Potentially painful

[Larry Brasfield]

"Robert Monsen" <rcsurname@comcast.net> wrote in
message news:B-adnbQgnM3fx7LfRVn-rQ@comcast.com...

Larry Brasfield wrote:
"John Larkin" <jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote
in message news:sa3t2113mifbj3e38eovdn74qc546dni4i@4ax.com...

On Tue, 8 Mar 2005 12:44:56 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...

No, as I recall, kinetic energy is 1/2 * m * v^2.

I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).

At low velocities, a rocket is a very inefficient source of
propulsion; at near-zero velocity, it's using its usual amount of fuel
but hardly delivering any kinetic energy to the vehicle. As velocity
increases, efficiency improves (or rather becomes less terrible) and
vehicle energy accumulates faster. That trend continues until you run
out of fuel.

In a system that goes from extremely inefficient to only rather
inefficient, it's not hard to shape the efficiency curve into a
quadratic. A cog railway can be nearly 100% efficient, so it will need
increasing amounts of fuel if it accelerates at constant gees, but
anywhere on the path it will be a lot more efficient than a rocket.

That make sense?


Yes. I should have known this little puzzle would
not take you folks long to sort out. The key is to
consider that starting condition. All the energy ends
up in the exhaust at the limit of zero rocket speed.
The problem gets a little easier to see if it involves
chucking cannonballs from a railway car.

Ok. I guess I'm feeling thick today, but it sounds like your solution is to attach a reference frame to the accelerating rocket. I
don't know enough physics to determine if that's reasonable.

If your surmise about my choice of reference frame
was correct, I would not have had to mention "the
limit of zero rocket speed" since that speed would
be zero at all times. I don't know a lot of physics,
but I know enough to declare that accelerating
reference frames are not reasonable!

The rate of change of kinetic energy of the rocket when the rocket isn't moving is still the time derivative of 1/2mv^2, which is

F^2*t/M

where M is the (assumed unchanging) mass of the rocket, and F is thrust. This is a function of t, meaning that the rate of energy
at t=0 is 0, so all the energy goes into the fuel. That is nice, but what happens at t=1? We are still increasing our 'rate of
change' of energy...

Gonna have to suppose some parameters and do
some math for that one. Not! (I've got enough of
that to do for pay at the moment.)

If you were to say that, for example, the rate of change of energy of the thrust (from the point of view of a stationary observer)
was decreasing with t, and the sum of these was constant, I'd buy that.

My reason for mentioning the infinitessimal time
during which the rocket velocity is zero (for the
observer implicit in speaking of E = M V^2 / 2),
is to agree with John's explanation and elaborate
that, over some very low range of velocities, one
must realize most of the power will be imparted to
the rocket's exhaust. Only by ignoring what goes
into the exhaust can that puzzle remain a puzzle.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Rich Grise]

On Wed, 09 Mar 2005 12:02:05 -0800, Robert Monsen wrote:

sergiometra wrote:
Here is a web page written by somebody who has been driven insane by
this problem:

http://nov55.com/enrgy.html

His solution is that kinetic energy is really mv, not 1/2 mv^2. He has
equations and everything. Pretty wierd stuff. Who knows?



I think he demonstrate the exact opposite;
if you jump from 2nd floor, speed at arrival is not double that a jumping
from 1st floor as it take less time from floor 1 to 0 than from 2 to 1 , but
energy is double as you need double energy to go up 2 floor than 1 , so
energy increase more rapidly than speed i/e Energy>mv;
regards
sergio



You didn't read the page. His thesis is that energy should be measured
as mv, not 1/2 mv^2. He claims that mv, and not 1/2mv^2 is conserved. He
uses rockets in a stationary reference frame to prove this (but doesn't
seem to take the kinetic energy of the fuel into account).

And nobody has yet mentioned the mass of the fuel itself. I'd think
that makes a difference, since m of the fuel/rocket assembly decreases,
not necessarily linearly. So the same force would cause more accelleration
toward the end.

I wonder what a 1,000,000 pound thrust plasma jet would look like. =:-O

Cheers!
Rich

 

[Rich Grise]

On Wed, 09 Mar 2005 10:47:17 -0800, Robert Monsen wrote:

sergiometra wrote:
The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'

if there is not resistance in circuit and cap, circuit will oscillate
forever;
otherwise energy dissipate in heat and oscillation eventually end
regards

Where is the oscillation coming from again?

