energy producing experiments

[pequaide]

The linear velocity of a tether ball is not changed by the winding or unwinding from the post.  The linear velocity is equal to the arch velocity no matter what size the circle. As the ball winds up or unwinds the arch velocity (the distance traveled around the arch of the circle) remains the same.

The angular momentum is constantly changing, on the winding up and unwinding, and angular momentum conservation is therefore a false concept because there is no application of outside force.

Using arch velocity in linear equations works without imaginary points of rotation or an imaginary radius. And why not; if the string is cut the linear velocity equals the arch velocity.

The block can give its motion back to the bullet; would you use the formula that always fails (1/2mv²) or the formula (F = ma) that always works.

[Furcurequs]

Hey pequaide,

How about I try to use the correct terminology to describe the scenario you were apparently trying to describe so that we can see what you got right and what you got wrong?

Before I do that, though, I would like to point out that "linear velocity" is a VECTOR quantity that has both a magnitude AND DIRECTION, so the term you might have meant to use in your description was "speed" - which, of course, at any given time is simply the MAGNITUDE of the instantaneous linear velocity.

So, in an ideal case of a tethered ball spiraling inwardly toward a pole or outwardly away from a pole as the tether either wraps around or unwraps from around the pole - and so where we can neglect air resistance and friction and stretching of the tether and whatnot - we should see the following:

The ball should maintain a constant SPEED as it moves throughout its spiraling path as its LINEAR VELOCITY CONSTANTLY CHANGES (in direction alone since its magnitude is the speed). Since the speed of the ball remains constant, its kinetic energy also remains constant or in other words is conserved.

The angular momentum of the ball around the center of the pole would also be constantly changing (as you were apparently pointing out) and this can be accounted for by seeing that the force the tether is exerting on the ball is not directed through the center of the pole but rather towards points around the circumference of the pole since the tether is wrapped around it and pulling on the pole tangentially. So, components of the force vector are in a direction that takes away from the angular momentum of the ball around the center of the pole.

That means you are correct about the angular momentum of the ball constantly changing, but since the force vector along the tether is not directed through the center of the pole we now must also include the pole and the mass of the earth in the total system when considering the conservation of angular momentum.

That means the ball must exchange angular momentum with the pole and the earth it's surely attached to in our scenario. Technically, then, the actual motion of the earth and pole are also affected, but the change in the motion of such a large mass would be so incredibly small and even undetectable that we normally just ignore it.

If the ball is released from the end of the tether it will continue moving at the same speed it had along the spiraling path - which I believe is what you were trying to point out - but with a velocity in the direction of the instantaneous velocity it had when it was released.

So, in summary, the speed of the ball is constant, the velocity of the ball is constantly changing (though in direction alone), the kinetic energy of the ball is constant (or conserved), the angular momentum of the ball around the center of the pole is constantly changing but the angular momentum of the system as a whole - which includes the earth and the pole - is conserved.

I hope that helps.

Now, I don't really know what you were talking about there concerning the block giving the bullet its motion. In the scenarios discussed in previous posts it seems the bullet ends up stuck inside the block.

I will point out that kinetic energy is a scalar quantity and so the V used in that equation is just the magnitude (the speed) of whatever velocity vector we are concerned with.

Some may think I'm nitpicking, but I'm surely not. Using the correct terminology and understanding the basic physics gives one incredibly powerful and time tested tools to model these simple problems.

Take care!

Dwayne

[Furcurequs]

Furcurequs quote "the angular momentum of the bullet moving along a linear path"

Wow. So it is: pick a point; any point--- unbelievable.

I will let you all argue about it, I know what is conserved.

Hey pequaide,

Sorry, but that's the way it works - and by definition.

Of course, you would want to try to choose your points wisely to make the math as easy as possible.

If I were standing next to a straight road along which cars were traveling, those cars would all have angular momentum relative to me no matter where they were along the rode. The numerical value would depend upon each car's mass, its speed and my distance away from the center of the lane of the road in which the car is traveling (assuming they are not a bunch of weaving drunk drivers, of course).

If I were standing in the center of the lane in the road in which a car was traveling and it was on a collision course with me or driving directly away from me, then the car would not have angular momentum relative to me.

Dwayne

ETA: In other words, if a moving mass is not headed directly toward or directly away from the chosen point, it has angular momentum relative to that point. ...but if the point is in line with the linear velocity of the mass, the moving mass has no angular momentum relative to that point.

[pequaide]

The positive direction is set by the experimenter or the person doing the math. I, the experimenter, can set the positive direction to be in any direction I choose. Instead of relating the direction to be where the object was I choose to set the positive direction to be where the object is. I can do the math more than once, I can tell the computer to set the positive direction to be the same as the path of the missile. The computer can do the problem faster than a change of direction can be detected. And now the computer is choosing the same path as the missile to be the positive direction. Therefore the linear momentum will always be the same.

It will always be the same until you are considering two or more objects, or there is an outside application of force.

The tether ball is the only object and there is no application of outside force. Whenever you ask the computer to tell you the momentum of the ball it will give you the same positive number. The computer is not living in the past.

Also: If you are getting an F = ma relationship for your acceleration then you know that the direction of applied force is in the same direction as the motion of the mass. If the motion direction of the mass and the applied force are skew then you will not get an F = ma relationship. Newton knew this of course and made it a requirement for the formula. I don't think I do anything that is not Newtonian.

An erroneous application of the meaning of vector is yet another stumbling block to keep people off of the proper path.

For a block/bullet to bullet motion transfer see 'smokin lamas'

[pequaide]

I am also posting this here so I can find it later.

We had an accident occurred at work on Monday of this week, no one was hurt but it will cost the corporation several thousand dollars. We had a tank lid that a tech was bringing back into the building on the forks of a hilo, he made a slight turn (his words) as he entered the building.

