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course Phy 201
9/10 2Here are some questions related to today's lab activities, and some things I want you to do using the materials you took home and the TIMER program. I have written fairly extensive class notes as well, but want to look over them before sending them tomorrow.
Link to TIMER (worth bookmarking): Timer_b
Ball off ramp to floor
If you weren't in class to do this:
We measured the landing positions of the ball after rolling down three ramps, one supported by a domino lying flat on its side (least steep), one supported by the domino lying on its long edge and one supported by the domino lying on its short edge (steepest). You have some dominoes, a ball and a ramp so if you didn't get to do this at the beginning of class, you should be able to repeat this.
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?
v=10 cm/.4 s= 25 cm/s
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How fast was it traveling in the horizontal direction when the domino was lying on its long edge?
v=21 cm/ .4s =52.5 cm/s
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How fast was it traveling in the horizontal direction when the domino was lying on its short edge?
v=33.5 cm/.4s =83.75 cm/s
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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?
40 cycles in a 30 second period
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What therefore is the period of motion of that pendulum?
The period of motion is .75 per second.
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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?
Using the given equation, T=.2 (sqrt of 10.5) = .65 cm/s. The period of motion of the pendulum is a little larger than using the equation.
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How well did the freely oscillating pendulum synchronize with the bouncing pendulum of the same length? Which was 'quicker'?
It was relatively difficult to synchronize the freely oscillating pendulum with the bouncing pendulum, if not impossible. The bouncing pendulum is much quicker than the freely oscillating pendulum.
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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?
The length of the pendulum was 10.5 cm. The ball hit the floor on the second “hit” of the pendulum at 1 meter.
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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?
The length of time between the release and second hit would be approximately .6s. I arrived at this answer based upon in 10 seconds, I can complete 4 counts of 8. 10/32=0.3 s/count
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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=100 cm/0.6 s =166.7 cm/s
that would be (correct) average velocity, not acceleration
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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?
To design an experiment to measure the velocities of v0, v_mid_x, v_mid_t and vf for different values of V0, we first must have a way to determine what the value of V0 is. One way to do that would be to use an additional ramp with the 30 cm length and adjust the height to arrive at different velocities when it hit the main ramp. After gathering that data, we can determine the vf. To determine v_mid_x, we would need to repeat the experiment and find the time it takes to reach the midpoint of the ramp. Using that data, we could find the velocity from V0 to v_mid_x. To find v_mid_t, we would again repeat the trial only this time, determine what time is the midpoint between the V0 and Vf, using a timing device, try to determine at which point the ball was on the ramp at the desired time.
Good plan. There are a few specifics to fill in, but with this plan I believe you could do so.
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How would you design an experiment to measure v0 and `dv for different values of v0?
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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.
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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:
12 cm
.609375, .71875, .671875, .625, .6875
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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.
.75, .6875, .6875, .625, .703125, .65625, .71875, .640625, .65625, .734375, .75
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For both sets of trials, how do your results compare with the prediction of the formula T = .2 sqrt(L)?
T= .2 (sqrt) 12=.69
The results are near or around the prediction using the formula. Actually, if I export the information to excel and average the time intervals of the freely swinging pendulum, the average is .68. Closer than I would have thought.
If I go back and average the time intervals of the bouncing pendulum, it is 0.66, so it is fairly close as well.
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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:
Using 2 dominoes, the ramp, and the ball, I formed a ramp similar to those in class. I chose 2 dominoes instead of more so the ball would be moving slower and v_mid_x would be easier to determine.
Good idea. The slower the motion, the easier to time it. You can get too slow, where the motion is no longer smooth, but at 1 or two dominoes that won't be a problem.
To calculate vf, I simply clicked Timer when I released the ball and when it hit the end of the ramp. I also placed a domino at the end of the ramp so I could also hear when it hit the bottom. Vf=30 cm/1.3s= 23 cm/s
To find v_mid_x, I simply again released the ball at the top and clicked Timer. When the ball reached 15 cm, I clicked Timer again. V_mid_x= 15cm/.95=15.8 cm/s.
This is the average of the velocity over the interval, not the velocity at the end of the interval. Fortunately that is easy to find from your result.
To find v_mid_t, I again released the ball at the top, only this time, I tried to stop the Timer at the midpoint of 1.3 sec and also judge the distance the ball had traveled. V_mid_t = 12 cm/.65 s=18.5 cm/s
the preceding note applies here as well
`dv=23cm/s (vf-v0)
It's not clear how this follows from your previous results or your data.
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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:
1 1308.859 1308.859
2 1309.422 .5625
3 1309.891 .46875
4 1310.344 .453125
5 1310.906 .5625
6 1311.531 .625
7 1313.078 1.546875
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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.
320, 384, 397, 320, 288, 116
To determine how many degrees per second the strap was rotating, I took 180 degrees from my interval and divided by the time interval of each rotation.
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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.
1308.859, 1309.1403, 1309.422
1309.422, 1309.6564, 1309.891
To find the midpoint clock time, I first took the time interval between the 2 corresponding clock times and divided by 2. Then I added that number to the first clock to arrive at the midpoint clock time.
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Very good. See my notes and let me know if you have questions.
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