course PHY 201
12:53 PMSunday, October 11, 2009
Class 090930
Horizontal range of a pendulum let loose at equilibrium
Hold a washer pendulum stationary at its equilibrium point and let it drop to the floor. Mark or measure the position at which it strikes the floor.
Now, holding the pendulum so its equilibrium point is the same as before, pull it back and release it so that it swings back toward equilibrium. The instant it reaches its equilibrium point, let go of the string and let it fall to the floor. Record the distance the pendulum was pulled back from its equilibrium position, its length, its height above the floor, and the position at which it strikes the floor.
Repeat for a few trials.
Report your raw data below:
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Pendulum length: 20cm
Table height: 87cm
Release height: 120cm
Pendulum released 10cm from equilibrium
Landing position from table:
Trial 1: 26cm
Trial 2: 31cm
Trial 3: 33cm
Avg: 30cm
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Starting with the event of letting go and ending with the event of first contact with the floor, we assume that the washer is in free fall, the only force acting on it being the force of gravity (air resistance on the washer and on the thread will actually be present, but will be insignificant compared to our uncertainties in measurement). We will analyze the data to determine its velocity at equilibrium.
If you hold the pendulum string in a fixed point, you can move the washer around a circle whose radius is equal to the length of the pendulum. So as long as you are holding the string in a fixed position the pendulum will move along an arc of a circle.
* As it swings back toward equilibrium, the direction of its motion is mostly horizontal but it also descends, so its motion has a downward vertical component.
* As it approaches equilibrium it moves faster and faster, but due to the shape of the circle it descends more and more slowly.
* After it passes equilibrium it continues for a time moving in the horizontal direction as it moves upward in the vertical direction.
At equilibrium the washer is at its lowest point, neither rising nor falling, so at that instant it is moving entirely in the horizontal direction.
* If you let go of the string before the washer reaches equilibrium, then its fall will begin with a downward vertical velocity.
* If you let go after equilibrium, then the initial vertical of its fall will be upward.
* If you let go exactly at the equilibrium point, then the initial velocity of its fall will be entirely horizontal and its initial vertical velocity will be zero.
* Of course you can't let go exactly at the instant the washer reaches the equilibrium position, and the initial velocity of the fall won't in fact be entirely vertical. The initial position of the washer won't be exactly at the equilibrium point, either.
* Since in our analysis we will assume that the initial event occurs at the equilibrium point, with zero vertical velocity, there are multiple sources of uncertainty in this experiment.
Using your raw data show how you find the following, for the interval between the event of letting go of the string and the event of the washer's first contact with the floor.
* The time required to fall your observed vertical distance, starting with initial vertical velocity zero and accelerating downward at 980 cm/s^2.
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`ds = 87cm, a = 980cm/s^2, v0 = 0cm/s
vf = sqrt( v0^2 + 2 a `ds )
= sqrt( 0cm/2 + 2 (980cm/s^2) (87cm) )
= sqrt( 170,520 cm^2 / s^2 )
= 412.941 cm/s
`dt = (vf - v0) / a
= (412.941 cm/s - 0 cm/s) / 980 cm/s^2
= 0.421 s
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* The displacement of the washer in the horizontal direction.
****
The average displacement of the washer from the 3 trials was 30cm from release.
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* The horizontal velocity of the washer.
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`ds = 30cm, `dt = 0.421s
Average velocity is the change in position with respect to clock time:
trial 1: v_ave = 26 cm / 0.421s = 61.758 cm /s
trial 2: v_ave = 31 cm / 0.421s = 73.634 cm /s
trial 3: v_ave = 33 cm / 0.421s = 78.385 cm /s
avg: v_ave = 30 cm / 0.421s = 71.259 cm /s
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For each of your trials, report the pendulum length, the pullback distance, and the horizontal velocity of the falling washer. Use one line to report the results of each trial:
****
20cm,10cm,61.758 cm /s
20cm,10cm,73.634 cm /s
20cm,10cm,78.385 cm /s
20cm,10cm,71.259 cm /s
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You need do this part for only one of your trials:
Suppose the pendulum was released a little early, so that the magnitude of its initial vertical velocity was 10% of the horizontal speed you just calculated. (If you didn't get that part you can assume a horizontal speed of 60 cm/s)
* How long would it then take to reach the floor? (you know the initial vertical velocity, vertical displacement and vertical acceleration)
****
Using the average velocity I found for the average of the 3 trials, 71.259 cm/s, I would begin by finding 10% of this velocity, which would be v0:
v0 = 71.259 cm/s * 0.1 = 7.126 cm/s
`ds = 87cm
a = 980 cm/s
vf = sqrt( v0^2 + 2 a `ds )
= sqrt( (7.126cm/s)^2 + 2 (980cm/s) (87cm) )
= sqrt( (7.126cm/s)^2 + 2 (980cm/s) (87cm) )
= sqrt( 50.780 cm^2/s^2 + 170,520 cm^2 / s^2 )
= sqrt( 170,570.780 cm^2 / s^2 )
= 413.002 cm/s
`dt = (vf - v0) / a
= (413.002 cm/s - 7.126 cm/s) / 980 cm/s^2
= 405.876 cm/s / 980 cm/s^2
= 0.414 s
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Estimate the vertical 'drop' as the pendulum swings to equilibrium
You need do this part for only one of your trials:
Draw a circle on your paper. The diameter of your circle should be at least half the length of your paper.
Sketch the pendulum, hanging at equilibrium, as follows:
* The center of the circle represents the point at which you held the pendulum.
* The radius of the circle represents the length of your pendulum.
* The 'lowest' point on the circle, vertically below the center, will represent the center of the washer.
* Sketch the washer, centered at this 'lowest' point.
* Sketch the string, which will extend from the center of the circle to the washer.
Now sketch the pendulum at its 'pullback' position (the 'held' end of the string will still be at the center of the circle). Keep your sketch reasonably to scale.
Presumably you have observed the pullback and the length of the pendulum. What is the pullback of the pendulum as a percent of its length? (e.g., if the length is 16 cm and the pullback 4 cm then pullback is 25% of the length).
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50%
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Since the radius of the circle represents the length of the pendulum, the number you just gave is also the pullback as a percent of the radius of the circle.
When the washer was pulled back in the horizontal, it was also raised in the vertical direction. Estimate the distance it was raised as a percent of the distance it was pulled back.
****
10%
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What therefore is your estimate of the distance the washer was raised as a percent of the pendulum's length?
****
2cm
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You measured the length. Based on your preceding estimate, how much was it raised?
****
2cm
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If a coin was dropped from rest, and allowed to fall a distance equal to your previous result, how fast would it be going at the end of its fall?
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vf = 412.941 cm/s
falling 2 cm vertically the velocity would be between 60 cm/s and 70 cm/s
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Homework:
Your label for this assignment:
ic_class_090930
Copy and paste this label into the form.
Report your results from today's class using the Submit Work Form. Answer the questions posed above.
Do q_a_9 and q_a_10 .
In case the links don't work, the full addresses of the qa's are given below:
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_3.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_4.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_5.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_6.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_7.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_8.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_9.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_10.htm
Note also that you will receive a subsequent document with some alternative materials, and that you will be asked to complete a short portion of that document.
This looks great. See my one brief note.