http://en.wikiversity.org/wiki/Torque_a ... celeration
At a certain distance r from the point of rotation a point mass has a certain moment of inertia I. If the mass of the point mass is halved then the moment of inertia is halved. Even in the debate between mr and mrr it is not the mass that is in question it is the r. If mass is halved inertia is halved.
If you take a close look at the aluminum rod experiment it is obvious that the longer setup is at 45° when the shorter radius setup is at 90°. With the same masses on the end the system is moving half as fast.
The formula at wikiversity is Torque T = I * angular acceleration. If I is halved angular acceleration double. If the end mass is halved then the angular acceleration in this experiment doubles. Half the mass on the end would have the same angular acceleration as the smaller radius setup. They would both reach 90° at the same time.
So according to wikiversity you can choose between two kilograms moving one meter per second or one kilogram moving two meters per second.
Two kilograms moving one meter per second has one joule of energy.
One kilogram moving two meters per second has two joules of energy. 1/2mv²
energy producing experiments
Moderator: scott
re: energy producing experiments
Where does the idea of mrr come from?
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When mathematicians chose to change F = ma into a rotational formula they multiplied F by its mechanical advantage. This was done by multiplying it by a real number r; which is the length of the radius from which the force now works.
Force itself did not change and it is still in newtons ; which is m/sec and the force still has to be applied in line with the previously existing motion and in line with the acceleration. So the left side is now Fr =
When mathematicians chose to change F = ma into a rotational formula they changed ‘a’ into Omega, where Omega is a/R. They arbitrarily divided the right side of the equation by R without dividing the left side of the equation by the same R. This a/R is arc velocity a divided by Radius, which is radians. For example; if the mass rotating is moving 10 m/sec in a 5 meter radius circle it is moving at two radian per sec.
Note: that the left side of the equation is in m/sec and the right side is now in radians per second. Apples and oranges.
The mass m is indeed harder to move when in rotation because it is on the end of a stick of radius R. That gives it a mechanical advantage of R. So mR is an appropriate expression of how much harder it is to move the mass. So the right side of the equation becomes = mR * a/R. Mathematically the Rs drop out of the equation; but m is indeed harder to move so this ma for a rotational equation is not a correct statement. So to compensate for the R that was placed under the a the mathematicians inserted yet another R in the equation. They placed the third R in the equation by inventing another concept called ‘moment of inertia’ and claimed that the moment of inertia for a point mass m is mR². One of the Rs in this mR² drops away because it will be divided by the R under the a; and we are left with a true statement, for rotation, on the right side of the equation. mRa
Check the books; moment of inertia 'I' is always multiplied by Omega; which is in radians.
On the left side of the equation the F is also being exerted at a radius r. This r is its mechanical advantage.
But this is not the same R as on the right side; so when you divide both sides by R the Rs do not cancel each other. So we have Fr = mRa. We can at this point divide both sides by R and we get F r/R = ma. This means that the linear acceleration of the applied mass is equal to the proportion of the radii times the force. The acceleration is equal to the proportion of the two mechanical advantages times the force.
Let’s check the F r/R = ma formula: If R is equal to r; then it would have the same acceleration as an Atwood’s. Atwood’s are F = ma so it checks.
Let F = 10 N; R = 5m; r = 5m and mass = 1kg: you would get 10m/sec/sec with both equations; . F r/R = ma: or Fr = mRR *a/R
 Let F = 10 N; R = 5m; r = 1m and mass = 1kg: you would get 2m/sec/sec with both equations.
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Let F = 10 N; R = 1m; r = 5m and mass = 1kg: you would get 50m/sec/sec with both equations.
So now lets go back to the aluminum tube experiment.
The applied force and the radius of that force remains the same in both arrangements Fr = Fr.
Because Omega is in radians one of the Rs in mRR drops out and you have Fr = mRa. When the mass is placed on the end of the aluminum tube the R doubles and a becomes half for the right side of the equation to equal Fr.
The second R in mRR is used to change radians back into m/sec.
I have done experiments like the aluminum tube experiment; and it is very useful to use the correct and easiest formula; Fr/R = ma.
Please; no comments are necessary.
 Â
When mathematicians chose to change F = ma into a rotational formula they multiplied F by its mechanical advantage. This was done by multiplying it by a real number r; which is the length of the radius from which the force now works.