Ringing.

I know that an LC circuit
near absolute zero will oscillate, but I didn't say anything about an
inductor.

Everything in real life has inductance - the wire from cap to cap is
the inductance here. (plus the self-inductance of the caps, and strays,
of course.)
....

Thus, instantaneous power through the resistor is

P(t) = V(t)^2/R

The integral of this from 0 to infinity is

C*V^2/4 = 1/2 * (1/2 C * V^2)

Doesn't time constant and e^xt have something to do with this?

Thanks,
Rich

 

[Robert Monsen]

Rich Grise wrote:

On Wed, 09 Mar 2005 10:47:17 -0800, Robert Monsen wrote:

sergiometra wrote:

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'


if there is not resistance in circuit and cap, circuit will oscillate
forever;
otherwise energy dissipate in heat and oscillation eventually end
regards

Where is the oscillation coming from again?


Ringing.

I assumed he meant no resistance + no inductance. He didn't specifically
mention inductance.

I know that an LC circuit
near absolute zero will oscillate, but I didn't say anything about an
inductor.


Everything in real life has inductance - the wire from cap to cap is
the inductance here. (plus the self-inductance of the caps, and strays,
of course.)
...

Of course, and everything has resitance. And capacitance...

Thus, instantaneous power through the resistor is

P(t) = V(t)^2/R

The integral of this from 0 to infinity is

C*V^2/4 = 1/2 * (1/2 C * V^2)


Doesn't time constant and e^xt have something to do with this?

Yes, it's the integral of that from 0 to infinity that gives the result.

For capacitance of C/2 with initial voltage V, discharging through R,
the voltage at time t is

V(t) = V * exp(-2*t/(R*C))

The power of the resistor at time t is thus

P(t) = V(t)^2/R = V^2*exp(-4*t/(C*R))/R

The indefinite integral of this is

-C*V^2/(4*exp(4*t/(R*C)))

evaluated at infinity, it's the constant, because of the exp(t) in the
denominator. Thus, the answer is the negative of that evaluated at 0,
which is

C*V^2/4

Thanks,
Rich

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Robert Monsen]

Rich Grise wrote:

On Wed, 09 Mar 2005 12:02:05 -0800, Robert Monsen wrote:


sergiometra wrote:

Here is a web page written by somebody who has been driven insane by
this problem:

http://nov55.com/enrgy.html

His solution is that kinetic energy is really mv, not 1/2 mv^2. He has
equations and everything. Pretty wierd stuff. Who knows?



I think he demonstrate the exact opposite;
if you jump from 2nd floor, speed at arrival is not double that a jumping
from 1st floor as it take less time from floor 1 to 0 than from 2 to 1 , but
energy is double as you need double energy to go up 2 floor than 1 , so
energy increase more rapidly than speed i/e Energy>mv;
regards
sergio



You didn't read the page. His thesis is that energy should be measured
as mv, not 1/2 mv^2. He claims that mv, and not 1/2mv^2 is conserved. He
uses rockets in a stationary reference frame to prove this (but doesn't
seem to take the kinetic energy of the fuel into account).


And nobody has yet mentioned the mass of the fuel itself. I'd think
that makes a difference, since m of the fuel/rocket assembly decreases,
not necessarily linearly. So the same force would cause more accelleration
toward the end.

I wonder what a 1,000,000 pound thrust plasma jet would look like. =:-O

Cheers!
Rich

The mass of the fuel causes the mass of the rocket to decrease as time
goes by. My old physics book has a section on dynamic mass systems,
where newton's laws must be modified to take this into account. They
talk about rockets and machine guns mounted on rail cars...

In the words of the immortal col kurtz, "The horror, the horror..."

http://www.nationstates.net/cgi-bin/index.cgi/-1/page=display_nation/nation=col_kurtz

All this means is that figuring out the actual velocity at a given time
for a given fuel burn is fairly hard. Thus, figuring the rate of change
in kinetic energy of both rocket and expended exhaust at an arbitrary
time t is also hard. However, we can guess that they add up to a
constant, which is related to the constant energy expended by the rocket
motor.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Don Kelly]

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:mbadnX4-scECF7PfRVn-gg@comcast.com...

