#$&*
** The RC Circuit_labelMessages **
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33ohms 4.0volts
0 4
3.57 3.5
6.26 3
10.67 2.5
16.85 2
23.41 1.5
34.13 1
41.11 0.75
50.21 0.5
61.28 0.25
units of table are voltage vs time in seconds they were obtained by voltmeter readings in a series ciruit
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17s
17s
18s
16s
Values were determined by looking at the line of best fit which appears to be exponential and finding times for corresponding voltages. These times were subtracted from one another to find time intervals
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0 150
7.1875 125
14.30469 100
22.78906 75
31.59375 50
40.35521 25
48.679 0
table is current in mA vs, time in seconds. The line of best fit for this data appears to be linear in nature.
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21s
18s
12s
15s
The determined graph appears to be linear and it portrays the current vs. clock time. I found a line of best fit and used its equation to find clock times for corresponding values and time intervals after that.
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The times reported above were not exactly the same, they differed from one another by a significant amount. The values are close to the time values reported for the voltages, but they are not completely accurate leading to the notion that experimental uncertainty must be associated with the values. This is most likely due to the fact that it is hard to tell whether the capacitor is really staying at 4 volts when the meter is connected in series.
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8.768s 0.120A 2.8V 23.3ohms
18.62s 0.09A 1.84V 20.44ohms
28.49s 0.06A 1.20V 20ohms
38.34s 0.03A 0.786V 26.2ohms
43.28s 0.015A 0.636V 42.4ohms
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slope:= -146.3 intercept: + 35.6865
V, A
R = -146.32I + 35.686
my graph is linear, but the fit is not very good, due to the fact that some of my data are outliers. I believe that this graph portrays the relationship between voltage current and resistance. Slope was determined by taking two points and taking rise over run. The y intercept was found at the point where the line of best fit crosses the y-axis. The equation was found by putting these numbers into the linear equation formula.
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10ohms
4 seconds
R = 19.058I - 2.9867
As apposed to my other graph, this one had a positive slope I believe this is because my other data might have had too much error associated with it to get a valid line of best fit. In terms of procedure, I attached a 10ohm resistor to in series with the capacitor because I do not have a 100ohm resistor, and I measure time intervals for voltage drop after the capacitor had been charged to 4 volts. I then performed the same procedure for the current. I used the graphs of current and voltage to find the corresponding resistance. I then made a graph of Resistance vs current.
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I had to reverse the cranking six times before I saw a negative voltage.
My estimate was very accurate because I counted accurately.
The bulb behaved in an interesting manner. At the point when I started doing the reversed cranking, the bulb would grow brighter when I would reverse the cranking and the voltage would drop significantly on the capacitor, then when I reversed back, the voltage increased slowly and the bulb glowed more dimly. My best explanation for this would be that somehow reversing the direction of current caused the capacitor to release much of its stored voltage in order to make the bulb glow brighter.
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When the voltage was changing most quickly, the bulb was glowing its brightest this would lead me to believe that the greater the rate of change of voltage of the capacitor, the brighter the bulb. This is, as stated before, because the capacitor is discharging a lot of its energy in a short amount of time causing the bulb to glow brighter.
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9 times
I believe this estimate is fairly accurate.
As is reversed cranking, the capacitor voltage dropped a great deal at first, and when I reversed back, the voltage increased at a much slower rate than when it decreased. However, when the voltage was close to zero, the voltage increase from the forward cranking was about the same as the decrease from the backward cranking.
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30 beeps, 20 seconds,
Voltage was changing more quickly as the 0 voltage was approached.
4.0 volts
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About 4 volts are produced at 1.5 cranks per second
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2 ,0.135,0.865,3.46V
The first value was determined by taking the 100 cycles divided by 1.5 cycles per second to get 66.7 seconds. 66.7 seconds divided by RC=2. e^-2=0.135. 1-.135=0.865. 4volts * 0.865=3.46 volts.
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3.46volts, 4.0 volts
0.54 volts, 86.5%
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1.59V, 2.54V, 3.12V
values found by performing the same process above.
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-4.0V, 4.0V, -8.0V, 20s
0.363V
v1(t)= 4.0V-8.0V*(1-e^(-t/33))
v1(t)=4.0V-8.0V*(0.4540
v1(t)=4.0v-3.63v
v1(t)=0.363V
The results mean that there is some error associated with my values because based on the numbers that I was using, the voltage at time 20 seconds should have been zero, but I got a number that was approximately 0.363V
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Q=4V(1c/v), so the number of Coulombs would be 4 because the capacitor is one Farad
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3.5Coulombs, 0.5Coulombs
The 3.5 was determined by the fact that the capacitor has a capacitance of 1.0F=1C/V
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2.5seconds, 0.2 Coulombs/second
The seconds were found by looking at a graph of voltage vs time. It was simply 0.5Coulomb/2.5seconds
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average current was approximately 110mA
I don't really understand how this is supposed to relate to the previous question.
