basics
https://scienceblogs.com/
enWhat is math?
https://scienceblogs.com/goodmath/2009/12/07/what-is-math
<span>What is math?</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p><img src="http://scienceblogs.com/goodmath/wp-content/blogs.dir/476/files/2012/04/i-e9144736d9d77ecd862e0bdef8b549a4-File:Braque.woman.400pix-thumb-72x130-23398.jpeg" alt="i-e9144736d9d77ecd862e0bdef8b549a4-File:Braque.woman.400pix-thumb-72x130-23398.jpeg" /></p>
<p> I've got a bunch of stuff queued up to be posted over the next couple of days. It's<br />
been the sort of week where I've gotten lots of interesting links from<br />
readers, but I haven't had time to finish anything!</p>
<p> I thought I'd start off with something short but positive. A reader sent<br />
me a link to <a href="http://www.reddit.com/r/AskReddit/comments/abiax/can_someone_explain_mathematics_to_me/">a post on Reddit</a>, with the following question:</p>
<blockquote><p>
Throughout elementary and high school, I got awful marks in math. I always<br />
assumed I was just stupid in that way, which is perfectly possible. I also<br />
hated my teacher, so that didn't help. A friend of mine got his PhD in math<br />
from Harvard before he was 25 (he is in his 40's now) I was surprised the<br />
other week when I learned he isn't particularly good at basic arithmetic etc.<br />
He said that's not really what math is about. So my question is really for<br />
math fans/pros. What is math, really? I hear people throwing around phrases<br />
like "elegant" and "artistic" regarding math. I don't understand how this can<br />
be. To me, math is add, subtract, etc. It is purely functional. Is there<br />
something you can compare it to so that I can understand?
</p></blockquote>
<p> This hits on one of my personal pet peeves. Math really is a beautiful<br />
thing, but the way that math is <em>taught</em> turns it into something<br />
mechanistic, difficult, and boring. The person who posted this question<br />
is a typical example of a victim of lousy math education. </p>
<p> So what is math? It's really a great question, and not particularly<br />
an easy one to answer.</p>
<!--more--><p> You'll get lots of different answers depending on just who you<br />
ask. It's a big enough thing that you can describe it in a lot of<br />
different ways, depending on your perspective. I'm going to give<br />
my own, and you can pipe in with your own in the comments.</p>
<p> To me, math is the study of how to create, manipulate, and understand<br />
abstract structures. I'll pick that apart a bit more to make it more<br />
comprehensible, but to me, abstract structures are the heart of it. Math<br /><em>can</em> work with numbers: the various different sets of numbers are<br />
examples of <em>one</em> of the kinds of abstract structures that we can work<br />
with. But math is so much more than <em>just</em> numbers. It's numbers, and<br />
sets, and categories, and topologies, and graphs, and much, much more.</p>
<p> What math does is give us a set of tools for describing virtually<br /><em>anything</em> with structure to it. It does it through a process<br />
of <em>abstraction</em>. Abstraction is a way of taking something<br />
complicated, focusing in on one or two aspects of it, and eliminating<br />
everything else, so that we can really understand what those one<br />
or two things <em>really</em> mean.</p>
<p> For example, look at topology. Topology is basically a way of<br />
understanding shapes. But it does it in a completely abstract way. It throws<br />
away everything except the concept of <em>closeness</em>. You have a<br />
collection of points, and you've got a concept of things that are<br /><em>close</em> to one another, defined in terms of <em>neighborhoods</em>. By<br />
playing with different notions of what things are close to each other, you can<br />
create any shape you can imagine, and some that you probably can't. But you<br />
don't really need numbers at all: you can just create and play with shapes in<br />
topology - as long as you've got the set of points, and you've got set<br />
relations, you can figure out what it really means for something to be a<br />
torus. You can see what's really strange about a moebius strip. You can<br />
take the moebius strip, and add a dimension to it, and see exactly how you<br />
produce a klein bottle. </p>
<p> For another example, look at category theory. It's a way of understanding<br />
function. What's a function? At it's core it's a <em>mapping</em> from one<br />
thing to another. But what does that really mean? What can you <em>do</em><br />
with that basic idea? What can you <em>make</em> with it? The answer is:<br />
virtually anything you can imagine. </p>
<p> But math is more even than just those abstract things. Why does music<br />
sound good to us? Because it's got an underlying structure. That structure<br />
can be described mathematically. Personally, I'm a huge Bach fan. I believe<br />
that he was the greatest composer of music that ever lived. His music<br />
is magnificently beautiful, and incredibly moving. But to really understand<br />
it, to really grasp all of what he was doing in his music, you need to understand<br />
that it's structure on structure on structure on structure. That structure<br />
is mathematical. If you're really understanding the structure of Bachs music -<br />
if you sit down and analyze it, <em>you're doing math</em></p>
<p> When you look at a cubist painting, you're looking at a strange kind of<br />
projection of something. The artist has taken the subject of the painting<br />
apart, viewed it from different perspectives, different points of views,<br />
different ways of understanding it or seeing it, and assembled them together<br />
into a single image. When you look at a cubist painting, and try to understand<br />
what the artist was seeing, how they were seeing it, and how the pieces<br />