Larry Brasfield wrote:
"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...

No, as I recall, kinetic energy is 1/2 * m * v^2.



I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp.

You've put that behind you then.

I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?


Beats me. May have something to do with the exhaust, as Don says. Might
also have to do with the heat generated in the rocket. Probably
something to do with reference frames, and the velocity of the fuel
exhaust changing with regards to a stationary observer. That is all it
really can be, since the closed system is rocket, fuel, exhaust. Inside
the system, the energy is constant. However, working it out is another
matter.
-----------------

Hold on here:
Acceleration constant- force constant- rate of consumption of fuel is
constant.
However acceleration is *rate of change of velocity* so with constant
acceleration the velocity is increasing.
v=vo +at increasing linearly. Lets assume initial velocity vo is 0
going for a given time T the velocity is aT and the KE is (0.5)*M*(aT)^2
Energy input over that length of time =integral of force times distance. The
force is Ma (constant) and the distance is the integral of velocity over the
time =0.5*a*(T^2)
so energy input is 0.5*M*a*(a)*(T^2) =(0.5)M (aT)^2

Newton lives!
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).


Energy is a funny thing. You have to be careful about how you count it up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Robert Monsen]

Don Kelly wrote:

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:mbadnX4-scECF7PfRVn-gg@comcast.com...

Larry Brasfield wrote:

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...


No, as I recall, kinetic energy is 1/2 * m * v^2.



I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp.

You've put that behind you then.


I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?


Beats me. May have something to do with the exhaust, as Don says. Might
also have to do with the heat generated in the rocket. Probably
something to do with reference frames, and the velocity of the fuel
exhaust changing with regards to a stationary observer. That is all it
really can be, since the closed system is rocket, fuel, exhaust. Inside
the system, the energy is constant. However, working it out is another
matter.

-----------------
Hold on here:
Acceleration constant- force constant- rate of consumption of fuel is
constant.
However acceleration is *rate of change of velocity* so with constant
acceleration the velocity is increasing.
v=vo +at increasing linearly. Lets assume initial velocity vo is 0
going for a given time T the velocity is aT and the KE is (0.5)*M*(aT)^2
Energy input over that length of time =integral of force times distance. The
force is Ma (constant) and the distance is the integral of velocity over the
time =0.5*a*(T^2)
so energy input is 0.5*M*a*(a)*(T^2) =(0.5)M (aT)^2

Newton lives!

The question is, why is the rate of change of kinetic energy of the
rocket linear in time, whereas the energy supply is only burning
chemical fuel at a constant rate? On a moment to moment basis, how do
you reconcile this? I think we've decided that the chemical energy is
changed into kinetic energy of the rocket *and* the fuel. The rate of
change of kinetic energy of rocket is linear with time, but the rate of
change of kinetic energy of the fuel is decreasing with time, owing to
the fact that the rocket is speeding up, so the difference in rocket
speed vs thrust speed is decreasing with time. The sum of the two equals
a constant.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[robin.pain@tesco.net]

"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote in message news:<d9oXd.34$UW6.2012@news.uswest.net>...

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...
No, as I recall, kinetic energy is 1/2 * m * v^2.


I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).

Learning from other people's mistakes is nearly as effective as
learning from one's own.

Your teacher had the opportunity to come back with "Oh no! I don't get
it, Help me out here!" but I suspect that teaching 3rd formers turned
him into a facist.

If kids hate school then let them leave at fourteen and be brickies,
plasterers or plumbers because:

1) The pay is better.
2) It's healthly.
3) The building scene is a natural brat camp.
4) Tax evasion is a snip.
5) You can retire at fourty-five owning two houses.
6) We need more tradesmen.

And then the teacher can stay liberal and the geeks get better
service. Everyone's a winner.

Cheers
Robin

 

[Robert Monsen]

robin.pain@tesco.net wrote:

"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote in message news:<d9oXd.34$UW6.2012@news.uswest.net>...

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...

No, as I recall, kinetic energy is 1/2 * m * v^2.


I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).



Learning from other people's mistakes is nearly as effective as
learning from one's own.