Force itself did not change and it is still in newtons ; which is m/sec and the force still has to be applied in line with the previously existing motion and in line with the acceleration. So the left side is now Fr =
When mathematicians chose to change F = ma into a rotational formula they changed ‘a’ into Omega, where Omega is a/R. They arbitrarily divided the right side of the equation by R without dividing the left side of the equation by the same R. This a/R is arc velocity a divided by Radius, which is radians. For example; if the mass rotating is moving 10 m/sec in a 5 meter radius circle it is moving at two radian per sec.
Note: that the left side of the equation is in m/sec and the right side is now in radians per second. Apples and oranges.
The mass m is indeed harder to move when in rotation because it is on the end of a stick of radius R. That gives it a mechanical advantage of R. So mR is an appropriate expression of how much harder it is to move the mass. So the right side of the equation becomes = mR * a/R. Mathematically the Rs drop out of the equation; but m is indeed harder to move so this ma for a rotational equation is not a correct statement. So to compensate for the R that was placed under the a the mathematicians inserted yet another R in the equation. They placed the third R in the equation by inventing another concept called ‘moment of inertia’ and claimed that the moment of inertia for a point mass m is mR². One of the Rs in this mR² drops away because it will be divided by the R under the a; and we are left with a true statement, for rotation, on the right side of the equation. mRa
Check the books; moment of inertia 'I' is always multiplied by Omega; which is in radians.
On the left side of the equation the F is also being exerted at a radius r. This r is its mechanical advantage.
But this is not the same R as on the right side; so when you divide both sides by R the Rs do not cancel each other. So we have Fr = mRa. We can at this point divide both sides by R and we get F r/R = ma. This means that the linear acceleration of the applied mass is equal to the proportion of the radii times the force. The acceleration is equal to the proportion of the two mechanical advantages times the force.
Let’s check the F r/R = ma formula: If R is equal to r; then it would have the same acceleration as an Atwood’s. Atwood’s are F = ma so it checks.
Let F = 10 N; R = 5m; r = 5m and mass = 1kg: you would get 10m/sec/sec with both equations; . F r/R = ma: or Fr = mRR *a/R
 Let F = 10 N; R = 5m; r = 1m and mass = 1kg: you would get 2m/sec/sec with both equations.
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Let F = 10 N; R = 1m; r = 5m and mass = 1kg: you would get 50m/sec/sec with both equations.
So now lets go back to the aluminum tube experiment.
The applied force and the radius of that force remains the same in both arrangements Fr = Fr.
Because Omega is in radians one of the Rs in mRR drops out and you have Fr = mRa. When the mass is placed on the end of the aluminum tube the R doubles and a becomes half for the right side of the equation to equal Fr.
The second R in mRR is used to change radians back into m/sec.
I have done experiments like the aluminum tube experiment; and it is very useful to use the correct and easiest formula; Fr/R = ma.
Please; no comments are necessary.
re: energy producing experiments
Sorry for the repost: I had trouble finding this experiment; it goes back beyond the new computer, and is therefore not on the hard drive.
Here is a simple but good experiment.
I went to a construction type department store yesterday and bought a six foot long piece of ½ inch square tubing.
I went across the street to a farm supply store and bought another 1 inch thick bolt 4 inches long with a nut and 20 washers; I already had one bolt with a nut. The bolt had a mass of 494 g, the nut 120.8 g and the washers have an average mass of about 90.5 g. I do not know the mass of the ½ inch square (hollow) tube.
I tied the tube in the center and suspended it; it balanced and it would hold a horizontal position. I tied a string to the tube and suspended one bolt with eight washers and the nut 3.5 inches from the right side of the tied center. I suspended another bolt with eight washers and a nut 3.5 inches out on the left side. I added 20 grams or so to the left side to get a perfect balance. The left side was either light or was a little closer to the center; but now they are balanced.
Leaving every thing else in place I removed the left side bolt and washers with the nut and replaced them with one washer at 35.8 inches from the center. I had to add 6 small magnets and a few BBs to make it balance again. When the bolt-washers and nut had a mass of 1334 grams then the washer magnets and bbs had a mass of 130 grams. A snap lock was used to secure the bolt so the tied position remained in place. Now we have a balanced system with 1334 grams at 3.5 inches and 130 grams at 35.8 inches.
I then added about 10 grams to the right side bolt and observed how long it took the end of the right side to descend about 6 inches. It took about 3 second.
I then replaced the 130 gram mass on the left end with the second set of bolt washers and nut combo at 3.5 inches from the center on the left. I observed how long it took the end of the right side to descend about 6 inches. It took about 3 second.