The 4' x 8' tank lid is made of HDPE and is slick material, and the hilo forks are polished steel.

As he made the turn the friction on the hilo fork provided centripetal force that tried to force the lid into a circular path; the previously existing centrifugal force was greater than the centripetal and the lid continued on its linear path. The lid slid off of the forks and broke up as it hit the floor.

Now hear is the question? How can centrifugal be a reactionary force to centripetal. There was enough centrifugal to equal the centripetal.

But there was not enough reactionary centripetal to equal the centrifugal.

The original force was from the fact that the lid was moving in a straight line; and the forks could not provide enough centripetal to make it do otherwise.

Centrifugal is not a mysterious force; it is a plain and simple fact that we all know. Objects will continue to move in a straight line unless compelled to do otherwise.

And maybe just a little harder to see; All points of mass in a moving circle are moving in a straight line that is tangent to the circle. But the points of mass are continually compelled to move out of that straight line by a constant application of centripetal force. And the centripetal force is a reactionary force to the previously existing straight line motion.

A proper conceptual understanding helps you make wheels that work; as opposed to dumping $2,000 lids on the floor.

[rlortie]

Now hear is the question? How can centrifugal be a reactionary force to centripetal. There was enough centrifugal to equal the centripetal.

But there was not enough reactionary centripetal to equal the centrifugal.

Obviously the lack of friction between lid and forks was not enough to create static inertia to hold the lid in place. There was no centripetal reactionary or otherwise to prevent inertia from maintaining the lid in straight trajectory. Centripetal is a physical fixed item, it does not change in value to meet or beat centrifugal forces applied against it, I have never heard of "reactionary centripetal". It is either there or it isn't.

Putting it in another perspective, the lid did not slide off the forks, the forks changed trajectory driving out from underneath it. The lid under the influence of inertia wished to travel straight and did so. Centrifugal force had no effect on the lid but the centripetal force on the tires of the hilo or forklift did.

Reminds me of the recently discussed cell phone sliding across the dash of an automobile while turning a corner. Its the car that's changing trajectory not the cell phone.

the previously existing centrifugal force was greater than the centripetal and the lid continued on its linear path.

This is contradictory, if the operator entered the building in a straight path, there was no existing centrifugal force nor centripetal, the latter did not enter the picture until the operator turned the steering wheel. Inertia assured that there was no centrifugal as the lid continued in a straight path.

Ralph

[pequaide]

Quote rlortie: Centripetal is a physical fixed item, it does not change in value to meet or beat centrifugal forces applied against it

But that is exactly what it does; it changes its value to meet centrifugal force. If you had a rope instead of friction the rope would have exactly the correct quantity of centripetal to balance centrifugal.

[rlortie]

A rope is as weak as its strongest strand, a chain is as strong as its weakest link. Neither change values no matter how much pull or centrifugal force is applied.

If you had a rope instead of friction the rope would have to have the quantity to maintain centripetal restraint, not balance centrifugal. The rope may be capable of holding stronger forces, if weak it will break. There is no change of value or balancing, only the stress imposed upon it.

Ralph

[pequaide]

When you tie the rope up the rope is limp. When the turn is made the lid applies force to the rope, the motion of the lid determines how much force there is in the rope. And the two forces are equal.

[rlortie]

The motion of the lid, its weight and radius of travel determines how much stress not force there is in the rope. The two forces are not equal but stress is, why? because the lid is doing the pulling the other end of the rope is only holding, it is not pulling equal with the stress created by the revolving lid.

when I push on an immovable wall its resistance does not change when I vary my pushing force.

Ralph

[pequaide]

Here is a simple experiment; take a yardstick and push a toy car with the side of the yardstick. Now tape a string to the car and tie the other end of the string near one end of the yardstick. Push the car at about 2 inches down with the side of the yardstick and if the string is tied at 34 inches that would give you a tether radius of 32 inches. Push the car forward with the side of the yardstick and then stop the yardstick. Take care not to move the yardstick once you have made it stop. *

I think the car will move forward in a 32 inch radius circle with the same speed with which you pushed it. The car will pull upon the string which then pulls upon the stick. This view would have an animated object pulling on the string.

To have the stick pulling upon the car would have an inanimate object originating a pull. No: it is obvious that CF is the action force and CP force is the reaction.

Rlortie quote “when I push on an immovable wall its resistance does not change when I vary my pushing force.

Of course it does; a wall can not push on you more than you push on it; and the animate object originates the push. The pushes are equal and opposite, and you started it.


*I actually did this experiment but with a puck on an air table.

[rlortie]

95 pages later, we are back to page one with a puck on an air table.

I am going to go have a talk with my wall and tell it not to push back as hard as I push it.

I give up! what ever you say is OK!

Ralph

[pequaide]

It is not just Okay; it is correct.

I propped a bathroom scales up against a wall. The scale read zero.

I applied a force of 10 pounds on the face of the scale and the wall pushed the bottom of the scales with a force of 10 pound. The scale read 10 pounds. I applied a force of 20 pounds on the face of the scale and the wall pushed the bottom of the scales with a force of 20 pound. The scale read 20 pounds.

The wall does not push we until I push it. And the wall pushes back with the exact same force as with which I pushed it. I am the action and the wall is the reaction.

Believe me; I does not bother me that Wikipedia has it backwards. Maybe they do not have a bathroom scale.

Yep 95 pages; apparently it is to simple. Atwood's makes extra momentum and a bolas through a pipe can use that momentum to make energy.

[justsomeone]

If you were to push a friend with 20lbs. of force and your friend did not move, you could say your friend pushed back with the same force. A solid wall don't push back.

[eccentrically1]

according to the 3rd, something has to push back, or the 3rd is wrong.