of the final image really fit together - <em>you're doing math</em>.</p>
<p> When a scientist tries to analyze something about the world, to understand<br />
how it works, and describe it in a way that tells us something important about<br />
how things behave - they're doing math. They're <em>abstracting</em> the<br />
world to come up with a precise, formal, descriptive way of stating what<br />
they've learned. </p>
<p> When you look at a road map, and figure out how to get from one place to<br />
another - you're doing math. The map is an <em>abstract</em> representation of<br />
the world that allows you to do certain useful things with it. (And frankly,<br />
this is one example that I've never been able to understand. I can't read<br />
maps. Quite literally a bit o' brain damage - some scar tissue in the left<br />
frontal lobe of my brain.)</p>
<p> When a jazz musician improvises, part of what they're doing is<br />
math. For an improvisation to make sense, for it to sound good, and fit<br />
with what's going on around it, there are a set of constraints on it:<br />
on pitches, pitch progressions, rhythm, chords. Those are all abstract<br />
properties of the music, which are mathematical!</p>
<p> Math is unavoidable. It's a deeply fundamental thing. Without math,<br />
there would be no science, no music, no art. Math is part of all of those<br />
things. If it's got structure, then there's an aspect of it that's<br />
mathematical.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/goodmath" lang="" about="https://scienceblogs.com/goodmath" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">goodmath</a></span>
<span>Mon, 12/07/2009 - 07:21</span>
Mon, 07 Dec 2009 12:21:11 +0000goodmath92777 at https://scienceblogs.comBasic lab stuff: pouring an agarose gel
https://scienceblogs.com/digitalbio/2009/04/21/basic-lab-stuff-pouring-an-aga
<span>Basic lab stuff: pouring an agarose gel</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>This is a video that a friend made that shows, very clearly, how to pour an agarose gel, load the samples and run it. I especially like the way he used a bit of time lapse photography to show the dyes separating as the gel ran.</p>
<object width="480" height="295"><param name="movie" value="http://www.youtube.com/v/QEG8dz7cbnY&hl=en&fs=1" /><param name="allowFullScreen" value="true" /><param name="allowscriptaccess" value="always" /><embed src="http://www.youtube.com/v/QEG8dz7cbnY&hl=en&fs=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="295"></embed></object></div>
<span><a title="View user profile." href="https://scienceblogs.com/author/sporte" lang="" about="https://scienceblogs.com/author/sporte" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">sporte</a></span>
<span>Tue, 04/21/2009 - 06:32</span>
Tue, 21 Apr 2009 10:32:40 +0000sporte69869 at https://scienceblogs.comBasics: Significant Figures
https://scienceblogs.com/goodmath/2009/03/04/basics-significant-figures
<span>Basics: Significant Figures</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p> After my post the other day about rounding errors, I got a ton of<br />
requests to explain the idea of <em>significant figures</em>. That's<br />
actually a very interesting topic.</p>
<p> The idea of significant figures is that when you're doing<br />
experimental work, you're taking measurements - and measurements<br />
always have a limited precision. The fact that your measurements - the<br />
inputs to any calculation or analysis that you do - have limited<br />
precision, means that the results of your calculations likewise have<br />
limited precision. Significant figures (or significant digits, or just "sigfigs" for short) are a method of tracking measurement<br />
precision, in a way that allows you to propagate your precision limits<br />
throughout your calculation.</p>
<p> Before getting to the rules for sigfigs, it's helpful to show why<br />
they matter. Suppose that you're measuring the radius of a circle, in<br />
order to compute its area. You take a ruler, and eyeball it, and end<br />
up with the circle's radius as about 6.2 centimeters. Now you go to<br />
compute the area: π=3.141592653589793... So what's the area of the<br />
circle? If you do it the straightforward way, you'll end up with a<br />
result of 120.76282160399165 cm<sup>2</sup>.</p>
<p> The problem is, your original measurement of the radius was<br />
far too crude to produce a result of that precision. The real<br />
area of the circle could easily be as high as 128, or as low as<br />
113, assuming typical measurement errors. So claiming that your<br />
measurements produced an area calculated to 17 digits of precision is<br />
just ridiculous.</p>
<!--more--><p> As I said, sigfigs are a way of describing the precision of a<br />
measurement. In that example, the measurement of the radius as 6.2<br />
centimeters has two digits of precision - two <em>significant<br />
digits</em>. So nothing computed using that measurement can<br />
meaningfully have more than two significant digits - anything beyond<br />
that is in the range of roundoff errors - further digits are artifacts<br />
of the calculation, which shouldn't be treated as meaningful.</p>
<p> The rules for significant figures are pretty straightforward:</p>
<ol><li> Leading zeros are <em>never</em> significant digits. So in "0.0000024", only the "2" and the "4" could be significant; the leading<br />
zeros aren't.</li>
<li> Trailing zeros are only significant if they're measured. So,<br />
for example, if we used the radius measurement above, but expressed<br />
it in micrometers, it would be 62,000 micrometers. I couldn't<br />
claim that as 5 significant figures, because I really only measured<br />
two. On the other hand, if I actually measured it as 6.20 centimeters, then I could could three significant digits.</li>