The ten gram extra mass could accelerate 130 grams at 35.8 inches just as easily as it could accelerate 1334 gram at 3.5 inches. With the same end deflection on the right the 130 g is moving 10 times as fast as the 1334 grams.
Apparently the ten grams of extra mass at 3.5 inches doesn’t know about the moment to inertia. The moment of inertia predicts that it would be ten times harder to rotate the 130 grams at ten times the distance. But the ten grams did not care; it rotated both arrangements at the same rate
Here is a simple but good experiment.
I went to a construction type department store yesterday and bought a six foot long piece of ½ inch square tubing.
I went across the street to a farm supply store and bought another 1 inch thick bolt 4 inches long with a nut and 20 washers; I already had one bolt with a nut. The bolt had a mass of 494 g, the nut 120.8 g and the washers have an average mass of about 90.5 g. I do not know the mass of the ½ inch square (hollow) tube.
I tied the tube in the center and suspended it; it balanced and it would hold a horizontal position. I tied a string to the tube and suspended one bolt with eight washers and the nut 3.5 inches from the right side of the tied center. I suspended another bolt with eight washers and a nut 3.5 inches out on the left side. I added 20 grams or so to the left side to get a perfect balance. The left side was either light or was a little closer to the center; but now they are balanced.
Leaving every thing else in place I removed the left side bolt and washers with the nut and replaced them with one washer at 35.8 inches from the center. I had to add 6 small magnets and a few BBs to make it balance again. When the bolt-washers and nut had a mass of 1334 grams then the washer magnets and bbs had a mass of 130 grams. A snap lock was used to secure the bolt so the tied position remained in place. Now we have a balanced system with 1334 grams at 3.5 inches and 130 grams at 35.8 inches.
I then added about 10 grams to the right side bolt and observed how long it took the end of the right side to descend about 6 inches. It took about 3 second.
I then replaced the 130 gram mass on the left end with the second set of bolt washers and nut combo at 3.5 inches from the center on the left. I observed how long it took the end of the right side to descend about 6 inches. It took about 3 second.
The ten gram extra mass could accelerate 130 grams at 35.8 inches just as easily as it could accelerate 1334 gram at 3.5 inches. With the same end deflection on the right the 130 g is moving 10 times as fast as the 1334 grams.
Apparently the ten grams of extra mass at 3.5 inches doesn’t know about the moment to inertia. The moment of inertia predicts that it would be ten times harder to rotate the 130 grams at ten times the distance. But the ten grams did not care; it rotated both arrangements at the same rate
re: energy producing experiments
no comments are necessary
re: energy producing experiments
Thank you; comments are not necessary.
This very simple experiment shows the same thing as the aluminum tube experiment. That torque and inertia have the same principle; double the distance (r) doubles the inertia as it does the torque.
Halving the mass for the inertia is like halving the force for the torque; it is half the inertia and half the torque.
This experiment could be modified to be in a horizontal plane ; a lever; or a vertical wheel. They are all cheep and easy to make. There is no need to remain unenlightened. But it is your choice: there is no need to argue with you. Or for me to be insulted.
This very simple experiment shows the same thing as the aluminum tube experiment. That torque and inertia have the same principle; double the distance (r) doubles the inertia as it does the torque.
Halving the mass for the inertia is like halving the force for the torque; it is half the inertia and half the torque.
This experiment could be modified to be in a horizontal plane ; a lever; or a vertical wheel. They are all cheep and easy to make. There is no need to remain unenlightened. But it is your choice: there is no need to argue with you. Or for me to be insulted.
re: energy producing experiments
http://www.pa.msu.edu/~hallstein/pdfs/L ... ation2.pdf
note this portion of their site
“To illustrate we will calculate the moment of inertia for a mass of 2 kg at the end of a massless rod that is 2 m in length (object #1 above):
I = mr2 = (2 kg) (2 m)2 = 8 kg*m2
If a force of 5 N were applied to the mass perpendicular to the rod (to make the lever arm equal to r) the torque is given by:
τ = Fr = (5 N) (2 m) = 10 N*m
By equation (1) we can now calculate the angular acceleration:
2 sec2
1.25
8 *
10 * rad
kg m
N m
I
α = τ = =
NOTE: The moment of inertia of a complicated object is found by adding up the moments of each�
pequaide: The last equation got lost in the copy paste process. But you can find the paragraph in question on the site.