Your teacher had the opportunity to come back with "Oh no! I don't get
it, Help me out here!" but I suspect that teaching 3rd formers turned
him into a facist.

If kids hate school then let them leave at fourteen and be brickies,
plasterers or plumbers because:

1) The pay is better.
2) It's healthly.
3) The building scene is a natural brat camp.
4) Tax evasion is a snip.
5) You can retire at fourty-five owning two houses.
6) We need more tradesmen.

And then the teacher can stay liberal and the geeks get better
service. Everyone's a winner.

I completely agree with this. At my local HS, the focus is on science
and math. Everybody has to pass algebra and science. Thus, you end up
with algebra and science classes full of bored senior stoners who could
care less, along with freshmen and sophmores who want to learn. You also
end up with people who can't pass the exit exams, and so can't get a
diploma even if they would be good at non-technical things like plumbing
or plastering.

Trade schools are a great solution. Do the French still do this?
Unfortunately, at least here in California, the 'trade schools' are
really set up for the kids who are troublemakers, and have been tossed
out of the normal high schools. Thus, even if they excel, they feel like
losers, because they couldn't do algebra.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Don Kelly]

I gave the solution for constant acceleration and constant mass. It is
messier for a rocket.

Initial mass =Mo=Mr +Km where Mr =mass of empty rocket and Km is the
initial fuel mass
mass of fuel at time t is m(K-t) where m is the mass per second of the
exhaust. Assuming the exhaust is at constant speed c with respect to the
rocket, then the force is cm ={Mo-mt)dv/dt
solving gives velocity =c*ln{Mo/(Mo-mt)} {See Artley -Fields and
configurations p262} which is not varying linearly with time so the KE is
not changing quadratically with time.
Integrate this to find the distance travelled in time t and multiply by the
force cm to get the total KE at time t

After t=K/m the fuel is exhausted and velocity is constant at 2c*ln(Mo/Mr)
and the mass is Mr.
so final KE appears to be 2*(c^2)*Mr*(ln(Mo/Mr))^2
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer


"Don Kelly" <dhky@peeshaw.ca> wrote in message
news:vCMXd.623741$8l.511855@pd7tw1no...

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:mbadnX4-scECF7PfRVn-gg@comcast.com...
Larry Brasfield wrote:
"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...

No, as I recall, kinetic energy is 1/2 * m * v^2.



I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp.

You've put that behind you then.

I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?


Beats me. May have something to do with the exhaust, as Don says. Might
also have to do with the heat generated in the rocket. Probably
something to do with reference frames, and the velocity of the fuel
exhaust changing with regards to a stationary observer. That is all it
really can be, since the closed system is rocket, fuel, exhaust. Inside
the system, the energy is constant. However, working it out is another
matter.
-----------------
Hold on here:
Acceleration constant- force constant- rate of consumption of fuel is
constant.
However acceleration is *rate of change of velocity* so with constant
acceleration the velocity is increasing.
v=vo +at increasing linearly. Lets assume initial velocity vo is 0
going for a given time T the velocity is aT and the KE is (0.5)*M*(aT)^2
Energy input over that length of time =integral of force times distance.
The
force is Ma (constant) and the distance is the integral of velocity over
the
time =0.5*a*(T^2)
so energy input is 0.5*M*a*(a)*(T^2) =(0.5)M (aT)^2

Newton lives!
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer



I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).


Energy is a funny thing. You have to be careful about how you count it
up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Don Kelly]

<robin.pain@tesco.net> wrote in message
news:bd24a397.0503080128.77ba467e@posting.google.com...

Biking is much more fun than electronics, especially as summer is
icumen in, but do you youngsters *know* about it?

E.g. you know that its a lot more fun discharging a small 370V
capacitor than a *huge* 5V one because energy is proportional to the
voltage *squared* (because if you double voltage then the current
doubles too and energy is their product)because
Energy = CV^2

I
-------------
| |
| |
V | Energy |
| |
| |
| |
-------------

The same thing applied to speed. If you double the speed you quaduple
the energy because Energy = MV^2

So if you hit a massive artic' while doing 1 mph (the lorry is
stationary) on yer bike then (assuming you weight "one" unit) you will
dissipate one unit of energy.

If you hit the same vehicle at 101 mph when it is doing 100 mph then,
ignoring the lorry's slight speed increase, you will dissipate the
difference between you doing 101 and 100 i.e.