<li> Digits other than zero in a measurement are always significant<br />
digits.</li>
<li> In multiplication and division, the number of the significant<br />
figures in the result is the <em>smallest</em> of the number<br />
of significant figures in the inputs. So, for example,<br />
if you multiple 5 by 3.14, the result will have on significant<br />
digit; if you multiply 1.41421 by 1.732, the result will have<br />
four significant digits.</li>
<li> In addition and subtraction, you keep the number of<br />
significant digits in the input with the smallest number of<br /><em>decimal places</em>.</li>
</ol><p> That last rule is tricky. The basic idea is, write the numbers<br />
with the decimal point lined up. The point where the last significant<br />
digit occurs first is the last digit that can be significant in<br />
the result. For example, let's look at 31.4159 plus 0.000254. There<br />
are 6 significant digits in 31.3159; and there are 3 significant digits in 0.000254. Let's line them up to add:</p>
<pre>
31.4159
+ 0.000254
-------------
31.4162
</pre><p> The "9" in 31.4159 is the significant digit occuring in the<br />
earliest decimal place - so it's the cutoff line. Nothing<br />
smaller that 0.0001 can be significant. So we round off<br />
0.000254 to 0.0003; the result still has 5 significant<br />
figures.</p>
<p> Significant figures are a rather crude way of tracking<br />
precision. They're largely ad-hoc. There is mathematical reasoning<br />
behind these rules - so they do work pretty well most of the time. The<br />
"right" way of tracking precision is error bars: every measurement has<br />
an error range, and those error ranges propagate through your<br />
calculations, so that you have a precise error range for every<br />
calculated value. That's a much better way of measuring potential<br />
errors than significant digits. But most of the time, unless we're in<br />
a very careful, clean, laboratory environment, we don't really<br /><em>know</em> the error bars for our measurements. Significant digits<br />
are basically a way of estimating error bars. (And in fact, the<br />
mathematical reasoning underlying these rules is based on how<br />
you handle error bars.)</p>
<p> The beauty of significant figures is that they're so incredibly<br />
easy to understand and to use. Just look at <em>any</em> computation<br />
or analysis result described anywhere, and you can easily see if<br />
the people describing it are full of shit or not. For example, you<br />
can see people claiming to earn 2.034523% on some bond; they're<br />
not, unless they've invested a million dollars, and then those last<br />
digits are pennies - and it's almost certain that the calculation<br />
that produced that figure of 2.034523% was done based on<br />
inputs which had a lot less that 7 significant digits.</p>
<p> The way that this affects the discussion of rounding is<br />
simple. The standard rules I stated for rounding are for<br />
rounding <em>one</em> significant digit. If you're doing a computation<br />
with three significant digits, and you get a result of<br />
2.43532123311112, anything after the 5 is <em>noise</em>. It doesn't<br />
count. It's not really there. So you don't get to say "But<br />
it's <em>more</em> than 2.435, so you should round up to<br />
2.44.". It's <em>not</em> more: the stuff that's making you think it's<br />
more is just computational noise. In fact, the "true" value is<br />
probably somewhere +/-0.005 of that - so it could be slightly more<br />
than 2.435, but it could <em>also</em> be slightly less. The computed<br />
digits past the last significant digit are <em>insignificant</em> -<br />
they're beyond the point at which you can say anything accurate. So<br />
2.43532123311112 is <em>the same</em> as 2.4350000000000 if you're<br />
working with three significant digits - in both cases, you round off<br />
to 2.44 (assuming even preference). If you count the trailing digits<br />
past the one digit after the last significant one, you're just using<br />
noise in a way that's going to create a subtle upward bias in<br />
your computations.</p>
<p> On the other hand, if you've got a measured value of 2.42532, with<br />
six significant figures, and you need to round it to 3 significant<br />
figures, <em>then</em> you can use the trailing digits in your<br />
rounding. Those digits are <em>real</em> and <em>significant</em>.<br />
They're a meaningful, measured quantity - and so the correct rounding<br />
will take them into account. So even if you're working with<br />
even preference rounding, that number should be rounded to three sigfigs as 2.43.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/goodmath" lang="" about="https://scienceblogs.com/goodmath" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">goodmath</a></span>
<span>Wed, 03/04/2009 - 14:55</span>
Wed, 04 Mar 2009 19:55:07 +0000goodmath92694 at https://scienceblogs.comRounding and Bias
https://scienceblogs.com/goodmath/2009/03/01/rounding-and-bias
<span>Rounding and Bias</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p> Another alert reader sent me a link to a YouTube video which is moderately interesting.<br />
The video itself is really a deliberate joke, but it does demonstrate a worthwile point. It's about rounding.</p>
<object width="480" height="295"><param name="movie" value="http://www.youtube.com/v/MbLRxl3Rj7k&hl=en&fs=1" /><param name="allowFullScreen" value="true" /><param name="allowscriptaccess" value="always" /><embed src="http://www.youtube.com/v/MbLRxl3Rj7k&hl=en&fs=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="295"></embed></object><!--more--><p> The overwhelming majority of us were taught how to round decimals back in either elementary or middle school. (I don't even recall exactly when.) The rule that most of us were taught is:
</p><ol><li> If the first digit after the rounding point is 0, 1, 2, 3, or 4, then round the previous digit down;</li>