I did the math my way to see if their answer was correct; and it was. Their 1.25 radians per second is correct. I was a little surprised because they are using the classic moment of inertia formula of mrr. But a closer look reveals why it is correct; Their answer is in radians not meters per second.
They divided torque Fr by I 'moment of inertia' and left angular acceleration of the opposite side. By doing this the second R in mrr has no yet been canceled by the R that is in the radians formula.
Radians is 'linear acceleration / Radius'; and a newton N is linear; so in a sense the equation is not yet complete; there is one more function to be performed in the equation. The step or function left out is to divide the second R in mRR by the R in a / R (Radians). This returns us to the same units as found in a newton. Apples to apples.
Here is how I did the equation; F r/R = ma
The R and r are equal so they drop out. 5 N = 2 kg * a : a equals 2.5 m/sec/sec
In a 2 meter radius circle that would be 2.5 m/sec/sec / 2 m = 1.25 radians /sec/sec.
The left side of the equation is in arch velocity 'newtons'; and the right side is in radians. You ask a question in linear terms and answer it in angular terms. So to get the correct answer you have to have an extra R in mr so that it cancels with the denominator R in radians.
Or maybe the way to look at it is; that you need another R (they use m) in the denominator (of their arrangement of the formula) to be in radians.
This answers a very big question for me; which is how can they miss this. It is not 100 time harder to rotate a 10r mass; how can they miss this; how can they anticipate such a slow rotation? But they are expecting the same rotation as I would expect. They know it is not 100 times harder to rotate; it is just that they are leaving the answer in radians.
Well: then on the other hand maybe they don't know. But I do.
But anyway this is a good find; they expect the same rotational acceleration as I do.
note this portion of their site
“To illustrate we will calculate the moment of inertia for a mass of 2 kg at the end of a massless rod that is 2 m in length (object #1 above):
I = mr2 = (2 kg) (2 m)2 = 8 kg*m2
If a force of 5 N were applied to the mass perpendicular to the rod (to make the lever arm equal to r) the torque is given by:
τ = Fr = (5 N) (2 m) = 10 N*m
By equation (1) we can now calculate the angular acceleration:
2 sec2
1.25
8 *
10 * rad
kg m
N m
I
α = τ = =
NOTE: The moment of inertia of a complicated object is found by adding up the moments of each�
pequaide: The last equation got lost in the copy paste process. But you can find the paragraph in question on the site.
I did the math my way to see if their answer was correct; and it was. Their 1.25 radians per second is correct. I was a little surprised because they are using the classic moment of inertia formula of mrr. But a closer look reveals why it is correct; Their answer is in radians not meters per second.
They divided torque Fr by I 'moment of inertia' and left angular acceleration of the opposite side. By doing this the second R in mrr has no yet been canceled by the R that is in the radians formula.
Radians is 'linear acceleration / Radius'; and a newton N is linear; so in a sense the equation is not yet complete; there is one more function to be performed in the equation. The step or function left out is to divide the second R in mRR by the R in a / R (Radians). This returns us to the same units as found in a newton. Apples to apples.
Here is how I did the equation; F r/R = ma
The R and r are equal so they drop out. 5 N = 2 kg * a : a equals 2.5 m/sec/sec
In a 2 meter radius circle that would be 2.5 m/sec/sec / 2 m = 1.25 radians /sec/sec.
The left side of the equation is in arch velocity 'newtons'; and the right side is in radians. You ask a question in linear terms and answer it in angular terms. So to get the correct answer you have to have an extra R in mr so that it cancels with the denominator R in radians.
Or maybe the way to look at it is; that you need another R (they use m) in the denominator (of their arrangement of the formula) to be in radians.
This answers a very big question for me; which is how can they miss this. It is not 100 time harder to rotate a 10r mass; how can they miss this; how can they anticipate such a slow rotation? But they are expecting the same rotation as I would expect. They know it is not 100 times harder to rotate; it is just that they are leaving the answer in radians.
Well: then on the other hand maybe they don't know. But I do.
But anyway this is a good find; they expect the same rotational acceleration as I do.
re: energy producing experiments
Sorry but I am having a problem finding a 2m massless rod, in fact my spell checker does not like the word "Massless".
Could you please tell me where I can find one?
Could you please tell me where I can find one?
re: energy producing experiments
Oh: I think maybe the trick is that they adjusted the force to work at the same radius. Obviously the force in the experiment is not working at the same radius. In that way they get the radii to drop out. And probably the formula is defective unless you make this adjustment.
At any rate they know what speed they are looking for and they find some way to adjust the formula to make it work.
