i = mrr vs i=mr experiment

[pequaide]

Wubbly your disk experiment is really quit amazing.

It is amazing that you get as close as you do moving so slow. Here is why slow counts.

Your thrown mass drops off of the disk at 3 o'clock, you can see motion away from the circle shortly after 3 o'clock. What do you think happens between three and six o'clock? I know that the thrown mass is off the wheel now but in this general area what do you think is happening?

Gravity is pulling on the thrown mass and that pulls the wheel in the opposite direction between 3 and 6 o’clock. I know that the overall acceleration is still clockwise but gravity pulling on the thrown mass in this area is pulling it counter clockwise. I would guess that you have a significant quantity of motion lost right here. I went to the lab and drop a mass at 3 o'clock and the motion was counter clockwise. I think this is gravity removing motion from the wheel.

More important is the backward rotation of the wheel before the thrown mass reaches maximum height. Any transfer of motion back to the wheel removes large quantities of rise, because rise is a square of velocity. And this backward motion occurs with the drive mass still on the right side. This means that the thrown mass (before maximum height) accelerates the wheel backwards and lifts the drive mass. This would remove large quantities of motion from the thrown mass. The motion the drive mass and wheel acquire is lost by the thrown mass (before maximum height).

And what is gravity doing as soon as the thrown mass begins to rise. This is not part of the accumulated motion of the wheel. Before the wheel has transferred all its motion to the thrown mass gravity is kicking it in the head. I know rise is how we measure the energy but this is during the transfer of motion. Much of the motion is going to be removed by gravity before the transfer is even complete.

Ideally you want to transfer all the motion to the small mass and then throw; with gravity having no effect on any of it. Thereby comes the use of the horizontal throw.

Throwing fast can minimize these effects.

[Tarsier79]

Gday Wubbly,

Yes the atwoods was an unexpected progression in this thread, but Peq stated that it doesn't matter if the weights are attached at radius, or in an atwoods configuration. As I thought the linear movement of the weights may act differently to the mrr relationship of the fixed radius weights. The good thing about the atwoods, is the mass is applied at the exact radius of the pulley, which may reduce the effect of mrr, being that the outside of each weight has more inertia than the inside of it on a fixed wheel.

The relation to the conservation of energy is that if you double the radius of a weight its inertia is 4x. If you double the radius at a particular angular velocity the KE of the high inertia weight is also 4x. So it is 4 times more difficult, and takes 4 times more energy due to inertia to accelerate the high inertia weight. This probably isn't news to anyone well versed in physics or even the theory of it, but I am not so well versed. It is also this relationship that helps explain why your flinger didn't gain PE, any why it may be impossible any flinging device will.

[pequaide]

Suspend 120 grams from the 15 cm diameter and then suspend 480 from the 7.5 cm diameter on the other side. You say that it is just as easy to rotate the one side as the other so it must be balanced: right.

Apparently the obvious is to obvious. One side is more easily moved than the other or it wouldn't.

[pequaide]

Make a top (a device for spinning with a point coming down) with 480 grams at 7.5 m and 120 grams at 15 cm. You could take your wheel off the shaft and put a pointed rod in the center and spin it like a top.

It is not going to sit there and spin is it. Gravity is perpendicular to the motion now and not involved in the spin. It is mass against mass and it is still not balanced.

One side is harder to move than the other. When spinning the one can rip the other out of its place. And the side in control is not your so called high energy side.

[nicbordeaux]

Just get an old turntable from the dump, remove the belt drive, hey presto. If it's a semi-decent turntable, the "wheel" is heavy, the bearings are absolutely brilliant (as long as you don't start sticking 1 kg weights on it then speeding it to 600 rpm...).

[nicbordeaux]

Just get an old turntable from the dump, remove the belt drive, hey presto. If it's a semi-decent turntable, the "wheel" is heavy, the bearings are absolutely brilliant (as long as you don't start sticking 1 kg weights on it then speeding it to 600 rpm...).

