Building a self looping "SMOT"

[TinselKoala]

@MH
I was thinking the same thing,in regards to setting up a test exit ramp,and measureing how much the ball is slowed when makeing the 90* turn.By useing my HD camera,and having a 3 decimal point timer behind the ramp,i could calculate the speed loss of the ball after the turn. If the ball lost half it's speed,that would mean the ball lost half it's kinetic energy + friction losses.

Wrong.
Kinetic Energy == 1/2 (mass x velocity SQUARED)  so half the speed = 1/4 the KE.

And don't try to say that this error doesn't matter! If you have been thinking all this time that KE is directly proportional to velocity, rather than to the square of the velocity... that's a pretty big conceptual error and will "resonate" through your whole world-view.

http://en.wikipedia.org/wiki/Kinetic_energy
http://www.physicsclassroom.com/class/energy/u5l1c.cfm
http://formulas.tutorvista.com/physics/kinetic-energy-formula.html

[tinman]

Wrong.
Kinetic Energy == 1/2 (mass x velocity SQUARED)  so half the speed = 1/4 the KE.

And don't try to say that this error doesn't matter! If you have been thinking all this time that KE is directly proportional to velocity, rather than to the square of the velocity... that's a pretty big conceptual error and will "resonate" through your whole world-view.

http://en.wikipedia.org/wiki/Kinetic_energy
http://www.physicsclassroom.com/class/energy/u5l1c.cfm
http://formulas.tutorvista.com/physics/kinetic-energy-formula.html

Yes,i was just comeing back to correct that after doing some reading,but i see(as usual) you have quickly jumped on my mistake. But it was a mistake that is in our favour. So the correction would be,if our ball slows by 1/2 of it's speed makeing that 90* turn,then we have lost 4x the kinetic energy that ball had-Did i get that right?.
Now another mistake i made (some where back in the thread)was to say mass is directly related to weight,which is not the case. But who is going to the moon or another planet anyway.

You are good at picking up on mistakes of others TK,so maybe you can watch the video,and find the potetial energy gain we could get from the system?-MH has missed it so far. Can you spot it as well as you spot peoples mistakes?.

[Newton II]

@TK

You seem to be intelligent in answering qustions.  I have got one question:

The motion of earth around the sun, motion of electrons around nucleus and several such cases are accelerated motions.   For accelerated motions you have to apply force continuously inturn supply continuous energy since these bodies are in motion.  For bodies in uniform motion in space you need not supply energy unless you intend to accelerate it.

Who is supplying energy to keep these bodies in continuous accelerated motion?   Is it God or field?   If gravitational field is supplying  continuous energy to keep earth in accelerated motion then should gravitational field be considered as force field,  energy field or God field or just a space time curve in space?

The following wiki page says that psuedo-forces can do work.

http://en.wikipedia.org/wiki/Fictitious_force

[TinselKoala]

Tinman asked,
"So the correction would be,if our ball slows by 1/2 of it's speed makeing that 90* turn,then we have lost 4x the kinetic energy that ball had-Did i get that right?."

Well, that's not quite the way I'd put it in words. If you lose 1/2 the speed, you wind up with 1/4 the KE remaining. 

For example, if you have a 1 kilogram mass and you are first going 2 m/sec, your KE is (1kg x 2 m/sec x 2 m/sec)/2 or 2 Joules. If you slow down to 1 m/sec your KE is (1kg x 1 m/sec x 1 m/sec)/2 or 0.5 Joule. Half the speed = a quarter the KE.

So you are losing 3/4 of the KE the ball had. Or you could say, if you reduce (or increase) your speed by a factor of 2, you reduce (or increase) your KE by a factor of 2², or 4. If you reduce speed by a factor of 3 (leaving 1/3 of what you started with) you reduce KE by 3², or a factor of 9, leaving 1/9 of what you started with.

Now, linear _momentum_ is just "p = mv" so maybe that's what you were thinking of at first. But in momentum, the "v" is a vector, that is, it has a direction as well as a magnitude, and momentum is also always conserved.

http://en.wikipedia.org/wiki/Momentum

I'll watch the video _yet again_ with an eye to what you are looking at, but I still think some of your basic assumptions are wrong, and so it is likely that you and I will interpret what we are seeing differently.

(Please check my math, I'm notorious for misplacing decimals or forgetting to divide by two in the KE calculation).

[TinselKoala]

@TK

You seem to be intelligent in answering qustions.  I have got one question:

The motion of earth around the sun, motion of electrons around nucleus and several such cases are accelerated motions.   For accelerated motions you have to apply force continuously inturn supply continuous energy since these bodies are in motion.  For bodies in uniform motion in space you need not supply energy unless you intend to accelerate it.

Who is supplying energy to keep these bodies in continuous accelerated motion?   Is it God or field?   If gravitational field is supplying  continuous energy to keep earth in accelerated motion then should gravitational field be considered as force field,  energy field or God field or just a space time curve in space?

The following wiki page says that psuedo-forces can do work.

http://en.wikipedia.org/wiki/Fictitious_force

In the accelerated motions you mention, uniform circular motion, the direction of the applied force is always at a right angle to the circular motion. The accelerated object does not travel in that direction, though, so there isn't any work done, therefore no energy is "expended".

From your Wiki reference:

Fictitious forces can be considered to do work, provided that they move an object on a trajectory that changes its energy from potential to kinetic.

The acceleration due to gravity that keeps planets in circular orbits does not change the planet's energy from PE to KE, because it always acts at a right angle to the motion. No work is done since the force does not act over a distance (the radius of the orbit isn't changing). For closed elliptical orbits the inward work is balanced by the outward work so over a full orbit, again no net work, no energy expended.