At any rate they know what speed they are looking for and they find some way to adjust the formula to make it work.
re: energy producing experiments
W please remove your post from this thread;
re: energy producing experiments
On a beam you can place 4 N at .1 m; or 2 N at .2 m; or 1 N at .4 m; and torque on the beam will remain the same. Each one of these torques would accelerate a mass on a frictionless plane at the same rate. The rate of acceleration would depend upon the placement, along the beam, of the string attached to the mass; it would depend upon the radius of the mass. But at a particular radius for the mass all three forces would accelerate the mass at the same rate. Both the mrr formula and the mr formula say that this is true.
But now let us move the position of the mass, lets start with 1 kilogram at .1 m; with the force of 4 N at .1 m. Your formula gives you 40 radians, which is 4 m/sec. My formula (F r/R = ma) gives you 4 m/sec. The linear momentum is 4 units.
But now let us move the position of the mass and reduce the mass so that the sum of the mass and the radius remain the same; as does Fr and Laws of levers. Lets halve the mass and double the radius; .5 kilogram at .2 m; with the force of 4 N at .1 m. Your formula gives you 20 radians, which is 2 m/sec at the position of the drive force and 4 m/sec for the .5 kg. And you have lost 50% of your energy. My formula (F r/R = ma) gives you 4 m/sec at the position of the drive mass. This is 8 m/sec at the .5 kg and we have an energy gain to 200% of the original energy. The linear momentum is 4 units.
But now let us move the position of the mass and reduce the mass so that the sum of the mass and the radius remain the same; as does Fr and Laws of levers. Lets take one twentieth the original mass and multiply the original radius by 20; .05 kilogram at 2 m; with the force of 4 N at .1 m. Your formula gives you 2 radians. This is .2m/sec for the position at the drive mass and 4 m/sec for the .05 kg. And you have lost 95% of your energy. My formula (F r/R = ma) gives you 4 m/sec at the position of the drive mass. This is 80 m/sec at the .05 kg and we have an energy gain to 2000% of the original energy. The linear momentum is 4 units.
All experiments that I have found on the internet have confirmed F r/R = ma. And all experiments (the square tube; the double 12 in. and 18 in. pulley, the big blue wheel) I have done confirm F r/R = ma.
But now let us move the position of the mass, lets start with 1 kilogram at .1 m; with the force of 4 N at .1 m. Your formula gives you 40 radians, which is 4 m/sec. My formula (F r/R = ma) gives you 4 m/sec. The linear momentum is 4 units.
But now let us move the position of the mass and reduce the mass so that the sum of the mass and the radius remain the same; as does Fr and Laws of levers. Lets halve the mass and double the radius; .5 kilogram at .2 m; with the force of 4 N at .1 m. Your formula gives you 20 radians, which is 2 m/sec at the position of the drive force and 4 m/sec for the .5 kg. And you have lost 50% of your energy. My formula (F r/R = ma) gives you 4 m/sec at the position of the drive mass. This is 8 m/sec at the .5 kg and we have an energy gain to 200% of the original energy. The linear momentum is 4 units.
But now let us move the position of the mass and reduce the mass so that the sum of the mass and the radius remain the same; as does Fr and Laws of levers. Lets take one twentieth the original mass and multiply the original radius by 20; .05 kilogram at 2 m; with the force of 4 N at .1 m. Your formula gives you 2 radians. This is .2m/sec for the position at the drive mass and 4 m/sec for the .05 kg. And you have lost 95% of your energy. My formula (F r/R = ma) gives you 4 m/sec at the position of the drive mass. This is 80 m/sec at the .05 kg and we have an energy gain to 2000% of the original energy. The linear momentum is 4 units.
All experiments that I have found on the internet have confirmed F r/R = ma. And all experiments (the square tube; the double 12 in. and 18 in. pulley, the big blue wheel) I have done confirm F r/R = ma.
re: energy producing experiments
You can see what comments I get; nothing but insults.
If you have a scientifically based comment or question go ahead.
If you have a scientifically based comment or question go ahead.
re: energy producing experiments
Yes;
An inner wheel that could throw a mass: or a lever that could flip a massive end, etc.
But it is best to not limit your arrangements to Bessler's perimeters. There is far more energy here than just enough to gently turn a wheel.
An inner wheel that could throw a mass: or a lever that could flip a massive end, etc.
But it is best to not limit your arrangements to Bessler's perimeters. There is far more energy here than just enough to gently turn a wheel.