- #1
swbluto
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Ehhhhh... this really isn't a "homework" problem but something derived out of curiosity(But I guess it could be considered homework-esque).
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 \theta from the horizontal?
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 = \int F ds
(With the Y-axis perpendicular to the projectile's motion and X-axis parallel to it.)
F_x = Area*P_x -mgsin(\theta)
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 = \int_L,0 F ds =
\int_L,0 Area*P_x -mgsin(\theta) dx =
Area*\int_L,0 P_x dx -mgsin(\theta)\int_L,0 dx=
Area*\int_L,0 V_0*P_0/(V_0+x*Area) dx -L*m*g*sin(\theta)= (Area is now A)
V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin(\theta)
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(\theta))^(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.
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 \theta 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 = \int F ds
(With the Y-axis perpendicular to the projectile's motion and X-axis parallel to it.)
F_x = Area*P_x -mgsin(\theta)
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 = \int_L,0 F ds =
\int_L,0 Area*P_x -mgsin(\theta) dx =
Area*\int_L,0 P_x dx -mgsin(\theta)\int_L,0 dx=
Area*\int_L,0 V_0*P_0/(V_0+x*Area) dx -L*m*g*sin(\theta)= (Area is now A)
V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin(\theta)
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(\theta))^(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.
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