<li> If the first digit after the rounding point is 5, 6, 7, 8, or 9, then round the<br />
previous digit up.</li>
</ol><p> Here's the problem: those rules are <em>wrong</em>.</p>
<p> The problem is that if the first digit after the rounding point is zero, you're<br /><em>not</em> really rounding - that is, you're not doing anything that <em>changes</em> the value of the data point. But if the first digit after the rounding point is 5,<br />
then it's exactly halfway in-between; it's <em>not</em> closer to the either the rounded up value or the rounded down value - it's exactly between them. Always rounding 5 up will create a bias, because it's taking the point at the middle, and shifting it as if it were closer<br />
towards the upward side.</p>
<p> To demonstrate, let's try an easy example. Suppose we've got the following set<br />
of numbers: {0, 0.5. 1, 1.5. 2, 2.5, 3, 3.5, 4, 4.5}. Let's compute the mean<br />
of those numbers: 22.5/10 = 2.25.</p>
<p> Now, let's round them off: {0, 1, 1, 2, 2, 3, 3, 4, 4, 5}; and then compute the mean: 25/10 = 2.5.</p>
<p> With the standard rounding rule, we've biased the numbers upwards enough to create a significant error!</p>
<p> The correct way to round is to randomly round 5s either up or down. The standard rule, used in most scientific settings, is to pick either odd or even as the "preferred" outcome, and to always round 5s towards the preferred outcome. If we try that with our example, using<br />
preferred even, the rounding is {0, 0, 1, 2, 2, 2, 3, 4, 4, 4}. Taking the mean of that, we get 22/10 = 2.2 - which is significantly closer to the mean of the original numbers than the<br />
mean rounding 5s up. The practice of rounding up adds a systematic bias to the data. It's a very small systematic bias, but it's a real one.</p>
<p> Does it matter? Not usually. As the commentary to the video points out, over the space of a couple of years, that systematic error in rounding gas prices amounts to about a dime. For most things in our daily experience, the difference between random rounding and upward rounding for 5s is just not significant. But if you're doing statistical analysis of<br />
large quantities of data, or you're doing computations that rely on a high degree of<br />
precision, then it can introduce enough error to foul your results. If you're doing statistical analysis, it can do things like make an insignificant result appear to be statistically significant. If you're doing high precision computations for things like<br />
navigation of a space probe through a gravitational slingshot, it can introduce enough error<br />
to crash your probe.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/goodmath" lang="" about="https://scienceblogs.com/goodmath" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">goodmath</a></span>
<span>Sun, 03/01/2009 - 15:32</span>
Sun, 01 Mar 2009 20:32:28 +0000goodmath92693 at https://scienceblogs.comUnit conversions for mere mortals
https://scienceblogs.com/dotphysics/2009/01/20/unit-conversions-for-mere-mortals
<span>Unit conversions for mere mortals</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>So, you are taking a college science course. Maybe it is physics, maybe it is chemistry, maybe its a lab. Either way, you always end up with these problems that involve unit conversions. You think you have the hang of it, but sometimes you make some mistakes. Here is my explanation for converting units. </p>
<p>Convert units? Me? Why? I have <a href="http://www.google.com">google</a>. Yes, that is true, google (for the most part) does an excellent job at unit conversions. But....I doubt your instructor will let you use google on your test. Don't you think you should have a good idea of how to do it? Don't worry. Unit conversion only involves 1 thing.</p>
<h3>Unit conversion is multiplication by 1</h3>
<p>Yes, really. The key thing to realize is that units are an important part of a number and we don't want to change what that number represents, just its units. Take for instance the number 5. Suppose I do the following:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-c7b2d3d778ec86c6b0da431a87bf668c-page-0-blog-entry-15-1.jpg" alt="i-c7b2d3d778ec86c6b0da431a87bf668c-page-0-blog-entry-15-1.jpg" /></p>
<p>What would I get? Let me just work it out.</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-cbdaa98b887afe7e975b325e15f9c984-page-0-blog-entry-15-2.jpg" alt="i-cbdaa98b887afe7e975b325e15f9c984-page-0-blog-entry-15-2.jpg" /></p>
<p>It is the same thing as what I started with. You might say "duh - you multiplied by 2/2 which is one". And you would be correct. What if I do the following: </p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-c7a240167300566f5ef7b64a6f93bc1b-page-0-blog-entry-15-3.jpg" alt="i-c7a240167300566f5ef7b64a6f93bc1b-page-0-blog-entry-15-3.jpg" /></p>
<p>Well, that will still be equal to 5 because 6/3 is still 2. In this case, I am still multiplying 5 by 1. If you are ok with this, then you are ready for a real unit conversion. </p>
<!--more--><p>First, suppose I measure the length of my desk and I get: </p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-9c3cc54f82d7e8f72e7e999e4c63f15d-page-0-blog-entry-15-4.jpg" alt="i-9c3cc54f82d7e8f72e7e999e4c63f15d-page-0-blog-entry-15-4.jpg" /></p>
<p>Note the the units are important. If I measured the desk and got a value of just 55, that is meaningless. 55 what? 55 chickens? 55 gobstoppers? 55 golden gate bridges? Nonetheless, suppose I want the length of the desk in inches instead of cm? I only need to multiply by "1" in this case, my "1" will be created from a a fraction with 0.394 inches and 1 cm since 1 cm = 0.394 inches. That will give: </p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-72bf502e23ba5cf5ee1bb87c681b69e0-page-0-blog-entry-15-5.jpg" alt="i-72bf502e23ba5cf5ee1bb87c681b69e0-page-0-blog-entry-15-5.jpg" /></p>
