Because you are moving the position of the inertial masses I decided to do the same with the 18 inch wheel. I kept the drive mass in the same location.
I placed 1180 grams at 8.91 (r) inches and accelerated it, and the wheel, with 118.2 grams.
I added the mass by suspending 590 grams from both sides of the wheel at the 17.82 inches diameter. This makes a 1180 gram Atwood's which is attached to about a 3000 gram wheel. The 118.2 grams was attached to one of the 590 gram units. The 590 grams was a 1 inch bolt with washer and the 118.2 grams was a nut.
After the third rotation of the wheel, as previously described, the flag tripped the (26 mm apart) gates at .0631, .0631, .0628 second.
I calculated what mass would be 17.82/12.05 ths as much as the 1180; which was 283.5 grams for each side.
I then added 283.5 grams to each side. This new 1747 gram Atwood's was placed at 12.05 inches. The 118.2 gram drive mass was given its own string and it remained at 17.82 inches. Without moving the gates or the flag I did five runs. The flag tripped the gates at .0625, .0630, .0624, .0629, .0625 second.
This is a near perfect mr relationship. I prefer to call it laws of levers.
So it doesn't mater if you move the position of the drive mass or the position of the inertial mass it is all laws of levers.
I'm not sure I completely understand what you are doing in your experiment.... Adding two different weights at different radii in an atwoods configuration? I=mrr refers to one or more masses attached to a wheel accelerated by torque, either negatively or positively.
Anyway, following my last experiment, which admittedly did not prove I=mrr, I had a chance this afternoon/evening to perform a better and more controlled test. Again I have not measured, or taken into account the inertia of the bar, because we are attempting to maintain the same overall inertia. As the bars inertia remains the same, If the different weights inertia is the same at different radii then the entire systems inertia will remain constant, and provide an equal resistance to the acceleration of the drive weight.
I purchased some approx 3kg fishing line, which is a lot lighter much better suited to this experiment. I also reduced the pulleys to 1 single pulley, by mounting my bar assembly on top of a stand, that may or may not look familiar. This allowed a 115cm drop of the drive weight, hopefully adding to accuracy.
I am using the same bar assembly, and the same weights, with the addition of weights 1/4 that of the large ones. You will notice instead of using their exact weight, which isn't that important, I have used their relative weight to eachother, and called them 2w, 1w, and 1/2w.
I performed the test in 4 configurations, in order to confirm the previouse results, and under the assumption that I=mrr, 2 new configurations:
1: 2W at 24.5cm radius. This is our controll
2: 1W at 49cm radius. If I=mr, this weight at this radius would have the same inertia, and therefore the same drop time.
3: 1/2W at 49cm radius. According to I=mrr, this will have the same inertia as our controll
4: 1W at 34.6cm radius. This is the radius, according to I=mrr, where this weight will also have an equivalent inertia as our controll.
I performed each of these tests in rotation, so I set the weight at its radius each time for each test. Due to the unsymmetrical distribution of the 1W I flipped the weight 1/2 way through the test, then took the average of the results.
Below is a picture of all 4 setups. Sorry, No vid this time. I did the test right on sunset, and took some of the following pictures in the dark.
Attachments
Last edited by Tarsier79 on Wed Oct 26, 2011 10:19 am, edited 1 time in total.
3:(I=mrr with 1/4 weight of controll) 11.05 seconds
4:(I=mrr with 1/2 weight of controll) 10.92 seconds.
I have attached below a picture of the 3 different weights. The 1/4 weights are a bracket, with bolts and washers to conform as closely as possible to the correct weight. Below is also my original results.
Conclusion: According to this setup, I=mrr, confirmed with 2 different confugurations. I=mr is proved false by comparing any 2 results.
Firstly, I must admit I was wrong about the relationship between weights and their radius on an atwoods. weights hung over different sized pulleys do have a relationship that directly relates to inertia.
Below is a diagram of my interpreted replication of Peq's atwoods test.
The red drive weight falls a fixed distance, just as in the last test. As in Peq's description, it is on its own pulley, which has a diameter of 20cm.
The inner diameter is 15cm, half that of the outer diameter. So the purple weights below add a certain amount of inertia to the atwoods assembly. So by using the same testing method as before, I timed the drop of the drive mass over a certain distance. So, when we double the diameter and replace the purple weights with the blue, to conform with I=mrr, then the weight must be 1/4 that of the purple. If it conforms to the "law of levers", then the weight will be 1/2.
So which one will have the same acceleration/ drop time?
Before you watch the vid...
I am using the same weights as the previouse experiment.
In the .5W at r15, it the weights themselves are moving a lot faster, but if you ignore them and look at the drive mass, the drive mass is still moving the same as the 2W at r7.5.
The vid is a bit disjointed, but hopefully you can glean the information from this thread as well.
I lack super precise measuring equipment, so there may be some margin of error. The 3 values of inertial weights were measured comparitively, using a balance. I also calculated the drive mass with the balance.
What bothers me about this type of atwoods experiment, I how the wheel differentiates between the drive mass, and the inertial mass.
In my test, the drive mass is at a fixed radius for both tests. But the wheel feels an imbalance of mass x radius, and the drive mass accelerates in the same ratio as the inertial weight on the same side... So it doesn't differentiate where the acceleration is coming from.
When I did this test, I assumed that the drive masses inertia wasn't to be taken into account, because it is doing the accelerating. There is considerable difference betwen the drive mass, and the inertial mass, and I guess the drive weights inertia doesn't change between tests.
So, I guess, if you change the position of the drive weight, and adjust its weight in the ratio of m x r, then theoretically you should measure a difference in acceleration dependant on the ratio between the amount of balanced weight compared with the difference in weight at that radius.
So.... Is the drive weight inertia taken into consideration? / Will the right hand atwoods accelerate slower than the left?
Attachments
Drive weight in mr ratio, with inertial weights at mrr ratio
Im not that bright but the way i see this from a laymans perspective is that the "mass" of the "tension" in the string "accelerates" the wheel under the influence of gravity based on the "tensions mass" at the given radii.
IMHO, a fundamental building block
Dave
Last edited by FunWithGravity2 on Tue Nov 01, 2011 12:15 pm, edited 1 time in total.
Si mobile in circumferentia circuli feratur ea celeritate, quam acquirit cadendo ex
altitudine, quae sit quartae parti diameter aequalis ; habebit vim centrifugam suae
gravitati aequalem.
You can google "atwoods machine" and come up with websites that explain atwoods.
The acceleration of an atwoods is a = g * (m1 - m2)/(m1 + m2) and should be independant of the radius according to
Sorry Wubbly, I should have been more specific, although I think Dave got me. I was specifically referring to the acceleration of the atwoods pulley itself.
I relationship to each other .25 kg and .30 kilograms have the same radius. So the r in mr or mrr is one. In the .30 kg - 0.25 kg = .05 kilograms driving and .55 units of inertia. .05/.55 * 9.81 m/sec/sec acceleration .8918m/sec/sec.
The 1.0 and 1.1 has .1 units of drive and 2.1 units of inertia. .1 / 2.1 * 9.81 = .467 m/sec/sec acceleration. Slower.
The .25 and .30 have mechanical (2/1) advantage over the mass of the wheel than 1.0 and 1.1. But they have half the drive. so this would be the same.
