MythBusters’ Merry-go-round.
NoEinstein
foot-pounds (sic) of force. A nominal six foot diameter, human-
powered merry-go-round—such as are found in most public parks—requires
about five pounds of continuous force to cause rotation. For safety,
the friction of the ball pivot is intentionally kept that high so that
the merry-go-round won’t keep spinning for long once the motive force
stops.
The ‘myth’ that was to be verified was whether or not the impacts of
bullets would be enough to keep a playground merry-go-round spinning.
The initial shots penetrated the pipe rails, which were then
reinforced with what looked to be 6” square x ¼” thick steel plates.
Subsequent shotgun blasts easily indented the steel plates at least
1/16”, but rotated the merry-go-round only about a foot per shot.
This was after the bearings had been replaced with those of lower
friction. The show was correct to cite the inertia of the merry-go-
round for its resistance to rotation from impulse forces.
Finally, after about thirty shotgun blasts were fired at the steel
plates in quick succession, the merry-go-round rotated at a slow
walking speed of about two miles per hour. The MythBusters crew
seemed thrilled. None of those ’experts’ were perceptive enough to
realize that they had just put a supposed 200,000 foot-pounds of
’force’ against that merry-go-round. That amount of force would have
pulverized the device, but clearly it did not. The reason: The 5,000
foot-pound (sic) of force is never more than about 40 pounds! Hear
THIS, MythBusters: Newton’s third Law of Motion states that “for every
force, there is an equal and opposite reactive force.” That means
that the maximum impact force of the bullet can’t exceed the estimated
forty pound recoil of the shotgun! Also, consider this: No 5,000
pound force would indent a ¼” thick steel plate that has a minimum 20
kip per square inch strength. Yet, the plates were easily indented.
Why? *** Because all material have varying strengths depending on the
speed of application of the loads.
The errant inclusion of both force and distance in the misunderstood
units of kinetic energy has never correctly solved a single problem
regarding impact force or an impact’s destruction—heretofore, wrongly
considered to be 100% the result of the impacting object. No one
before yours truly has realized most of the destruction potential
resides in the materials being hit. Because slugs, like bullets, are
spinning rapidly, the materials respond not just to the 40 pound
force, but to that spin. It is primarily the devastating effect of a
slug’s spin that indents the plate. I’ll leave the determination of
what is happening to the steel at an atomic scale to future
researchers.
I’ve repeatedly urged MythBusters to use the 1830 Coriolis kinetic
energy equation, KE = ½ mv^2, to invalidate itself. By varying both
mass and velocity, that equation should be able to determine the
height of drop to cause two same-size, different weight balls to
impact with the SAME kinetic energy. Having matching KE for two same-
size balls should cause identical impediments of the balls into soft
clay. But such is NOT the case! I invite the ‘light’ scientists on
MythBusters, and any of you readers, to perform my $40.00 KE test.
Email me, directly, and I will send you a color photo of the balls not
imbedded even close to the same depth.
Respectfully Submitted,
— NoEinstein —
AKA John A. Armistead
KE = 1/2mv^2 is disproved in new falling object impact test.
http://groups.google.com/group/sci.physics/browse_thread/thread/51a85ff75de414c2?hl=en&q=
Jim
needed, but then there would be no need
for a how that makes millions.
"NoEinstein" <noein...@bellsouth.net> wrote in message
news:3d1ef1c7-2462-4283...@d19g2000yql.googlegroups.com...
Sam Wormley
> According to MythBusters, a 12 gage shotgun fires a slug with 5,000
>
>
> +------------+ +---------------------------------------------+
> | PLEASE | | BEST TO IGNORE ATTENTION SEEKING TROLLS |
> | DO NOT | | LIKE JOHN ARMISTEAD -- THEY DRY UP AND BLOW |
> | FEED | | AWAY WITHOUT FEEDBACK |
> | DA | | |
> | TROLLS | | http://www.angelfire.com/space/usenet/ |
> +------------+ +---------------------------------------------+
> | | | | | |
> | | | | | |
> | | | | | |
> | | | | | |
> | | | | | |
> `\ '/ / ' / `\ '/ / ' / `\ '/ / ' /
Frank
but actual number is about 2,500 ft/lb. They also used a 50 cal BMG
which is over 13,000 ft/lb and barely budged the merry go round. The
Barrett rifle that handles this round weighs over 23 lbs.
