i = mrr vs i=mr experiment

[Tarsier79]

from Wiki:

"Angular acceleration is the rate of change of angular velocity over time"


The I=mmr formula is in relation to angular velocity.... ie, how difficult it is to turn the wheel/bar/pendulum through a certain amount of degrees.

What angle is the 20m pendulum moving through compared to the 1m pendulum?

[pequaide]

18.19° for the 20 m and 90° for the 1 m, so what is your point?

[Tarsier79]

The point is the entire purpose of this thread.

How difficult is it for you, or gravity to rotate both pendulums through the same degrees. Think reference frame.

[pequaide]

Well: lay the two pendulums in the horizontal and apply a constant perpendicular force of one g to each bob. They both will have 4.429 m/sec velocity after traveling one meter. That is 57.32° for the one meter pendulum and 2.865° for the 20 meter pendulum. There is a twenty to one; but it is not harder to move they are both moving 4.429 m/sec. The mass does not obtain a harder to move capacity because it is at the end of a longer string. It takes longer to get around the circle but it is not harder to move.

[Tarsier79]

It takes longer to get around the circle but it is not harder to move.

Angular acceleration!.... This is measured in degrees, not distance, that is the important part. Wubbly explained to us that rotational motion is different to linear. I believe this is true if you change your reference frame. You can't use maths designed for degrees per second, and apply it to meters per second, unless you understand the relationship between them.



Add: Also, I was under the opinion that in the atwoods, the drive weights "ïnnertia" was irrelevant as it drives, rather than retards, but now believe I understand its addition to the equation. Basically, "force and inertia are diffferent entities."

[Tarsier79]

An example of the relationship between mass velocities and angular velocities are as follows in the formulas for CP. (from wiki: http://en.wikipedia.org/wiki/Centripetal_force)

CP = mvv/r (actual velocity of mass, as r decreases, CP increases.)

CP = mrww (angular velocity of mass, as r increases, CP increases.)


Actual velocity and angular velocities have a relationship, but are different.

[Wubbly]

... then how is the twenty meter pendulum bob 20 times harder to move than the one meter pendulum bob?

This thread is about I=mrr, so it would actually be 400 times harder to rotate. Since it is 400 times harder to rotate, the angular acceleration will be lower.

Perhaps a horizontal example would be clearer.

Let's suppose you have a horizontal system, with a point mass at a radius of 1 meter. It starts at rest and rotates through a full circle (360 degrees) in one second. That mass will have travelled 2 PI meters in one second.

Now move the point mass from 1m to 2m (you doubled the radius). If it starts from rest and rotates through a full circle (360 degrees) in one second, that mass will have travelled twice as far (4 pi meters in one second).

If you look at it from a linear perspective, you went twice the distance in the same time, so you had twice the acceleration. (d= 1/2 a t^2, if "t" = 1 and "d" doubled, then "a" doubled).

Since f=ma, and your mass remained constant, if your acceleration doubled, then the force had to have doubled.

But work is force x distance. If your force doubled, and that force acts over twice the distance, then your work went up by a factor of 4. It was 4 times harder to move the same mass, over twice the distance, and at double the acceleration. Remember, force x distance measures what it takes to move the mass, not force x time.

If you look at it from a rotational perspective, you accelerate an object by applying a torque, which is a force acting at a distance from the center of rotation. To accelerate the mass at 1 meter takes a certain amount of force acting at 1 meter from the center of rotation. To accelerate the same mass at double the radius (2 meters), takes twice the force, but that force is acting at twice the radius. So to rotate the exact same mass at twice the radius requires 4 times the torque.

If you do the same exercise at three times the radius, you would see the exact same mass requires 9 times the torque, to travel through that same full circle (360 degrees) in that same one second. Each case has the same angular acceleration.

[pequaide]

But a string can not apply torque, torque requires a rigid radial distance, the force of the string is directly applied to the mass.

Newton said it was force x time. So the debate continues.

[Wubbly]

lol. Shirly you can't be serious. Of course a string can apply a torque. The masses in an atwoods apply a torque. In the cylinder and sphere's they also apply a torque. There is a force along the string, the string connects to a rotating system so there is an overall net torque being applied to the center of rotation even though the initial force is not rigidly attached to the center of rotation.

[pequaide]

Suppose you have an electric motor with a 3 cm (dia) pulley attached on the motor shaft. You have a 10 kilogram block, on dry ice on a table,
that is attached to a string that is wrapped around the pulley. The distance from the block to the pulley and the length of the string is 204 cm. Then you are asked something like “How much torque does the motor have to apply to get the block to accelerate at 2 m/sec/sec�.

What information given above is useless information for solving the problem?
  
Answer: the Length of the string.

When you are making a power transfer with gears you don’t need to know the length of the chain.

The string is attached to the surface of the cylinder, The cylinder or wheel has a rigid radius and that is what applies the torque. The string could be 5 cm or 5 m but the cylinder's torque would remain constant.

In your Atwood’s the same string with the same mass can be suspended from two different radius pulleys and you would have two different torques. The torque is supplied by the radius of the pulley. The string could be ten times as long and the radius would supply the same torque.

The string supplies a force but it does not supply a radius. You are using the length of the string as if it supplied a radius. But a string does not supply a radius and its length can not be used in the mrr equation.

[nicbordeaux]

Edited to oblivion.

[rlortie]

It is not often I involve myself with these pages of non-productive quibble, but occasionally there is something stated that catches my attention.

Pequaide wrote:

the same string with the same mass can be suspended from two different radius pulleys and you would have two different torques.

Either way the mass falls the same distance; so big pulley or little pulley the end product in the form of 'work' is the same.

What happened to reading torque in inch, foot,or meter? I cannot make a motor produce more constant power by simply using a bigger pulley. The radius of the pulleys is irrelevant, all you are doing is dividing the output by farther or smaller distance, the end result will be the same. Where is the equation for 'work' which is what the end product will deliver.

Oh how I wish I were wrong! I could take a 1/4 horsepower motor and by using bigger gears make it into a 20 horsepower motor using the same amount of energy input. I will measure the torque of my invention with a Prony Brake 40' feet in length. The fantastic claim of 480 inch pounds of output will make it sell like hotcakes.

Ralph

[Tarsier79]

I do agree Ralph. As far as I am concerned, this thread has finished serving its purpose.

"Pointless Quibble?" Anyway, I have already made myself clear on the importance of this relationship, and the implications it has for certain arrangements of weights.

Once I have some free build time, I may build a fixed input horizontal flinger and attempt to measure its output power, if I don't have anything more interesting at the time.

Peq, I hope you gain a clear understanding of inertia and torque. After these experiments and the following discussion, I believe I have a much better understanding of it.

And thanks Wubbly for the flinger vid and the positive addition to the discussion.

Nuff said?

[FunWithGravity2]

Very well said,

Kaine good luck with finding an interesting direction.

Dave