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course Phy 241
9/13 2
Everyone should submit the following:Assuming that the ball fell to the floor in .4 seconds, after leaving the end of the ramp, and that after leaving the ramp its horizontal velocity remains constant:
How fast was it traveling in the horizontal direction when the domino was lying flat on its side?
The ball traveled 7 cm while domino was lying on its side, in comparison to the ball just being dropped from the table’s distance from rest instead of using the ramp. Therefor ball was traveling at 17.5 cm/s if traveling at constant rate.
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How fast was it traveling in the horizontal direction when the domino was lying on its long edge?
The ball traveled 18.2 cm while domino was lying on its side, in comparison to the ball just being dropped from the table’s distance from rest instead of using the ramp. Therefor ball was traveling at 45.5 cm/s if traveling at constant rate.
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How fast was it traveling in the horizontal direction when the domino was lying on its short edge?
The ball traveled 28.8 cm while domino was lying on its side, in comparison to the ball just being dropped from the table’s distance from rest instead of using the ramp. Therefor ball was traveling at 72 cm/s if traveling at constant rate.
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Pendulum count:
What was the length of the pendulum you counted, and how many counts did you get in 30 seconds?
Exactly 12 cm was length from magnet to center of plastic bead. I received counts of 36, 38 and 43 when count was preformed within 30 sec.
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What therefore is the period of motion of that pendulum?
Ave of counts was 39. With aver. of 39 oscillations in 30 sec.
T = 30/39 = approx.7692 (where T is period in sec.)
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How does your result compare with the formula given on the board, T = .2 sqrt(L) where T is period of oscillation in seconds and L the length in centimeters?
T = .2`sqrt(L) = .2`sqrt(12) = approx. .6928
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How well did the freely oscillating pendulum synchronize with the bouncing pendulum of the same length? Which was 'quicker'?
When experiment first began the two pendulums would seem to be synchronize almost every time. Very quickly though the pendulum which was bouncing off of steel bracket, seemed to lose momentum which caused it’s period to shorten and throw the synchronization off between two pendulums. The pendulum which was bouncing off of steel bracket always ended up being quicker.
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Ball drop
From what height did the drop of the ball synchronize with the second 'hit' of the pendulum, and what was the length of the pendulum?
Between 130 and 131 cm was ht of ball, with the length of the pendulum being 8.5 cm.
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How long should it have taken the pendulum between release and the second 'hit'? On what do you base this answer?
Using formula, T = .2`sqrt(L)
We plug in our values,
T = .2`sqrt(8.5)= approx. .5831(which is period, or time in sec to preform one oscillation)
We must mult this value by 1.5 because T value is for one oscillation and to find time at 2nd hit
we must calculate for T*1.5. We get…..
.5831*1.5 = .87465 sec.
Good, but should be only T * .75. So the time of fall was about .43 second.
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Given you answer to the preceding, you know the time required for the ball to fall from rest to the floor, and you know how far it fell. What therefore was its acceleration?
a = (vf-vo)/`dt = (297.2618-0)/.87465 = approx. 339.8637 cm/s^2
calculated vf knowing `dx = 130cm and time = .87465
v Ave = (vf+v0)/2, and vAve*.87465 = 130
vAve = 130/.87465 = approx. 148.6309
Using vAve = (vf+v0)/2,
148.6309 = (vf+0)/2
vf = 148.6309*2 = 297.2618 cm/s
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Ball down long ramp
How would you design an experiment to measure the velocities v0, v_mid_x, v_mid_t and v_f for different values of v0?
??????I’m not sure I completely understand the question???????, but I believe for example when using the ball on the ramp , you could use different starting points on ramp where the ball has already been in motion with different various v0. Another example would be throwing the ball up in the air at different forces.
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How would you design an experiment to measure v0 and `dv for different values of v0?
As long as you know either v0 or vf you could calculate the other as long as time of experiment was kept track of and distance object traveled during experiment was known or could be measured. You could change the force or altitude of a ramp to see effects it had on v0 and vf.
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Rotating strap
For the strap rotating about the threaded rod, give your data indicating through how many degrees it rotated, how long it took and the average number of degrees per second. Report one trial per line, with a line containing three numbers, the number of degrees, the number of seconds, and the average number of degrees per second, separated by commas.
2640, 10s, 264 deg/s
5490, 15s, 366 deg/s
9270, 24s, 386.25 deg/s
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To do with the materials you took home:
Using the TIMER program with the materials you took home:
Bracket pendulum:
Shim the bracket pendulum until the 'strikes' appear to occur with a constant interval. Click when you release the bead, then click for alternate 'strikes' of the ball on the bracket pendulum (that is, click on release, on the second 'strike', on the fourth 'strike', etc., until the pendulum stops striking the bracket). Practice until you think you think your clicks are synchronized with the 'strikes'. Report the length of the pendulum in the first line, then in the second line report the corresponding time intervals below, separated by commas:
????I can’t get my pendulum to stay bouncing consistently for any extended period of time.???????