I've wondered about spin. A 1500 fps slug through a rifled barrel of
one turn every 24 inches would spin at about 45,000 rpm. OTOH a Foster
slug through a smooth bore would not spin. Wonder what differences
would be on target?
Eric Gisse
[snip all, no interest]
Imagine that. People stopped responding to your previous stupid
threads and here appears another one.
NoEinstein
> If myth-busters used simple engineering none of these experiments would be
> needed, but then there would be no need
> for a how that makes millions.
>
Jim: If you are a clear-thinking engineer (or type), welcome! I
believe MythBusters would still have a show that makes millions if the
science was clearly understood and explained. As with the 1887
Michelson-Morley interferometer experiment which failed to show fringe
shifts, it was the inability of over a century of scientists to
realize this: That experiment was designed WITHOUT a control, or
unchanging light course. When a merry-go-round rotates at only
walking speed after having 200,000 "foot-pounds" of force applied, why
don't those people ask why? The reason is that they keep using an
easily disproved 1830 equation of Coriolis, KE = 1/2 mv^2. All of the
'g' forces stated about pilots in flight greatly exaggerate the
forces, because they have used Coriolis's wrong formula! If you have
any other thoughts, hang around! — NoEinstein —
NoEinstein
Dear Sam: If you suppose you know things about physics that better
explain that merry-go-round experiment, please enlighten us. Note
that is is YOU who sought me out, not the other way around. — NE —
NoEinstein
>
> I saw that show and remember the 5,000 ft/lb claim for a shotgun slug
> but actual number is about 2,500 ft/lb. They also used a 50 cal BMG
> which is over 13,000 ft/lb and barely budged the merry go round. The
> Barrett rifle that handles this round weighs over 23 lbs.
>
> I've wondered about spin. A 1500 fps slug through a rifled barrel of
> one turn every 24 inches would spin at about 45,000 rpm. OTOH a Foster
> slug through a smooth bore would not spin. Wonder what differences
> would be on target?
Dear Frank: You are MY type of scientist! I would bet that a non-
spinning bullet would do much less harm to the metal. Since part of
the energy of the recoil reaction is put into spinning the bullet, no
one can claim that there is extra KE, there. Something about metal
doesn't like combined stresses, like compression and torque.
Unfortunately, I'm spread too thin, time wise, to perform any
experiments myself. I do enjoy watching that new TV show about the
company that custom modifies guns for customers. As a kid, I was
always impressed by anything that could produce an effect at a
distance. Because I lived inside of a town where guns were
prohibited, I became a crack shot with a slingshot. It's still my
weapon of choice, because those never go off... accidentally! —
NoEinstein —
NoEinstein
Eric: That's their loss, not mine. — NE —
NoEinstein
Correction: 'it' is... not is is. — NE —
Eric Gisse
Then why can't you stop being such an attention whore for just a week?
Try it. Stop posting here for a week. See if anyone out of your
thousands of fans asks you to come back.
jbriggs444
It's sad -- the premise here had potential before NoEinstein mangled
it.
Why are "force" or "energy" inappropriate measures for this situation?
Why does the public confuse force and energy?
What quantity does the "foot-pound" measure?
Where does the energy go?
What's the relationship to the efficiency of a rocket motor?
If Newton's third law holds good, how can there be an asymmetry
between the force of the pellets on the target and the force of the
shotgun on the shooter's shoulder?
Why does stupidity sell better than intelligence?
Puppet_Sock
[snip]
> Why are "force" or "energy" inappropriate measures for this situation?
They are not inappropriate. They were used inappropriately.
Mostly by numstein.