<p>Notice that the cm canceled since that unit was on the top and the bottom. Really, you can consider the unit to be like a variable (in essence it is). But HEY! you may say - I have been doing this anyway. Why can't I just say "to convert cm to inches, multiply by 0.394". Yes, this is what happens, but not how it happens. With this way, it is easier to do all sorts of conversions. All you need to know is one relationship (like 1 cm = 0.394 inches - OR - 2.54 cm = 1 inch). Let me do a couple of useful examples: </p>
<p>Convert 3 feet per minute to meters per second: </p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-1835db0645e03b9a847526e433457bdb-page-0-blog-entry-15-6.jpg" alt="i-1835db0645e03b9a847526e433457bdb-page-0-blog-entry-15-6.jpg" /></p>
<p>You see, it is possible to do a series of conversions in a row. The key thing to remember is to multiply by fractions that have equivalent things in the top and bottom and make the units cancel (notice I multiplied by 0.305 m over 1 ft so the ft cancel).</p>
<p>Next example - (this one is tricky for many students): Convert 345 cm<sup>2</sup> into m<sup>2</sup>:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-4d68cc28d25f89438cf1c65f4593ea5c-page-0-blog-entry-15-7.jpg" alt="i-4d68cc28d25f89438cf1c65f4593ea5c-page-0-blog-entry-15-7.jpg" /></p>
<p>I did this one wrong (sort of wrong) on purpose. This is similar to the mistake many students make. They say there is 1 m<sup>2</sup> in 100 cm<sup>2</sup>. This just is not the case. Here is a simulation of that: </p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-af6108898109082614c9225c0e96ec15-page-0-blog-entry-15-8.jpg" alt="i-af6108898109082614c9225c0e96ec15-page-0-blog-entry-15-8.jpg" /></p>
<p>This is a square that is 10 smaller squares by 10 smaller squares. Although there are 10 small squares in one side, there are 100 total small squares in this big square. So, for the above problem, it really should be: </p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-ab7b7f09099a1491a56b6509e3bbd196-page-0-blog-entry-15-9.jpg" alt="i-ab7b7f09099a1491a56b6509e3bbd196-page-0-blog-entry-15-9.jpg" /></p>
<p>Last example: How fast is 20 furlongs per fortnight in meters per second? Solution: what in the world is a forlong? I know a fortnight is two weeks. Type the following in google:</p>
<p><a href="http://www.google.com/search?hl=en&q=20+furlongs+per+fortnight+in+m%2Fs&btnG=Google+Search">20 furlongs per fortnight in m/s</a></p>
<p>Yes, google calculator is pretty awesome. Here are some other fun google conversions to try:</p>
<ul><li>the mass of the sun in slugs</li>
<li>2 kg in solar masses</li>
<li>3 cm^3 in ft^2*m</li>
</ul><p><br /></p>
<p>Does that last one even make sense? Sure, I will do this one:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-48e4432b0ae2ec720112559d998eb27e-page-0-blog-entry-16-5.jpg" alt="i-48e4432b0ae2ec720112559d998eb27e-page-0-blog-entry-16-5.jpg" /></p>
<p>The number from google is slightly different (check it out) perhaps due to rounding.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Tue, 01/20/2009 - 04:18</span>
Tue, 20 Jan 2009 09:18:30 +0000rallain107802 at https://scienceblogs.comBasics: Collisions (Interactions between two objects)
https://scienceblogs.com/dotphysics/2008/12/12/basics-collisions-interactions-between-two-objects
<span>Basics: Collisions (Interactions between two objects)</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p><strong>Pre Reqs:</strong> <a href="http://scienceblogs.com/dotphysics/2008/10/basics-forces-and-the-momentum-principle.php">momentum principle</a>, <a href="http://scienceblogs.com/dotphysics/2008/09/basics-what-is-a-force.php">forces,</a><a href="http://scienceblogs.com/dotphysics/2008/10/basics-work-energy.php">energy</a>, <a href="http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-addition.php">vectors</a></p>
<p>Really, there is not much new here. This is an introduction to objects that interact. To describe this, I will need to pull several different ideas together (that you have probably already looked at). Let me start with a simple case. Suppose I have two objects, maybe they are two asteroids in space. I will call them asteroid A and B:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-dc0d0631de7b5ac39162b134b649d679-interaction-1.jpg" alt="i-dc0d0631de7b5ac39162b134b649d679-interaction-1.jpg" /></p>
<p>In this situation, the two objects have different momentums but one interaction between them. Notice that the gravitational force on asteroid A is the same magnitude but opposite direction as the force on asteroid B. This is because there is just one interaction, the gravitational force between the two asteroids. (notice that I am using the notation F<sub>AB</sub> to indicate the force from object A acting on object B) Anyway, what does this force do to the two asteroids? Well, forces change the momentum. If I want to look at the momentum of the two asteroids after some very short time, I can write:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-4617904937fc3e2f791c43b9aba9ff68-deltap-1.jpg" alt="i-4617904937fc3e2f791c43b9aba9ff68-deltap-1.jpg" /></p>
<p>However, I can write a relationship between the two forces (since they are really the same force):</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-179261bdd2535ac607754e96ac1bc802-fa-fb.jpg" alt="i-179261bdd2535ac607754e96ac1bc802-fa-fb.jpg" /></p>
<p>I didn't say it earlier, but the time that the force acts on asteroid A is the same as the time the force acts on asteroid B. So, I can re-write the two momentum equations as:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-fdde993db9000c2e39415dece040b8d6-deltap-2.jpg" alt="i-fdde993db9000c2e39415dece040b8d6-deltap-2.jpg" /></p>
<p>So, now I can relate the two changes in momentum as:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-2f5a04f334010bba25b47f505937411e-comparep.jpg" alt="i-2f5a04f334010bba25b47f505937411e-comparep.jpg" /></p>
