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amistre64
We used vector to find wind speed. A balloon is timed as it rises inside, assume a *vertical component only The balloon is taken outside, and timed. The angle it makes is also recorded. Using the vectors, calculate the wind speed.
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i used the law of cosines to determine the magnitude of the resultant. Was that correct to do?
Teacher wanted us to use pythag thrm to determine the wind speed. But the data collected gave a sqrt(-#) which would not allow the pythag thrm to be used. I decided to take the more general Law of Cosines.
i did try it, and since the hypotenuse is smaller than the "leg", it results in a non-real value.
his assumption is that the wind never blows things downward i spose
speed = distance/time speed up = 1.05 m/s speed along the angle = .97 m/s |dw:1348235842574:dw|
i followed per instructions; he noted on the lab report that the time of the outside balloon cannot be greater than the inside record
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Vy is the same for both
Im wondering if windspeed is only a horizontal measure? or does directionality not matter?
"assume a horizontal component only"
there was noting stated prior in the lab to indicate that assumption tho.
the data recorded showed that it rose slower outside, and the note i got on the lab report afterwards was that the data cannot show that. But he also says that we cannot manipulate the data ... i get the feeling that im getting contradictory advice from him :/
i wasnt. but if we only assume the wind blows things from a horizontal, or upwards ... then the data would show it to rise faster. Rise is prolly a bad term for the outside part.
the velocity along the angle is relatively quicker. just like the boat crossing a stream moves quicker than the boats speedometer measuers
the lab was designed to get you working with vectors. The speed of the wind could have been calculated from the outside data alone if we were to assume a horizontal wind speed only. .97*sin(23)
did you really get 23°?
well, it was the best i could do with the little plastic protractor :)
if you were getting <11° you could use the small angle approximation
the teacher suggested the pythag thrm as a means of overcoming the inherent inaccuracies in using the protractors. There was no way to tell if it was "level" and if the person catching the balloon to read the thing hadnt moved it.
i really dont think i understand this experiment , what results did you get /
\[\sqrt{.97^2+1.05^2-2(.97)(1.05)~cos(23^o)}\]
the wind speed itself was immaterial. The lab was designed to get us familar with using vectors and vector math.
was it a helium ballon ?
yes, we used 3 balloons filled with helium. those shiney things you get at a party supply place. I think one of them had Dora the Explorer on it :)
reminds me some of walter lewin 's experiments
what were you timing exactly ?
i held the balloon on the ground, released it, and timed how long it took it to tighten the string it was attached to.
so we were calculating the speed along a vector path
using both vectors (inside and outside) we would have been able to calculate the speed (magnitude) of the resultant vector - which in turn would have been the wind speed
so your vectors are ; \(\{ (1.05,0°) , (0.97, 23°) \} \) ?
those are more like the polar coordinants of the vectors
so converting to cartesian co-ordinates \[\{\quad(0,1.05)\quad,\quad\left(0.97 \cos(23°), 0.97\sin (23°)\right)\quad\}\]
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the theta is in the wrong spot
spose to be 23 degrees west of north
so theta is 90-23 ?
you can find the resultant right?
spose we just use this setup, and focus on the vectors |dw:1348240769620:dw| the resultant can be determined by the law of cosines
i thought you were ment to use pythag. ?
he said we could use whatever method we wanted to find the resultant, and when I used the law of cosines, he said I was wrong for using it.
do you get the same answer both ways /
which leads me to wonder, is windspeed only a horizontal measurement? if so, then the use of 2 vectors is a bad setup
no, you dont get the same answer both ways
|dw:1348241029039:dw| this would have been the resultant using pythag thrm; but my data does not present itself as a right triangle
there was no "answer" to compare to. So the answer itself is irrelevant. It was the process that I used to find it that he said was wrong. and I dont know why
how can i use the pythag thrm with the data collected? i cant.
\[0.97^2-1.05^2=-0.1616\] \[\sqrt{-0.1616}=nonReal~\#\]
lol, i just noticed a typo in the original post :) assume a *vertical* component only
im not sure what to make of demitris reply ...
the data wasnt "given". I went out and collected the data by performing the experiment
the 23 im using of course is from a faulty memory.
but not by use of pythag
you cant have a hypotenuse shorter than a leg ...
does wind only have horizontal component?
experiment: i dont know :/
the lab just wanted us to determine the resultant vector using the data we collected.
well .. maybe the wind is blowing from |dw:1348242000106:dw|
well, magnitude of the resultant that is
i agree, if the wind has a downward component, than the pythag will not be useful
nothing stated in the lab said to find a horizontal component ....
still even we consider wind, this is not just Pythagoras problem ...
any info given on acceleration?
no acceleration given. just direction and speed to define the vectors with
it was noted when i got the lab report handed back to me, that the time outside needed to be quicker than the time inside. But my data did not reflect that.
of course it does...
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the speed recorded outside needed to be quicker than the speed inside
the pythag thrm was suggests as a means to circumvent the inaccuracy of degree measurement. But that only works out if the data collected does not have the wind pushing down on the balloon outside.
|dw:1348242525949:dw| we can calculate the acceleration on balloon due to wind. \[{d^2x \over dt^2} = -k {dx \over dt} \]
thanks for reading it, but my results were calculated already using the cosine rule. I am wondering if it was appropriate to use it in the first place.
the acceleration seems to be a bit outside the scope that the lab was intended for
yes, becasue I use the cosine rule, and included it in the doc that a wrote up
not sure ... if this is mathematics |dw:1348243032602:dw| you can simply get that component using cosine rule.
experiment, I agree :)
lol ... anyway this was getting interesting. :)
we got a new printer in the office, and this ine allows scans to flash drives :) this is the instruction for the lab, just in case you can spot a place that I may have missed.
this is just a physics 1 class, very basic stuff so far
seems plausible. I've been having problems with physics quite a lot lately.
here in florida, temp outside is usually higher than inside :)
thats a good notice on the outside only part. the balloons were already a part of the lab in class; i had no control on prepping them. the inside part would have been an instruction from the teacher, since its not explicitly stated anywhere in the handout. My only idea as to why we would do the inside part would be to better establish the vertical part of the experiment, but it seems to have added more trouble than it was worth :) Thanks
thats what i had in mind at the start i prolly could have included that thought in the "results" section of the lab report.
@demitris, please enable private messaging so i may message you about a post that you'd tagged me in a few minutes ago