The quantity that is actually wanted is impulse. This can be
related to energy and force, but it takes some work.
> Why does the public confuse force and energy?
Their level of understanding is not adequate to confuse the two.
> What quantity does the "foot-pound" measure?
The amount of energy involved in applying a force of
one pound through one foot. Clever name, what?
> Where does the energy go?
Florida, for its vacation.
> What's the relationship to the efficiency of a rocket motor?
None.
> If Newton's third law holds good, how can there be an asymmetry
> between the force of the pellets on the target and the force of the
> shotgun on the shooter's shoulder?
By the time the bullet (or pellets) get out of the gun,
lots of energy is going off in the form of hot gas.
Then, there's some small loss to air friction between
the gun and the impact. For pellets, this might be an
appreciable fraction of the total.
As well, the force is not what is important here. What
is important here is impulse. The impulse that gets
transferred to the merry-go-round is more complicated
than the energy in the bullet, or the shell before it is
fired for that matter.
> Why does stupidity sell better than intelligence?
It's the Phillip J. Fry principle:
Clever things make people feel stupid.
New things make people feel frightened.
This is why soap operas can stay on broadcast
television for decades, while Firefly lasted only
one short season.
Socks
NoEinstein
Eric: Obviously, you can't get enough "attention" without going where
I, an attention getter, post. Jealously isn't science, Eric.
Androcles' investigation of you shows that you were a career student
who finally flunked out of your physics major. What do you have to
offer any discussion of science other than the jealously of not being
able to contribute at all? Note: If you don't like being exposed as
a loser, don't reply on my posts for a week and see if my readers give
a damn. — NE —
NoEinstein
>
>
>
>
>
>
>
>
>
> > On Apr 11, 12:10 am, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > > On Apr 10, 8:35 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > > On Apr 10, 12:38 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
> > > > [snip all, no interest]
>
> > > > Imagine that. People stopped responding to your previous stupid
> > > > threads and here appears another one.
>
> > > Eric: That's their loss, not mine. — NE —
>
> > Then why can't you stop being such an attention whore for just a week?
>
> > Try it. Stop posting here for a week. See if anyone out of your
> > thousands of fans asks you to come back.
>
> It's sad -- the premise here had potential before NoEinstein mangled
> it.
>
> Why are "force" or "energy" inappropriate measures for this situation?
Force is the appropriate measure. Since the merry-go-round has high
inertia, only continuous forces, like falling water on a water wheel,
or a rocket motor, tell the force needed to cause a given rate of
rotation of the wheel.
> Why does the public confuse force and energy?
Energy is the capacity to do work. And work is directly proportional
to the force, in pounds, needed to do the work. So, energy is
actually the pounds of force that are available to do work. Note: By
looking at the force BEFORE the work is performed, no one could make
the mistake of assuming that the "work" is measured by the distance
moved, rather than by the force. For all continuously-applied forces,
the accrued coasting distances within the total distance of travel at
any datum point require ZERO force to cause that coasting distance.
> What quantity does the "foot-pound" measure?
In Engineering, foot-pounds measures moment—the tendency to rotate
about any particular point—and torque, the tendency to rotate about a
known center of rotation. Foot-pounds has NEVER been a valid measure
of impact delivery potential. The correct units for KE is pounds!
> Where does the energy go?
All energy in a closed system is preserved. If objects collide, the
resultant mass-velocity-direction vectors will correspond to the KE
that was in the main object doing the hitting. But part of that
energy will be a frictional heating of the many parts. However, the
total of all the mass-energy remains constant—as required by the Law
of the Conservation of Energy-Mass.
> What's the relationship to the efficiency of a rocket motor?
Solid fuel rockets, in particular, have a rated thrust in kips or
pounds, and a rated burn time. You will have to define what you mean
by... "efficiency" — compared to what?
> If Newton's third law holds good, how can there be an asymmetry
> between the force of the pellets on the target and the force of the
> shotgun on the shooter's shoulder?