<p>If you think about this, since they have the same magnitude of force acting on them (because they are interacting) for the same time, they have the same change in momentum. If I think of the two asteroids as one system, then I can write:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-4962786f4a0e7e976298c2ef911aac71-conservation-p.jpg" alt="i-4962786f4a0e7e976298c2ef911aac71-conservation-p.jpg" /></p>
<p>This is what people usually think of when they think of collisions - conservation of momentum. In this case, this essentially means that the total momentum is the same before and after something happens. Is this always true? No. This is true if there is no external forces on the system. Suppose I had these two asteroids near a star, but the star was not in my system (why would I do that? I don't know, just pretend). Then both of these asteroids would have another force on them such that after some time delta t, the total final momentum would be different than the total initial momentum.</p>
<p>Another important thing to realize is that this does not have to be a gravitational force acting between the two objects. It could be anything. Suppose these two asteroids collided. There would again be a force between them that would be something (probably the coulomb force). As long as this was an internal force, the total momentum would not change. Even if this force changed the internal structure of the two objects, momentum would still be conserved.</p>
<p>Now, for more details.</p>
<!--more--><p>Really, the above stuff is all there is. All that is left is to look at a couple of special cases. I think it is important to NOT get bogged down in the particular solutions of particular cases, but instead realize that conservation of momentum is just an application of Newton's laws.</p>
<p>In the following stuff, I am mostly going to talk about things that "collide". These are interactions that are very quick with very large internal forces. For these cases, the external forces can usually be ignored. Take for instance two baseballs colliding in mid-air. If I look at the momentum of the two balls right before they collide and right after they collide, I can say that momentum is conserved. If I wait too long, clearly momentum will not be conserved because the balls will have an external gravitational force on them. So, I can cheat since the time for the collision is so short.</p>
<p><strong>Special Case: Objects stick together</strong></p>
<p>If the two objects are stuck together after the collision, this is called an inelastic collision (the name really isn't that important). Here is a before and after picture of two objects in an inelastic collision:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-4ff5e541650f789e29600e929d1c96a5-inelastic-1.jpg" alt="i-4ff5e541650f789e29600e929d1c96a5-inelastic-1.jpg" /></p>
<p>So, here the initial momentum is due to two objects. After the collision, there is essentially just one object (I used the subscript T to indicate the total). Since momentum is still conserved, I can write:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-6a72e3b9a2daddaf045a7030a337e7cd-inelastic-2.jpg" alt="i-6a72e3b9a2daddaf045a7030a337e7cd-inelastic-2.jpg" /></p>
<p>How do you deal with vector equations? One way is to break them up into components. If the collision takes place in just two dimensions, you could break this above equation into an x-direction and a y-direction. Not sure if it is worth given further details on this special case. If you know the masses and the two initial velocities (or momentums), you could easily find the vector value for the final velocity after the collision. Another interesting aspect to look at is the change in energy. What should be greater, the total kinetic energy before or the total kinetic energy after the collision? If this is just a plain-jane collision, the final kinetic energy should be less than the initial kinetic energy. This lost energy would go into deforming the two objects and increasing their thermal energy (plus maybe making some sound). If were some type of explosion, the kinetic energy afterwards would be greater. In either case, momentum is still conserved.</p>
<p><strong>Special Case: One dimensional perfectly elastic collision between two similar objects where one is initially at rest.</strong></p>
<p>Yes, this is a little specialized, but I want to show you an example where kinetic energy is conserved. Here is a before and after picture:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-8ed23f9402c8c7da4755ad93c1e89d7f-samecollision-1.jpg" alt="i-8ed23f9402c8c7da4755ad93c1e89d7f-samecollision-1.jpg" /></p>
<p>So, ball A and B have the same mass and ball B is initially at rest. After the collision, they do something. I just randomly put that they are both going to the right with a lower momentum. Also, for simplicity, I will call the right direction the positive x direction. This means that the following is true from the conservation of momentum:(since both balls have the same mass, I will use m for both ball A and B) (double-also: I left off the sub-x notation to indicate that these are the components of velocity in the x-direction)</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-cfa82b28c6f30e985fd6a21bd782bae5-collisionsame-2.jpg" alt="i-cfa82b28c6f30e985fd6a21bd782bae5-collisionsame-2.jpg" /></p>
<p>Clearly, the masses cancel. However, I can not determine the velocity of the two balls after the collision because there are too many unknowns and only one equation. I can get another equation by looking at the kinetic energy. If KE is conserved then:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-7434b48a4b6150dbdc277c5986d570ac-kesame.jpg" alt="i-7434b48a4b6150dbdc277c5986d570ac-kesame.jpg" /></p>
