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course Phy 201
10/12 1Here are questions related to the lab activities for the 10/04 class:
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?
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1 descent=2 full 8 counts plus 7. So 23 counts. My counts are approximately .2 s each so 23*.2=4.6 s
2 descent= 1 full 8 count plus 2, so 10 counts. 10*.2s= 2s
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`qx002. What were the two accelerations?
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average velocity=`dx/`dt vf=2*Vave since v0=0 acceleration=`dv/`dt
Descent 1=50 cm/4.6s=10.9 cm/s vf=2*10.9 cm/s=21.8 cm/s
a=21.8cm/s/4.6s=4.74 cm/s^2
Descent 2=50 cm/2s= 25 cm/s vf=2*25 cm/s= 50cm/s
a=50 cm/s/2s= 25 cm/s^2
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`qx003. Why did the systems accelerate?
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The systems accelerated because of the mass of the paperclips pulling downward on the string.
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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?
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Fn=maA
Descent 1=Fn=30g*4.74cm/s^2=142.2 g*cm/s^2
Descent 2=Fn=31g*25 cm/s^2=775 g*cm/s^2
good, but there are six large clips in the system, and the entire mass of the system is being accelerated
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.. if uncertainty +-1%
`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?
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Descent 1=142.2 g*cm/s^2 +.01 g*cm/s^2 or 142.2 g*cm/s^2 -.01 g*cm/s^2
Net Force =0.02 g*cm/s^2
Descent 2= 775 g*cm/s^2 +.01 g*cm/s^2 and 142.2 g*cm/s^2 +/- 0.01 g*cm/s^2
net Force= 632.8 g*cm/s^2 +.02 g*cm/s^2 or 632.8 g*cm/s^2
this should have been calculated based on the 10 g mass of each large clip, and the 1 g mass of each small clip
note that the 142.2 g cm/s^2 result has nothing to do with the second trial
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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?
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The ratio has to be slightly greater than the 10:1 ratio because of the acceleration of the system when the system had the same number of paperclips on each side.
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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?
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Fnet=3M 1m-3M
A=Fnet/6M 1m
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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?
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Yes, it would sufficient to explain the acceleration because there would a slight difference in the collective mass of the paper clips.
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... sample the accelerations for random divisions of the six large clips ... predict what the distribution of masses would look like ...
Magnet and Balance
Everyone was given a small magnet and asked to achieve a state where the balance was in an equilibrium position significantly different from that observed without the magnet. It was suggested that the length of the suspended clip beneath the surface of the water should differ by at least a centimeter.
... assuming 1 mm diam ...
`qx009. Describe in a few lines your efforts to achieve the desired result. What worked, what didn't, what difficulties presented themselves, etc.?
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The main difficulty was getting the magnet close enough to the balance without it attaching itself to the balance. Also, finding a way to accurately measure the movement instead of just assuming movement. The best way we discovered was for one person to hold the magnet and the other to hold a measuring device (it was a half meter stick). The first person slid the magnet down the measuring device until motion was detected. Then we determined at what position it was. Also, in the beginning, we measured the initial placement of the balance.
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`qx010. How much difference was there in the length of clip suspended in the water? If you didn't actually measure this, give a reasonable estimate.
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While we didn’t actually measure this, I would estimate it to be between .5 cm and 1 cm because we found the slightest distance it took to move the system.
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`qx011. How did you adjust the magnet? If you wanted to quickly increase or decrease the length of the suspended paper clip beneath the surface by 1 millimeter, using only what you had in front of you during the experiment, how would you go about it?
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To quickly increase or decrease the length of the suspended paper clip beneath the water, it would be feasible for us to move the magnet slightly closer to increase the length. I am not sure we could have sufficiently decreased the length due to the fact we went to the slightest distance to detect movement.
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`qx012. Assuming the diameter of the suspended clip to be 1 millimeter, by how much did the buoyant force on the suspended clip change? How much force do you therefore infer the magnet exerted? If you have accurate measurements, then use them. Otherwise use estimates of the positions of various components as a basis for your responses.
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using my estimates of .5 cm and 1 cm of displacement, the buoyant force would change by .5 cubic cm or 1 cubic cm.
you have to multiply the length of a section of the clip by its cross-sectional area to get the volume
the cross sectional area is less than .01 cm^2
each cm^3 of water displaced has mass 1 gram, and weight equal to mass * accel of gravity
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&#Your work looks good. See my notes. Let me know if you have any questions. &#
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