The recoil of a gun against one's shoulder may not be perfectly axial
with the path of the bullet or pellets. But you can be sure that the
impact force of that bullet (or pellets) will match the COMPONENT of
the recoil that IS in the direction of the those. Is that what you
mean by asymmetry?
> Why does stupidity sell better than intelligence?
It doesn't! MythBusters knows the show has "credibility" with people
who like destruction, but don't like science. The cast seems
powerful, because they get to debunk things done in movies or taken
from urban legend. Their bubble of perfection will get busted if
those people have to admit that a term... "force, in foot-pounds" is
wrong. They have used that term, on average, once per week for 12
years. It was a major bit of science truth when the show admitted
that it erred to say that a truck hitting a concrete wall has double
the impact effect. Hasn't everyone heard Newton's law that says: "For
every action there is an equal and opposite reaction? The impact
FORCE (in pounds) of the truck could never have the concrete wall
pushing back with twice that force! — NoEinstein —
NoEinstein
>
Dear P. S.: There is no person on Earth who uses force and energy more
correctly! Hell, I've even authored the equation that replaces
Einstein's E = mc^2 and Coriolis's KE = 1/2 mv^2; it's: KE = a/g (m)
+ v / 32.174 (m). Any questions? — NoEinstein —
Sam Wormley
> On Apr 11, 2:27 pm, Puppet_Sock<puppet_s...@hotmail.com> wrote:
>> >
> Dear P. S.: There is no person on Earth who uses force and energy more
> correctly! Hell, I've even authored the equation that replaces
> Einstein's E = mc^2 and Coriolis's KE = 1/2 mv^2; it's: KE = a/g (m)
> + v / 32.174 (m). Any questions? — NoEinstein —
>
>
>>
>>
Eric Gisse
> On Apr 11, 3:51 am, Eric Gisse <jowr...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
> > On Apr 11, 12:10 am, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > > On Apr 10, 8:35 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > > On Apr 10, 12:38 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
> > > > [snip all, no interest]
>
> > > > Imagine that. People stopped responding to your previous stupid
> > > > threads and here appears another one.
>
> > > Eric: That's their loss, not mine. — NE —
>
> > Then why can't you stop being such an attention whore for just a week?
>
> > Try it. Stop posting here for a week. See if anyone out of your
> > thousands of fans asks you to come back.
>
> Eric: Obviously, you can't get enough "attention" without going where
> I, an attention getter, post. Jealously isn't science, Eric.
Self awareness is a beautiful thing.
NoEinstein
NoEinstein
Sam: Your graphics is an attack on "the messenger". You do that
because you, nor anyone, can attack my correct New Science! — NE —
NoEinstein
>
Eric: Attempting to flatter me, or by sarcasm, to flatter yourself,
isn't discussing science. Long before you flunked-out in physics, you
couldn't hold up your end of a conversation on science. Hell, you
can't even paraphrase that "air hockey table" confirmation experiment
(sic) of KE = 1/2 mv^2. Though you've been gone for a while, you
never have been in this group—as far as delivering on your false
claims about science. Remember, Eric, folks! He's the one who lies
about science, and then lies in a vain attempt to cover-up his errant
claims. — NoEinstein —
NoEinstein
Where are you, Eric: Is "the truth" too painful for you to hear? —
NE —
NoEinstein
Folks: The next time MythBusters refers to "foot-pounds of... force"
are any of you readers going to laugh your heads off? If not, you
should re read my original post on this subject. — NoEinstein —
Eric Gisse
> On Apr 14, 10:28 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
> > On Apr 13, 7:35 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
[...]
Who are these 'folks' you imagine you are talking to? All I see is you
talking to yourself.
NoEinstein
Eric go away. People are busy with Spring business and with Spring
Vacations. They will be back, soon. They’ll see that you are still
my jealous groupie. Aren't you catching enough fish in Alaska? — NE —
alie...@gmail.com
> foot-pounds (sic) of force. A nominal six foot diameter, human-
> powered merry-go-round—such as are found in most public parks—requires
> about five pounds of continuous force to cause rotation. For safety,
> the friction of the ball pivot is intentionally kept that high so that
> the merry-go-round won’t keep spinning for long once the motive force
> stops.