<p>Again, the masses cancel. The "1/2" cancels also. If I take the equation from conservation of momentum, I can solve for the final velocity of object A squared:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-f2e71be06fd31b1caba2b37b03954ecc-sdfsdw.jpg" alt="i-f2e71be06fd31b1caba2b37b03954ecc-sdfsdw.jpg" /></p>
<p>And now plugging this into the kinetic energy equation:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-e9b8affea8f2f494c2878763c5513bef-vba2.jpg" alt="i-e9b8affea8f2f494c2878763c5513bef-vba2.jpg" /></p>
<p>So, the final velocity of the "target ball" that was originally stationary is the same as the initial velocity of "shooting" ball. What about the final velocity of the shooting ball (Ball A)? I can plug this solution for the final velocity of ball B into the other equation and I get:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-2bc6af0788a538cc1ddecd9ff4402a25-piwer.jpg" alt="i-2bc6af0788a538cc1ddecd9ff4402a25-piwer.jpg" /></p>
<p>I cheated. There is actually another valid solution. This works also:</p>
<p><img src="http://scienceblogs.com/dotphysics/wp-content/blogs.dir/342/files/2012/04/i-c7b2a30c59c18fba581ced5307fd6168-possiblevelocty.jpg" alt="i-c7b2a30c59c18fba581ced5307fd6168-possiblevelocty.jpg" /></p>
<p>So, the first solution has Ball A come in and hit ball B. Ball B leaves with the initial velocity of ball A and ball A stops. The other solution is that Ball A continues on with the same speed and ball B stays stationary. This second case would be what would happen if they "missed" but it would still satisfy the momentum and kinetic energy equations.</p>
<p>One application where you sort of see something like this example is in billiards (or pool, I don't know which is the proper term). If you shoot the cue ball straight at a stationary ball, you COULD make the cue ball stop and the other ball continue. This doesn't usually happen because the cue ball is rolling not just moving. If you could make a non-rolling collision between these two balls, you would achieve the result from this example.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Fri, 12/12/2008 - 11:29</span>
Fri, 12 Dec 2008 16:29:50 +0000rallain107776 at https://scienceblogs.comBasics: Potential Energy
https://scienceblogs.com/dotphysics/2008/11/04/basics-potential-energy
<span>Basics: Potential Energy</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>**Pre Reqs:** [Work-Energy](<a href="http://scienceblogs.com/dotphysics/2008/10/basics-work-energy.php">http://scienceblogs.com/dotphysics/2008/10/basics-work-energy.php</a>)</p>
<p>You need to be familiar with work and energy to understand this. If you are not familiar, look at the pre requisite link. Ok? Now, let's begin.</p>
<p>Suppose a ball moves from point A (3 m, 3 m) to B (1 m, 1 m) at a constant speed as shown in the diagram below:</p>
<p>![Screenshot 31](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-31.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>Suppose there is some other force (like my hand) also exerting a force on this ball to make it move along this path at a constant speed. What is the work done by gravity? [If you recall](<a href="http://scienceblogs.com/dotphysics/2008/10/basics-work-energy/">http://scienceblogs.com/dotphysics/2008/10/basics-work-energy/</a>) work is:</p>
<p>![Screenshot 32](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-32.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>Where θ is the angle between the force and the displacement. In this example, θ is 45 degrees. This means that the work done by gravity is:</p>
<p>![Screenshot 33](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-33.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>I will leave it like that, but note that this is a positive value since the gravitational force is mostly in the same direction as the displacement. Ok - now let me do the same thing, but take a different path:</p>
<p>![Screenshot 34](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-34.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>What is the work from A to B done by the gravitational force along this path? I can break this into two parts (1 and 2). The work done by gravity along path 1 is zero. This is because the angle between the gravitational force and the displacement is 90 degrees. The cosine of 90 degrees is zero. For path 2, the angle is zero so the total work is:</p>
<p>![Screenshot 35](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-35.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>I can relate L and L<sub>2</sub> using the right triangle by combining the first and second path (L was the length of the first path)</p>
<p>![Screenshot 36](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-36.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>This is the SAME as the work done by gravity along the first path. So, what does this mean?</p>
<!--more--><p>
This means that the work done by gravity does not depend on the path, but only the two points (the starting and ending point). I know what you are saying. But what if it is some wacked-out path? Why would someone do that? I don't know, but they could.</p>
<p>![Screenshot 37](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-37.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>Well, how do I calculate the work for this path? The angle between gravity and the displacement keeps changing. Also, I don't know the total distance. To do this, I will do what you should always do when faced with a difficult problem: break it into a bunch of small steps. If I break the path into many small pieces, each piece is approximately a constant displacement - I can do that. Here is a sample small piece (blown up for viewability):</p>
<p>![Screenshot 38](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-38.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>Here I have shown two small pieces. For the whole thing there would be many many pieces. Now let me look at just one piece. I can do two things to calculate the work. First, I could just find the angle between the small piece and gravity, or I could break that piece down even further like this:</p>