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5 hours
@&
I can't speculate on why the current appears to change linearly while the voltage changes exponentially. That does affect some of your conclusions.
In any case, accepting your data, your analysis is good.
*@
phy 201
Your 'conservation of energy on an incline' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Conservation of Energy on an Incline_labelMessages **
short version
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This short version is currently a working draft. Please ask for clarification when that is necessary. Modifications will be made to this document in response to questions, which will also be answered in the usual manner.
Note that this is an alternative short version of the experiment intended for
• Principles of Physics students
• General College Physics students whose lab goal in only a passing lab average
This is not intended for University Physics students.
See also the short video at
http://www.vhcc.edu/dsmith/genInfo/qa_query_etc/EnergyConIncline_DialUp200.wmv
which demonstrates a finger delivering a quick impulse to the ball, which coasts to a stop as it travels up the ramp, then coasts with increasing speed down the ramp. The finger 'pokes' the ball at the very beginning of the video.
Note: You may assume for the purposes of this writeup that the ball has a mass of 40 grams. This probably isn't accurate, but it will serve the purpose of the experiment.
Report a preliminary run of the experiment
I suggest that before actually running the experiment you read through the instructions, set up the system, get a few preliminary timings and submit them with a brief description of what you did. I'll be able to tell you if your results make sense, and might make a suggestion or two. No need to do any calculations, and no need for a detailed description. 15 minutes should do it for the preliminary observations. I'll also be glad to clarify anything you think requires clarification.
Goal of the experiment
The experiment concerns a ball which coasts up a ramp, stops, and coasts back down. We are trying to detect the difference between the magnitude of the acceleration going up and the acceleration going down. It is suggested that to get an initial 'feel' for the system you take the
ball, the ramp and a domino, set the ramp up with a fairly small slope, and use your finger to 'bump' the ball in this manner. Again, the ball
just has to go up and come back. This will give you a point of reference for further instructions.
Basic instruction
The basic instruction is this:
There is a difference between the time required for a ball, given an initial velocity at the bottom of the ramp, to roll up a ramp and the
time required for it to roll back down to its initial position.
• This is easiest to detect on a ramp whose slope is just enough that the ball will 'turn around' and roll back down.
• It is necessary to start the ball with an initial impulse in the form of a 'bump' from your finger or from an object, as opposed to a prolonged push. This is because the presumably uniform acceleration up the ramp does not begin until the ball loses contact with the source of the impulse.
• You can set the system up with too much slope, which will make the difference in time up and time down undetectable, or with too little slope, which will not result in uniform acceleration. It is up to you to determine the optimal slope, but if your time intervals are less than a couple of seconds you won't be able to time them with sufficient accuracy.
• You also need to 'bump' the ball hard enough, but not too hard. You won't always get it right; simply disregard the trials that don't result in sufficient time intervals.
• If the 'bump' isn't strong enough the ball won't go very far and you'll be timing a short interval.
• If the 'bump' is too strong the ball will keep going right off the high end of the ramp.
More specific instructions
More specifically:
Too little slope causes problems:
• If the ramp has too little slope it will be difficult to give the ball an initial impulse that causes it to travel most of the length of the
ramp without rolling off the 'high' end.
• If you don't have quite enough ramp slope, once the ball comes to rest it might simply stay at rest.
Also, if there is too little slope, the small irregularities in the track interfere with the uniformity of the acceleration and throw off the results.
Too much slope causes problems:
• There are unavoidable uncertainties in timing. If the slope is too great the time intervals will be short, and the resulting percent uncertainty will be too high to make an accurate distinction between the 'up' and 'down' intervals.
• A ramp with a rise of a single domino is probably steeper than necessary. It is suggested that you use coins or shims. You won't need an accurate measurement of the slope.
Suggestions for experimental technique:
• You want to start the ball rolling up the ramp, using a sudden impulse rather than a sustained push, giving it enough velocity that it travels
20 cm or more before coming to rest for an instant and then traveling back down.
• At the instant of the 'bump' you need to start the TIMER,
and you need to operate the TIMER in such a way as to determine as accurately as possible the time up the ramp and the time back down.
• You also need to observe, with reasonable accuracy, the point at which the ball comes to rest before rolling back down.
• In the event of a mis-strike (e.g., too hard or too easy) it's simple enough to try again. After a couple of minutes' practice it's not difficult to do this at least well enough to get a good trial every minute or so.
• To get good results it's important to avoid backspin and overspin; if the ball is struck just a little higher than the middle, the ball will start out with about the right amount of spin.