Foot-pound is a unit of energy, not force. That's why it's commonly
used in ballistic tables for a given gun-bullet combination's "muzzle
energy".
What do you mean by "five pounds of continuous force"? In the
relevant systems, the force unit is the "pound-force". It's called
that to differentiate it from the mass unit "pound" (unless you prefer
systems that use the "slug").
The pound-force is a static unit; it isn't measured over a distance.
Apply a specific force over a specific distance, and you are back to
measuring energy.
> The ‘myth’ that was to be verified was whether or not the impacts of
> bullets would be enough to keep a playground merry-go-round spinning.
First, they wanted to *start* it spinning.
And of course it wasn't exactly a "myth" as they usually do, just a
trick pulled in a movie.
> The initial shots penetrated the pipe rails, which were then
> reinforced with what looked to be 6” square x ¼” thick steel plates.
> Subsequent shotgun blasts easily indented the steel plates at least
> 1/16”, but rotated the merry-go-round only about a foot per shot.
> This was after the bearings had been replaced with those of lower
> friction. The show was correct to cite the inertia of the merry-go-
> round for its resistance to rotation from impulse forces.
What does "impulse forces" mean?
> Finally, after about thirty shotgun blasts were fired at the steel
> plates in quick succession, the merry-go-round rotated at a slow
> walking speed of about two miles per hour. The MythBusters crew
> seemed thrilled. None of those ’experts’ were perceptive enough to
> realize that they had just put a supposed 200,000 foot-pounds of
> ’force’ against that merry-go-round. That amount of force would have
> pulverized the device, but clearly it did not. The reason: The 5,000
> foot-pound (sic) of force is never more than about 40 pounds! Hear
> THIS, MythBusters: Newton’s third Law of Motion states that “for every
> force, there is an equal and opposite reactive force.”
So what?
> force, there is an equal and opposite reactive force.” That means
> that the maximum impact force of the bullet can’t exceed the estimated
> forty pound recoil of the shotgun!
"impact force", "forty pound recoil"... meaningless.
> Also, consider this: No 5,000
> pound force would indent a ¼” thick steel plate that has a minimum 20
> kip per square inch strength. Yet, the plates were easily indented.
Did you mean "kiloPascal"?
> Why? *** Because all material have varying strengths depending on the
> speed of application of the loads.
I have to stop you here. Muzzle *energy* is the important thing to
focus on, not "pounds of force" or anything else.
Getting the thing spinning requires a minimum amount of *energy* to
overcome the static drag of the bearing, and keeping it spinning
requires supplying just enough energy to overcome the dynamic drag of
the bearing.
Both can be done discretely or continuously. One thing you
definitely got right is mentioning the ability of the structure to
*accept* the energy as it is presented. Bullets aren't *designed* to
deliver all their kinetic energy as momentum transfer, they are
designed to do what they did, to deform and penetrate a target.
Muzzle energy is the product of the slug's mass and velocity. A
slow, heavy bullet can deliver the same energy to a target structure
as a light, fast bullet if their mass x velocity product is the same.
Point is that the rate of energy delivery is also strongly dependent
on how long the energy is available, and how fast the structure can
absorb it *without deformation*.
If you or I stand next to the merry go round and apply the exact
same amount of energy as the .50 rifle did, but much slower, our hands
won't punch holes in the pipes or deform them; instead we can make it
spin.
Consider intermediate cases like a black powder rifle throwing
large, slow lead balls, or even throwing bowling balls at it; same
energy, delivered slower, you get spin (and maybe a bent rail or two
but no holes).