101*101 - 100*100 == ~200

That's nearly 200 hundred times more dangerous.

Wazzat? ya don't believe it? (The local boy racer is reading this over
my shoulder, he is snorting disbelief). Well if you prefer a belief
system to a mathematical system then look at it this way: How much
energy do you have to put it to get ya bike up to 1 mph? Not much, you
just start pushing it and after a couple of strides both of you are up
to 1 mph but what about 100 to 101? First of all you have to *be*
running at 100 *before* you start pushing and *then* you must push and
get it to 101. The ground you have to cover *while* pushing is
evidently a hellofalot more. Does that not seem like a *lot* more
stored energy to you?

Cheers

Robin Pain (ouch)
In this case the interaction is between the lorry and the cyclist. The

change in KE is related to the change of the relative velocity. The KE as
measured with respect to the ground or to the galactic center is not germane
to the problem. If it were, then walking into a wall would have deadly
consequences. You have to look at the system that is involved. Ground is
not involved in this particular collision.
Part of the problem is that the KE expression that is most commonly used
deals with a single moving object with respect to a stationary reference.
This KE expression was originally developed for objects and is an
integration of force*distance.

If you grab the lorry not a problem, if you miss *and* fall, the next thing
you hit is the stationary ground at a velocity between 100 and 101 mph -
that is a different and far more painful thing. This is where you expend
nearly all of of the energy that you put in in getting up to 101 mph.

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Me]

Larry Brasfield wrote:

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...

No, as I recall, kinetic energy is 1/2 * m * v^2.



I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).

Actually, v=at only applies if F = mass * acceleration, and both mass
and acceleration are constant. So relativity is out, and so are rockets
(mass changes, as well as the fuel loss eventually means the rocket
thrust stops). The atmosphere angle makes the problem harder accept
again, the force is velocity dependent, so acceleration is not constant
unless the rocket adjusts. So assume a simpler problem: A hammer is
dropped on the moon. After t seconds it hits the ground with velocity
v, and KE=1/2 mv^2. Why does the velocity increase linearly with time,
while the KE goes as t^2?

Answer: It doesn't go as t^2. KE is a function of v and not t, so you
have to use the chain rule to find dK/dt. So,

dK/dt = dK/dv dv/dt
= 1/2m (2v) dv/dt
= m (at) a
= ma^2t

Or, more fundamentally, the kinetic energy is increasing because work is
being done a the hammer (work-energy theorem). Work is defined as a
force exerted over a distance, so

dK/dt = dW/dt
= d(Fx)/dt
= F (dx/dt)
= mav
= ma(at)
= ma^2t

Craig

 

[Rich Grise]

On Wed, 09 Mar 2005 09:41:33 +0000, sergiometra wrote:

Energy is a funny thing. You have to be careful about how you count it up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'


if there is not resistance in circuit and cap, circuit will oscillate
forever;
otherwise energy dissipate in heat and oscillation eventually end
regards
sergio

How about a thought experiment - for each plate, use two plates that
slide over each other, and that can move, so that you can widen the
plates to twice their size, causing the capacitance to double? Does
the voltage change? And what happens to the energy? (the plates are
very thin, so the thickness difference between overlapping plates
is negligible)

And what if you have just a movable plate so that you can change
the spacing?

Thanks,
Rich

 

[Robert Monsen]

Rich Grise wrote:

On Wed, 09 Mar 2005 09:41:33 +0000, sergiometra wrote:


Energy is a funny thing. You have to be careful about how you count it up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'


if there is not resistance in circuit and cap, circuit will oscillate
forever;
otherwise energy dissipate in heat and oscillation eventually end
regards
sergio


How about a thought experiment - for each plate, use two plates that
slide over each other, and that can move, so that you can widen the
plates to twice their size, causing the capacitance to double? Does
the voltage change? And what happens to the energy? (the plates are
very thin, so the thickness difference between overlapping plates
is negligible)

The voltage changes, and you do mechanical work on the system to change
it. Electrostatics causes a force to be exerted, and you need to move
against or with that force, thus doing work.

And what if you have just a movable plate so that you can change
the spacing?

Same thing.

Thanks,
Rich

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.