speed of the projectile inside the launching barrel is of the order of the speed of sound in the gas being used under that pressure. I never thought that you wanted such high speeds. I feel in that case many other significant factors would have to be considered. You have to talk to experts.
 

1. How do you calculate the final velocity of a compressed-air launched projectile?

The formula for calculating the final velocity of a compressed-air launched projectile is:
Vf = √(2P/ρA),
where Vf represents the final velocity, P is the air pressure, ρ is the air density, and A is the cross-sectional area of the projectile.

2. What units should be used for the variables in the final velocity formula?

The units used for the variables in the final velocity formula should be consistent.
P should be in Pascals (Pa), ρ should be in kilograms per cubic meter (kg/m3), and A should be in square meters (m2).

3. Can the final velocity be calculated without knowing the air pressure or density?

No, the final velocity cannot be accurately calculated without knowing the air pressure and density. These variables greatly affect the final velocity of the projectile.

4. How does the angle of launch affect the final velocity of a compressed-air launched projectile?

The angle of launch does not directly affect the final velocity of a compressed-air launched projectile. However, it can affect the distance the projectile travels due to the trajectory and air resistance.

5. Is the final velocity of a compressed-air launched projectile affected by the mass of the projectile?

Yes, the final velocity of a compressed-air launched projectile is affected by the mass of the projectile. A heavier projectile will require more force from the compressed air to reach the same final velocity as a lighter projectile.

Similar threads

  • Introductory Physics Homework Help
2
Replies
53
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
954
  • Introductory Physics Homework Help
Replies
6
Views
887
  • Introductory Physics Homework Help
Replies
4
Views
776
  • Introductory Physics Homework Help
Replies
10
Views
863
  • Introductory Physics Homework Help
Replies
6
Views
1K
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
11
Views
689
  • Introductory Physics Homework Help
Replies
16
Views
3K
  • Introductory Physics Homework Help
2
Replies
55
Views
1K
Back
Top