Mark L. Fergerson
NoEinstein
>
Dear Mark L. Fergerson: You, or anyone else, err if you suppose to
explain to me—the man who has rewritten-the-book on mechanics—what
forces and KE are. Mathematicians wrongly supposed that it would be
desirable for every calculation in physics to have different units so
that those air brains can tell things apart. If you will read back
into my thread for a week or so, you should get more than one clear
explanation of what is going on. The force of impact NEVER exceeds
the recoil of the weapon (Newton's 3rd Law). And the 1830 Coriolis
equation used to calculate the kinetic energy VIOLATES the Law of the
Conservation of Energy-Mass by getting out more 'hamburger' than the
'steak' that was ground up!
ALL of the stated KE values for all weapons in use, and ALL stated 'g'
forces that pilots supposedly experience in accelerating or turning
are WRONG! The correct kinetic energy equation is my own: KE = a/g
(m) + v / 32.174 (m). Such is an actual EQUATION, because the units
are POUNDS on the left and POUNDS on the right. As any engineer will
tell you, foot-pounds is a measure of moment, or its close cousin,
torque. Real world experience will trump 'physics' shallowness every
time! — NoEinstein —
Jacob
>On Apr 10, 12:38 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
>> Also, consider this: No 5,000
>> pound force would indent a ¼” thick steel plate that has a minimum 20
>> kip per square inch strength. Yet, the plates were easily indented.
>
> Did you mean "kiloPascal"?
http://en.wikipedia.org/wiki/Kip_(unit)
In the United States, a kip is a non-SI unit of force that equals
1,000 pounds-force, used primarily by architects and engineers to
measure engineering loads.
http://www.engineersedge.com/material_science/yield_strength.htm
Some examples of yield strength for metals are as follows.
Carbon steel 3.0 x 10^4 to 4.0 x 10^4 psi
"20 kip per square inch strength" seems low.
NoEinstein
Dear Jacob: I don't use ultimate strength design. I was trained to
use working stress design, and that provided just the safety factor I
liked. I use the AISC Steel Handbook for most flexural members.
Things like anchor bolts and connection details, I like the working
stress method. If the connections are done right, buildings will be
OK in most loading conditions. — NoEinstein —
NoEinstein
Jacob: Another thought: Why is it that blunt-ended 12 gage slugs
caused a dent in the likely mild steel plates comparable to the
pointed 50 caliber? I know for a fact that most of the penetrating
power of tank-busting shells, like the 25 mm Bradley anti tank gun, is
due to the very high hardness and high specific gravity of the point
of (or the entire) projectile. A principle of engineering is that the
pressure (concentrated force per square inch) goes up as the
projectile gets more pointed. Another huge factor is that such sudden
pressure is transferred in a microsecond through the entire thickness
of the metal armorment. That speed of application of the load turns a
single thickness of steel to 'butter', but not in direct proportion to
either the correct momentum, or the incorrect KE, if Coriolis's `1830
equation is used. Projectiles can be stopped by layering the
armorment with hard and soft materials so that the P/A pressure is
transferred only a short way. And the devastating spin of the
projectile can be countered by including in the softer material
abrasives. The latter bite into the projectile and stop the spin in
like an inch and a half of penetration. Lest I give away some of the
secrets for building bomb-proof bunkers, I'll stop. The important
thing about this post is that it is the recoil of the weapon that
determines the hitting power of the bullet, not the calculated KE by
any formula. — NoEinstein —
PD
> On Apr 17, 12:57 am, "n...@bid.nes" <alien8...@gmail.com> wrote:
>
> Dear Mark L. Fergerson: You, or anyone else, err if you suppose to
> explain to me—the man who has rewritten-the-book on mechanics—what
> forces and KE are. Mathematicians wrongly supposed that it would be
> desirable for every calculation in physics to have different units so
> that those air brains can tell things apart.
I think these two sentences capture everything one needs to know about
John Armistead's (NoEinstein's) state of mental health, and how that
should guide your expenditure of time regarding him.
NoEinstein
Folks: Whatever my state of mind, Paul Draper avidly seeks me out,
because he knows he can get a lot of exposure from my readership.
