Vertical Travel of a projectile 1

[John7474]

I have a threaded bar which could act as a sealed plug being forced vertically out due to pressure.


F = x-section area x pressure

I think I can calculate the initial accelaration as per,

F = ma - mg

I am trying to find its maximum vertical travel.

 

[IRstuff]

If a THREADED bar can be launched, you've got more problems than just the maximum height.



TTFN
faq731-376

 

[rb1957]

i think it's the initial velocity you need to consider; the acceleration would only be applied for an instant. then it's just like a high school physics projectile problem, how does the velocity change under the deceleration due to gravity ?

 

[zdas04]

You can get initial velocity by assuming the upward acceleration ends as soon as the projectile is free of the barrel. Integrate net initial acceleration for initial velocity. Then it is a simple dynamics problem.

David

 

[TheTick]

So....

F = ma
v = at
How much t?

 

[handleman]

F dot d = mgh where d = length of barrel

-handleman, CSWP (The new, easy test)

 

[Cockroach]

The pressure times the seal bore area will give you the force acting on the bar. You can consider the bar as a point mass at the centroid of mass in the calculation. The acceleration is simply the force (pressure X seal area) over mass of the bar.

So you have a mass with unbalanced force due to pressure upwards, fighting gravity. This is a general dynamics question, you can use various equations to get the height of the projectile.

But I think you got your acceleration wrong at the start of the problem.

Regards,
Cockroach

 

[MiketheEngineer]

I think I would wear a LOT of protective gear behind a very thick concrete wall and ceiling.

As noted - this is a simple physics problem.

I made a potato gun and my buddy and I wanted to find the potato speed. So - using basic and "rough" physics we decided to shoot it straight up and time it. Speed up will be about speed coming down or at least fairly close.

So that's what we did. Must have aimed it fairly well because after it hit apogee and started coming down - it was quite obvious we might get hit!!

We ran like scared school kids. Sure enough - it hit within 1' from the launch site!!

If memory serves - it was over 80 mph.

 

[hydtools]

Use work = energy for the solution.
Weight of the rod, W = mg
Force of the gas, F = PA
Xsec area of the rod, A
Pressure of the gas, P
Distance to travel to exit the bore, d
Height traveled to v = 0, h
Exit velocity, v

When rod exits the bore, (F-W)*d = 0.5*m*v^2, work done = kinetic energy at exit.

When rod reasches h, W*h = 0.5*m*v^2, potential energy at h = KE at exit.

therefore, W*h = (F-W)*d, or h = (F-W)*d/W

Assume pressure is constant during rod travel through distance d, then F = P*A

Then, h = (P*A-W)*d/W

Ted

 

[handleman]

That's what I said.

-handleman, CSWP (The new, easy test)

 

[John7474]

Thanks guys, I'm humbled. Yet, I do beleive that work energy method is the best approach.

This is not a potato gun, and some of you won't need a helmet while going to work. There are enough resistances to the force created by the pressurisation, besides there is a pressure equalisation design which takes the force created to a considerably lower value.

I have got my answer, thank you all.