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course Phy 241
September 10 around 12pm
Here 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?
0.4 seconds *9.5 cm=3.8 cm/s
I got 9.5cm from where I laid the paper on the floor and when the domino was flat on its side, the ball rolled off the ramp, onto the paper in the floor, and it was 9.5 cm from where I dropped the ball with my hand.
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How fast was it traveling in the horizontal direction when the domino was lying on its long edge?
0.4 seconds*20 cm=8cm/s
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How fast was it traveling in the horizontal direction when the domino was lying on its short edge?
0.4 seconds*30cm=12 cm/s
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Careful.
seconds * cm is s * cm, not cm/s
Velocity is the derivative of position with respect to clock time, dx/dt. Average velocity is `dx / `dt, the average rate of change of position with respect to clock time.
Pendulum count:
What was the length of the pendulum you counted, and how many counts did you get in 30 seconds?
11 cm, from below magnet to center of bead
In 30 seconds, I got 45 WHOLE oscillations.
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What therefore is the period of motion of that pendulum?
45 oscillations
or
22.5 half oscillations
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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?
Period: T=0.2*sq. rt of Length
T=0.2*sq. rt. of 11cm
T=0.663 seconds (over and back)
So, 0.663*45=30 seconds
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How well did the freely oscillating pendulum synchronize with the bouncing pendulum of the same length? Which was 'quicker'?
When observing this, I tilted both pendulums with a paperclip, measured both to 12cm (top of magnet to bead) and when I released them at the same time, one pendulum had 4 clicks, the other pendulum had 4 HALF oscillations. I did this several times, and each time seemed to give me the same number, therefore I think they DO synchronize well when they are of the same length. I didn’t find one to be quicker than another.
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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 approx. 12cm. When we tried several different measurements to determine when the ball would synchronize with the 2nd click, we measured it to be around 1 meter from the ground. Therefore we said when the ball was dropped at approx. 1 meter, it hit on the 2nd click of the pendulum.
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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?
T=0.2*sq. rt of Length
T=0.2*sq. rt. of 12cm
T=0.693 seconds, over and back or ONE OSCILLATION
I think it would’ve taken the pendulum approx. 1.04 seconds, since it takes 0.693 seconds to go over and back, when released, it goes over and back and over (1.5 oscillations), therefore 0.693+(0.693/2)=1.04 seconds. Good reasoning, but I believe that would therefore be 1.5 half-oscillations.
I based this answer on the formula given on the board, to find the period in seconds.
That result is consistent with the results of your count, to within +- a few percent.
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Given your 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?
Time required: 1.04 seconds (t) from dropped 1 meter (100cm) above floor
Acceleration: (1m/s) /1.04 seconds=0.96 m/s^2
Your average velocity would be about .96 seconds (close to 1 m/s). However average acceleration is not average velocity / time interval. Go back to the definition of average rate and be sure you have all the words right. Should be easy to correct.
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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 would set up a measuring system, such as a tape measure going from start of ramp to end. I would then use a timing method, either the pendulum, 8 counts, or the TIMER, and record a lot of data at different measurements. Instead of starting v0 at 0 (rest), place the ball at different measurements, higher and lower positions. We then can apply the equations and rules we know about the “quantities of velocity” and can figure out v0, v_mid_x, v_mid_t, and vf according to the different values we did for v0.
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How would you design an experiment to measure v0 and `dv for different values of v0?
Well if v0 is greater than 0, then your change in velocity (‘dv) will be less than vf, instead of v0=0 and vf=’dv. So you could do like we did in class and start the ball rolling at a higher length, but don’t start counting until it reaches 0cm, for example. In class, the ball got at the end in 6 counts of 8, instead of when released at rest, which were 7 counts of 8.
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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.
1080 degrees, 4 seconds, 270 degrees/sec (1080/4=270)
1170 degrees, 6 seconds, 195 degrees/sec
2340 degrees, 12 seconds, 195 degrees/sec
3060 degrees, 15 seconds, 204 degrees/sec
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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:
Length of pendulum: 11cm
Corresponding time intervals on TIMER (2nd column): 0.343, 0.672, 0.703, 0.891
(1st column): 3.969, 4.312, 4.984, 5.687, 6.578
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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.
HALF OSCILLATIONS/CLICK ON TIMER: 0.25, 0.344, 0.328, 0.328, 0.328, 0.328, 0.313, 0.328, 0.281, 0.328, 0.328
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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=0.2*sq. rt of 11
T=0.663 seconds FOR ONE OSCILLATION
Since I counted HALF OSCILLATIONS, the interval I got the most was 0.328, therefore if I multiply 0.328 by 2, it is VERY CLOSE to 0.663 (0.003 seconds difference between the two)!
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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 set my ramp up, using one domino on its side. I set a ruler parallel to the ramp, lining up 0cm with the start of the ramp. I used my pendulum (11cm in length, 0.663 sec period) to record some data. It took the ball 6 clicks on the pendulum to roll down the entire ramp, approx. 30cm. It took the ball about 3.5 clicks to go from rest to halfway down the ramp (15cm). Since I start at rest, my v0 is 0, therefore makes my vf=’dv. I also know that my v_mid_t has to come before v_mid_x. I did some calculations, knowing it takes 0.663 sec for one whole click, and multiplied my # of clicks by that. My final velocity (ball rolling from rest, down entire ramp) was 6 clicks (6*0.663=3.978 seconds). I said going from 0cm-15cm was midway and it took 4 clicks (4*0.663=2.652 seconds). Since v_mid_t is velocity at the halfway time, I said 3.978/2=1.989.
If I did everything correctly and understood right, my order of these quantities is:
V_mid_t (1.989 sec) < v_mid_x(2.652 seconds) < vf=’dv (3.978 seconds)
Very good experiment; however note that you are reporting time intervals rather than velocities
Can you calculate the final velocity for each of your distances?
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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:
Doing 360 degree intervals: 0.765, 0.735, 0.906, 1.078, 1.406, 1.907, 4.39
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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.
470.6, 489.8, 397.4, 333.95, 256.05, 188.8, 82.0
I did 360 degrees divided by the interval clicks I got in the answer above.
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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.
0.765, 0.3825, 0.735
0.735, 0.3675, 0.906
I don’t really understand this question, but what I did was take my 1st interval, divide it by 2 to get midpoint, then list the interval next in the column.
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Good work. A couple of things could use correcting. I expect you'll understand from my notes, but be sure to ask if you have questions.
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