Isn't PD a sad case? Ha, ha, HA! — NoEinstein —
Eric Gisse
What readership?
jbriggs444
> [snip]
>
> > Why are "force" or "energy" inappropriate measures for this situation?
>
> They are not inappropriate. They were used inappropriately.
> Mostly by numstein.
Fair enough.
> The quantity that is actually wanted is impulse. This can be
> related to energy and force, but it takes some work.
Just so. They can be related.
Impulse could be computed from energy, given figures for muzzle
velocity or figures for total pellet mass.
Impulse might be computed from force if you could identify
what force you're talking about and come up with a force
profile over time. "Force" in the sense of "sports
science" where a linebacker hits a quarterback with
3600 pounds of "force" is just a number with little or
no physical significance other than that it's big enough
for a viewer to say "wow" and precise enough for
a vewer to say "that's science".
[I can watch that show for at most 30 seconds
before being dumfounded by the fact that we can
as a species be competent enough to create a
television set and ignorant enough that somebody
can make a living producing shows like that]
> > Why does the public confuse force and energy?
>
> Their level of understanding is not adequate to confuse the two.
"Too ignorant to be confused". I suppose that's
a meaningful concept.
> > What quantity does the "foot-pound" measure?
>
> The amount of energy involved in applying a force of
> one pound through one foot. Clever name, what?
Indeed.
> > Where does the energy go?
>
> Florida, for its vacation.
Deformation of the pellets and of the target,
vibrational modes and the resulting heat
and sound. Kinetic energy in the deflected
pellets, if any.
And, of course, as you point out below,
heat and kinetic energy in the exhaust
gasses, sound and air resistance.
> > What's the relationship to the efficiency of a rocket motor?
>
> None.
The "efficiancy" of a rocket motor can be judged in terms
of the kinetic energy imparted to the payload as a fraction
of the chemical energy content of the expended fuel.
A motionless rocket has an efficiency of 0% in this sense.
A rocket moving at exhaust velocity has an efficiency of
100% in this sense.
Efficiency is directly proportional to vehicle velocity.
d (1/2 mv^2) = mv dv
dv is proportional to expended fuel in a rocket motor.
[At least until you stop holding mass constant]
Similarly, we may look at the kinetic energy imparted
to a carousel as a fraction of the chemical energy in
the powder in a shotgun round.
A motionless carousel picks up negligible kinetic
energy on impact from the pellets. The efficiency is
0%.
A moving carousel picks up some kinetic energy.
The efficiency is (to first order) proportional to the
velocity of the carousel at the point of impact.
To first order, the two situations are mirror images
of one another, reversed in time.
> > If Newton's third law holds good, how can there be an asymmetry
> > between the force of the pellets on the target and the force of the
> > shotgun on the shooter's shoulder?
>
> By the time the bullet (or pellets) get out of the gun,
> lots of energy is going off in the form of hot gas.
True, though that's not the main factor in play here.
> Then, there's some small loss to air friction between
> the gun and the impact. For pellets, this might be an
> appreciable fraction of the total.
True, though that's not the main factor in play here.
> As well, the force is not what is important here. What
> is important here is impulse. The impulse that gets
> transferred to the merry-go-round is more complicated
> than the energy in the bullet, or the shell before it is
> fired for that matter.
Not that much more complicated. If you make the
simplifying assumption that the pellets hit the
carousel and stick in, for instance, a block of wood
(inelastic collision) then the computation should
be well within the grasp of a first year physics
student.
NoEinstein
Eric: Oh, that's right... you can't read! Ha, ha, HA! —
NoEinstein —
NoEinstein
Briggs: Work is NOT needed to compute KE! The latter is the increase
in the force-delivery-potential of falling (or accelerating) objects.
Count the number of seconds the force of gravity acts on the falling
object and multiply by the static weight to find the KE, but with THIS
important proviso: ALL objects sitting at rest prior to being dropped
have a beginning downward force = the object's static weight. My
correct equation is: KE = a/g (m) + v / 32.174 (m). — NoEinstein —