[Wubbly]

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peqaide, you are confusing two rotational concepts, one of which you don't believe exists in the lab, so there's no point in explaining it to you.

Your belief that both sides should be balanced in the vertical plane is incorrect, and it is easy to understand why it is incorrect, but you seem to be unteachable.

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[pequaide]

Tarsier say that the mrr is why flingers don't work.

Flingers don't have radius changes, as you propose.

The flung missile goes way out on the end of a string. Your radius in mrr is for rigid levers that place force that is perpendicular to the length the the rigid lever. The string can place no force perpendicular to its length. So the concept of mrr has no meaning for a string. The 7.5 cm and 15 cm are ridged lever arms that apply perpendicular force in proportion to their length. A string can only apply force along its length and it would not mater if it were 7.5 cm or 50 m.

The only radius applicable is the radius of the wheel at the string's point of attachment. The string can only apply force along the length of the string and it does not mater one bit how long it is.

It would be like being concerned about the length of the string on a suspended Atwood's weight. Does it apply more force because the length of the string gets longer?

[Tarsier79]

Peq, I am making what I believe to be logical assumptions based on observations.

The only radius applicable is the radius of the wheel at the string's point of attachment. The string can only apply force along the length of the string and it does not mater one bit how long it is.

If you look at wubbly's vid, look at the direction of the string(and the force) while it is tight, compared to the direction of the weight travel. Obviousely there is a lot more at play here. The weight provides a constant varying load on the wheel as soon at it is released.

Re the mrr relationship with a string, get a weight on a string and swing it at different radii, a very short string will be very easy to accelerate/decelerate, while a 1.5m radius string will be much more difficult. Remember mrr refers to angular accleration.

[Wubbly]

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I would agree with Tarsier79's previous post.

Flingers don't have radius changes, as you propose.

As the string unwinds, the radius from the center of rotation does change. It is getting larger (before the release). After the release, the flung mass is no longer rotating about a central point.

... So the concept of mrr has no meaning for a string.

The radius from the center of rotation is getting larger and mass on the end of the string is getting harder and harder to rotate at the same angular velocity. mrr would definitely have meaning for a string situation.

The only radius applicable is the radius of the wheel at the string's point of attachment.

In a flinging situation, this is not correct. The flinging mass (while it is being accelerated) is being rotated about a center point, not rotated about the radius of the attachment point.

The string can only apply force along the length of the string and it does not mater one bit how long it is.
It would be like being concerned about the length of the string on a suspended Atwood's weight. Does it apply more force because the length of the string gets longer?

An atwoods is completely different from a flinging situation. In the Atwoods, the suspended weights are falling straight down and are not being rotated about a central point. In a flinging situation, during the acceleration phase, the masses are being rotated about a central point even though the force is being applied at a different radius than the rotational radius. After release, they are no longer being rotated.

pequaide, you are confusing linear and rotational motions. You look at everything as linear, which it is not.

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[FunWithGravity2]

Wubbly and Kaine, And Fletcher

Thank you.

I truly appreciate the amount of time that some of you spend to explain things to others. As well as tying to educate them the rest of us also get to go along for the free ride.Wubbly and Kaine have done a wonderful job at different times duplicating setups and helping he rest of us learn vicariously through them. I always feel that what i am bein g presented is honest and open and objective and i can learn from it without having to wonder what is hidden.

Fletcher who constantly lays out what is required to "prove" a point and tirelessly tries to educate others as to why they are not "proving" what they think is a huge asset as he also allows the rest of us to maintain a "bar" by which we need to keep within view so as not to make to much of a fool of ourselves.

So for those of us that don't like to hear what those of you that are trying to help say. Thanks, i for one appreciate all the rebuttals and effort that are put in to help others understand. Even when they refuse to accept the truth.

Mistakes make progress but only if we can see them. Some here are the glasses without which the rest of us would be blind.


