And remember there exist no perfect vacuums any more than there are
perfectly inelastic ball bearings.
On Tue, Dec 1, 2015 at 9:07 AM, Henry Vanderbilt <hvanderbilt@xxxxxxxxxxxxxx
wrote:
In an ideal perfectly rigid rocket or cannon projectile structure in
vacuum, that should be so. In the real world where the structures will be
to some degree elastic, removal of the accelerating force will cause the
structure to revert to its unstressed shape, which, depending on design
details, could then impart a force to your ball-bearing or mercury droplet.
There's also the elasticity of the ball-bearing (or even, given
surface-tension, of the mercury droplet) which could also cause a rebound
once the overall acceleration ends.
As for the specific case of a projectile within a cannon barrel, that
depends on whether it is being accelerated for its entire time within the
barrel. This depends on the quantity and burn-rate of the propellant
charge. Typically the answer is yes, accelerated the whole time, because
calculating the propellant charge to still be producing considerable
positive pressure on the base of the projectile as it reaches the muzzle
tends to make best use of a given length of barrel. It is however possible
to use a small fast-burning charge such that pressure has dropped off
enough that the projectile is decelerating by the time it reaches the
muzzle.
On 12/1/2015 8:35 AM, John Dom wrote:
Just checking assumptions, please advise.
Am I right assuming a mercury droplet or loose ball-bearing will not
move forward inside a rocket in vacuum the instant the motor stops? It
then only might start floating.
Nor will such droplet move forward inside a cannon projectile after it
leaves the muzzle since on that instant push force has stopped. As to
during the short residence time inside barrel, it might fly forward I
reckon.
jd