12cm
.5674, .6472, .6802, .5932
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Using the same length, set the pendulum so it swings freely back and forth. Click each time the bead passes through the equilibrium position. Continue until you have recorded 11 'clicks'. Report the corresponding time intervals below in one line, separated by commas.
.3359375, .4000244, .3120117, .3599854, .3599854, .3760986, .3199463, .3520508, .3359375, .3280029, .3280029
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For both sets of trials, how do your results compare with the prediction of the formula T = .2 sqrt(L)?
Using formula T = approx. .6928, with my pendulum it was a slightly lower value for the period.
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Ball down ramp:
Do your best to take measurements you can use to find vf, v_mid_x, v_mid_t and `dv using your ramp and ball, releasing the ball from rest. (You could use the TIMER to get decent data. If you wish you can use the fact that a ball falling off a typical table or countertop will reach the floor in about .4 seconds. Note: Don't let the ball fall on a tile or vinyl-covered floor. You don't want broken tile, and you don't want dents in your vinyl. You could put your book on a carpeted or otherwise protected floor and land the ball on the book.)
Briefly describe what you did and what your results were:
I did kind of the same experiment as what was done in class. I started with one domino, then two, etc. Starting from rest with one dmino it took approx. 2.328 sec from one end of ramp to second. With 2 dominoes it took approx. 1.4004 sce. With 3 dominoes it took approx.8881 sec.
Data collected
1st : vf and `dv = 26.2028 cm/s, a = 11.2555, v mid t = 13.1014, v mid x =18.5266
2nd : vf and `dv = 43.559 cm/s, a = 31.1047, v mid t = 21.7795, v mid x = 30.803
3rd : vf and `dv = 68.686 cm/s, a = 77.3404, v mid t = 34.343, v mid x = 48.5698
No problem, but you're mislabeled what you're doing. You're reporting conclusions here, not data. Your data were the times you reported immediately preceding your heading 'data collected'.
In any case your vf and `dv look good, though you should have briefly stated how you obtained those results. It wasn't any problem for me to do a couple of quick divisions, but in general you don't want to make your reader interrupt the narrative you are developing to figure out what you did.
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Rotating strap:
Let the strap rotate on the threaded rod, as before. Click the TIMER at the start, and then at 180 degree and/or 360 degree intervals (the latter if it's moving too fast to do the former). Copy the output of the TIMER program below:
Trail #2
42 3081.807 .6308594
43 3082.432 .625
44 3083.231 .7993164
45 3083.968 .7368164
46 3085.039 1.070801
47 3086.056 1.01709
48 3087.392 1.335938
49 3088.968 1.576172
50 3091.775 2.807129
Total 10.5991222
Trail #1
10 2989.808 .2875977
11 2990.167 .359375
12 2990.503 .3359375
13 2990.839 .3359375
14 2991.199 .3598633
15 2991.576 .3769531
16 2991.936 .3598633
17 2992.336 .4003906
18 2992.712 .3759766
19 2993.096 .3837891
20 2993.536 .4399414
21 2993.984 .4482422
22 2994.368 .3837891
23 2994.752 .3842773
24 2995.152 .3999023
25 2995.512 .3598633
26 2995.912 .3999023
27 2996.408 .4960938
28 2996.904 .4960938
29 2997.592 .6879883
30 2998.104 .512207
31 2998.752 .6479492
32 2999.696 .9438477
33 3000.76 1.063965
34 3001.912 1.151855
35 3003.488 1.576172
36 3005.632 2.144043
37 3007.696 2.063965
Total 17.8389641
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On the average through how many degrees per second was the strap rotating during each interval? Report in a single line, giving the numbers separated by commas. Starting in the second line explain how you did your calculations.
1st trail, ave was 305.6857 deg/s. 2nd trail ave was 544.8747
This was found by adding together all time intervals and finding total degrees(found by mult 360 by intervals). Then dividing total deg by total time.
good but I would have subtracted clock times (2d column) to get the total interval
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The second column of the TIMER output shows the clock times. For a given interval the 'midpoint clock time' is the clock time in the middle of the interval. Report clock times at the beginning, middle and end of your first interval in the first line below. Do the same for your second interval, in the second line. Starting in the third line explain how you got your results.
Trail #2
42 3081.807, 3082.1224, 3082.432
43 3082.432, 3082.7445, 3083.231
To find middle time you find `dt, divided that value by 2 and add to starting time of the interval.
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"
Very good. See my notes.
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