Dave

[pequaide]

Tarsier quote: The weight provides a constant varying load on the wheel as soon at it is released.

That’s correct but all the force is in line with the string. I observe that the velocity changes are low as the mass first leaves but the velocity changes quickly become very large and diminish as it flares out further.

My guess is that at the very first the speed of the cylinder and the thrown mass are about the same and the tug of war in the string is not that great.

Quickly the velocity of the sphere become greater than the cylinder or wheel and the tug of war becomes huge.

As the mass flairs out further the angular change diminishes along with the force in the string but the tug of war continues. If the string does not pass 90° (or is released) it will reverses the direction of the wheel.

The mass is actually moving in a straight line perpendicular to the string. Release the string and the mass will show you what line of travel it is taking. Only the force in the string forces the linear motion into circular motion. If the wheel is jammed this is a balanced force and it will not change the linear speed of the mass. It is eminently reticulated to say that a one kilogram mass moving at one meter per second on the end of a 5 meter string has more energy than the same mass moving at one meter per second in a smaller circle. The centripetal force and centrifugal forces are balanced and will not change the actual velocity of the mass itself. There is nothing magic about the length of the string. The mass does not have super power because it is on a super long string.

As the mass unwinds the inner circle becomes proportionally smaller than the circle of the mass. The physical arrangement makes it difficult to rotate a mass that is out there. But not because the mass obtained some superhero abilities.

The r in mrr (in the lab) is for a rigid lever arm, what could be more simple. Lets use a lever to push a rock up out of the ground. You have a short end and a long end and a fulcrum and it is all laws of levers. But you can't use a long wet noodle. And a string is a long wet noodle

And you could place a pin at any point along the string and the shorter radius will not change the motion of the mass. The mass will still be moving the same meters per second in a straight line. Cut the string and see if it is not moving in a straight line. The mass will never pick up speed or lose speed when the string is cut or interupted with a pin.

[Tarsier79]

It is eminently reticulated to say that a one kilogram mass moving at one meter per second on the end of a 5 meter string has more energy than the same mass moving at one meter per second in a smaller circle.

rediculous?

Remember mrr refers to angular accleration.

If both masses were moving at the same angular velocity, then the energy would be much different.

Of course, you are correct about some parts. A string cannot accelerate or decelerate a mass at 90 degrees, but I still feel the mrr relationship at each moment exists in a flinger. The force, as previousely stated, is a separate entity, and in this case also has different directions. But how difficult will it be for the disk to accelerate or decelerate the mass will increase as it changes its radius, while torque is applied to the disk. Another example of this is a pendulum bob, and the period increasing with the length of the string. http://en.wikipedia.org/wiki/Pendulum.

[pequaide]

Yes; ridiculous. Must be a spell check thing; and then I missed it on the proof read. Yep; it is the top choice on the spell check if you already have it butchered, and then don't pay attention. How ridiculous of me.

[pequaide]

Perhaps a pendulum would be a good example.

Let’s start with a one meter pendulum where the bob is dropped one meter. At 90° the gravitational force is full g. It will pick up speed quickly and it only covers 1.57 m to the down swing position and a total of 3.14159 meters to the other side. So it covers the distance quickly.

Now let’s go to a twenty meter pendulum that is dropped one meter. At 18.19° the gravitational force is .3122 * g. It will pick up speed slowly and it covers 6.7 meters to the low point and 13.39 meters to the other side. So it covers the distance slowly.

You could divide the swings into a thousand different equal angular parts, each part having a certain period of time, and a certain quantity of Force. I believe that the sum of all these products will equal the same Ft relationship in the 1 meter pendulum as in the 20 meter pendulum. This would seem obvious because the final down swing velocity of both is the same 4.429 m/sec.
 
If the same Ft produces the same final velocity then how is the twenty meter pendulum bob 20 times harder to move than the one meter pendulum bob?