You have a good trial when you have data that allows you to determine the acceleration of the ball up the ramp, and back down.
• You need at least half a dozen good trials.
You therefore need to get a reasonable number of trials, timing the ball from 'strike' to 'turnaround' then back to the original position. It's
important to try to detect and eliminate or correct for systematic errors in timing.
The goal is to try to detect the difference in acceleration between the ball as it travels up the incline and as it travels down. It is assumed
that this difference is independent of how far the ball travels, and also independent of the slope, as long as the slope is small.
It's up to you to find a slope that yields good results. As outlined above, too much slope is counterproductive, as is too little.
Analysis of data and interpretation
First give a synopsis of your setup and all relevant data.
Then, for each trial, determine the acceleration of the ball as it travels up the ramp, and as it travels down the ramp. Show, using a couple of representative sample calculations, how your results were obtained from your data.
From right
: 27, 2.2, 27, 2.2
27, 2, 27, 2.3
23, 2.1, 23, 2
30, 2, 30, 2.7
21, 2, 21, 2.3
acc
5.5, 5.5
6.75, 5
5.2, 5.75
7.5, 4.11
5.25, 3.9
21/2=v=10.5cm/s Dv=10.5-0=10.5 A=dv/dt=10.5/2s=5.25cm/s
@&
Be careful. You appear to be dividing average velocity, not change in velocity, by time interval.
*@
From left
26, 2.2, 26, 2.1
28, 2, 28, 2.4
23, 2.2, 23, 2.3
26, 2.5, 26, 2.4
25, 2.2, 25, 2.3
acc
5.37, 5.9
7, 4.9
4.75, 4.3
4.16,4.5
5.2, 4.7
The acceleration of the ball results mainly from two forces, one being the component of the gravitational force parallel to the incline, the other the force of rolling friction between the ball and the ramp. The difference in the accelerations is due to two facts:
• The gravitational component is in the direction opposite the ball's velocity during one phase of the motion, and in the same direction during the other phase.
• The frictional force is in the direction opposite the ball's velocity.
In the absence of friction the only force would be that of the gravitational component parallel to the incline, which is the same for motion up the incline as for motion down the incline.
• How then does rolling friction result in a difference in the two accelerations? It seems like the only thingthat could cause a difference would be that of different forces from the nudge or different slopes after switching sides
According to your data, what is the magnitude of the acceleration due to the frictional force on the rolling ball, and what is this acceleration as a percent of the acceleration of gravity? 4 quarters
.4*9.8=3.92
-3.92 and 3.92/9.8=40%
• More appropriately, you might choose to give upper and lower bounds for the magnitude of the acceleration due to friction (e.g., the magnitude of the acceleration is at least ___, and at most ___).
Energy conservation states that `dW_NC_ON = `dKE + `dPE. How do your results illustrate this law?
More detailed questions about interpretation
If you have answered the above questions with some degree of confidence you don't need to answer the following at this point. Depending on the answers you submit I might ask you to look at these questions.
You don't need to do any more calculations, except perhaps a couple of additions or subtractions, but answer the following. Don't spend hours thinking through your answers--just think about what is going on with this system and give the best answers you can in, say, 30 minutes. Many of the answers are pretty obvious. When I get your answers I'll be able, if necessary, to help clarify some of the more difficult points.
What forces act on the ball as it rolls freely up or down the ramp? List the forces.
Which of these listed forces are identical both for motion up the ramp and for motion down?
Which of the listed forces change as the ball reverses direction?
How does your answer to these questions help explain why the magnitudes of the two accelerations should be different?
What do you think the acceleration of the system would be in the absence of frictional forces?
Don't actually calculate any of the quantities in the subsequent questions related to work and energy:
What happens to the gravitational PE of the ball as it goes up the incline, and what happens as it goes down the incline?
What happens to the KE of the ball as it goes up the incline, and as it goes down?
How does the PE change up the ramp compare to the PE change down the ramp?
As it rolls up the ramp, how does the PE change of the ball compare with the KE change?
As it rolls down the ramp, how does the PE change of the ball compare with the KE change?
In which case is the magnitude of the ratio of PE change to KE change greater, and why?
Do any nonconservative forces act along the line of the ball's motion as it rolls up the ramp, and as it rolls down the ramp?
How does the action of the nonconservative forces explain the answers to some of these questions?
Between the end of the 'bump' and the ball's return to the same position
• Does its KE increase, decrease or remain the same?
• Does it PE increase, decrease or remain the same?
• Do nonconservative forces do positive work on the ball, negative work or no work?
In what ways do the results of this experiment and support the conservation-of-energy equation `dW_NC_ON = `dKE + `dPE?