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course phy 201
Atwood System:The Atwood system consists of the paperclips suspended over the pulley. A total of six large clips connected by a thread were suspended, three from each side of the pulley. The system was released and, one side being slightly more massive than the other due to inconsistencies in the masses of the clips, accelerated from rest, with one side descending and the other ascending. The system accelerated through 50 cm in a time interval between 4 and 6 seconds; everyone used their 8-count to more accurately estimate the interval. Then a small clip was attached to the side that had previously ascended. This side now descended and the system was observed to now descend is an interval that probably lasted between 1 and 2 seconds.
If you weren't in class you can assume time intervals of 5 seconds and 1.5 seconds. Alternatively you can wait until tomorrow and observe the system yourself; the initial observation requires only a couple of minutes.
`qx001. What were your counts for the 50 cm descent of the Atwood system?
4.6 seconds or 23 counts for three clips on each side
2.2 seconds or 11 counts for one side with 3 large and 1 small clip
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`qx002. What were the two accelerations?
50cm=.5*a*4.6s2
a=4.73cm/s2 for trial 1
50cm= .5*a*2.2s2
a=10.33cm/s2 for trial 2
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Your first acceleration looks OK, but since the second interval was less than half the first your second acceleration should be more than 4 times as great as the first.
`qx003. Why did the systems accelerate?
There was more mass on one side for gravity to accelerate the entire system in that direction.
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`qx004. Suppose the large paperclips all had mass 10 grams, the small clip a mass of 1 gram. What then was the net force accelerating the system on the first trial, and what was the net force on the second?
First trial= 0 g*cm/s2
Second trial= 100g*cm/s2
That is a force, not an acceleration.
You need to explain at least briefly how you got your result. For example on the last question, your line
50cm=.5*a*4.6s^2
was sufficient explanation. I could tell how you got your result.
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.. if uncertainty +-1%
99g*cm/s2 or 101g*cm/s2
`qx005. Given the masses assumed in the preceding, what is the force acting on each side of the system? What therefore is the net force on the system?
First trial= 3000g*cm/s2 in counterclockwise direction, and 3000g*cm/s2 in clockwise, so there is no net force.
Second trial= 3000g*cm/s2 in counterclockwise, and 3100g*cm/s2 in clockwise, so there is a net force of 100g*cm/s2 in clockwise direction.
Right idea, but I believe the numbers would be more like 30 000 g cm/s^2 than 3 000 g cm/s^2.
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`qx006. Based on your counts and the resulting accelerations, do you think the ratio of the masses of the large to small paperclips is greater than, or less than, the 10-to-1 ratio assumed in the preceding two questions?
According to my counts the ratio of the masses would be less than 10:1. It would be closer to 5:1.
you haven't explained the basis for your calculation
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`qx007. If the mass of each larger clip is M and the mass of a smaller clip is m, what would be the expressions for the net force accelerating the system? What would be the expression for the acceleration of the system?
Net force= mg
Acceleration= mg
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`qx008. If the mass of the each of the larger clips is considered accurate to within +-1%:, would this be sufficient to explain the acceleration observed when 3 large clips were hung from each side?
Yes, if two clips that were +1% were on one side and two that were –1% were on the other the acceleration would be almost 4m/s2
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... sample the accelerations for random divisions of the six large clips ... predict what the distribution of masses would look like ...
Not bad, but there are some easily-corrected errors, as well as some more complex points.
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