<p>![Screenshot 39](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-39.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>If you remember your [vector addition](<a href="http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-addition/">http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…</a>), then you remember that a vector can be broken into an x-component and a y-component. Doing it this way, I can calculate the work done by gravity for the two components. The work done along the x-component is zero because the gravitational force is perpendicular to the displacement in the x-direction. The work done along the y-component is simply -mgΔy (because in this piece, the y displacement is up, but gravity is down). So if I break this path into a whole bunch of pieces, each piece has an x- and y-component. The work along the ALL of the x-components is zero. Now I just have to add up all the works done on the y-components. Since the force of gravity is constant, the y-pieces add up to the change in the y-position from A to B. Ok, you didn't really need to know this - I just wanted to give further explanation as to why it does not matter about the path.</p>
<p>Great, what next? Well, if the work done by the force is path independent (technically called a conservative force), then I can write:</p>
<p>![Screenshot 42](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-42.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>Here, W<sub>stuff</sub> is work done by other forces on the object. You can see that all I did was to take the work done by gravity and move it to the other side (the energy side). Before, only the change in kinetic energy was there. If that work is on the energy side, I should call it some type of energy. I will call it gravitational potential energy:</p>
<p>![Screenshot 43](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-43.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>So, if I move an object up - then Δy is positive and ΔU<sub>grav</sub> would be positive also. This means that (if I want to think about gravitational potential energy) that as the object moves up, it increases in gravitational potential energy.</p>
<p>**Key Points**</p>
<ul><li>You CAN NOT have work done by gravity AND gravitational potential energy. This would be like having your cake and eating it too. You can either HAVE your cake, or you can EAT your cake.</li>
<li>Since the potential energy is due to work done by a force, and forces are interactions between TWO objects, the potential energy is for the system consisting of the two objects. In the example above, the system is the ball plus the Earth.</li>
<li>You see that it is the CHANGE in potential energy that appears. This means that (for this potential) it doesn't matter where U = 0 - so choose a position that makes you happy.</li>
<li>For this calculation, I assumed the gravitational force was a constant mg. It isn't. This form of gravitational potential energy is only valid close to the surface of the Earth (you know, normal everyday stuff close).</li>
</ul><p>**Elastic Potential Energy**</p>
<p>The force exerted by a spring is:</p>
<p>![Screenshot 44](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-44.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>This is a scalar equation and Δs is the stretch of the spring (or compression). The direction of the spring force is in the opposite direction as the way the spring is stretched. What if I want to calculate the work done by the spring in stretching the spring an amount Δs?</p>
<p>![Screenshot 45](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-45.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>![Screenshot 46](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-46.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>The work is negative because the force exerted by the spring is in the opposite direction as the displacement. There is a problem. The problem is that the force exerted by the spring is not a constant value. I can fix this though. The force by the spring when it is not stretched (at the beginning) is zero. The force at the final stretch is kΔs. Since the force changes linearly with displacement and the work depends on displacement, I can use the average force. This means that the work done by the spring is:</p>
<p>![Screenshot 47](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-47.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>So, if I move this work done by the spring to the other side, I have the elastic potential energy is:</p>
<p>![Screenshot 48](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-48.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>Notice that since the stretch is squared, there is energy stored in the spring system both with it is stretched and when it is compressed.</p>
<p>**Can't the work done by all forces be made into a potential?**</p>
<p>No. Here is an example. Suppose a I push a block on a table from point A to B.</p>
<p>![Screenshot 49](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-49.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>In order to determine the work done by friction, I need to know the frictional force. A pretty good model for friction is:</p>
<p>![Screenshot 50](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-50.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>This says the magnitude of the frictional force is product of some coefficient times the force of the table pushing up on the block. The direction of this force is always in the opposite direction that the object moves. (this is kinetic friction). Well, then this looks pretty easy. If I call the distance from A to B L<sub>1</sub> then:</p>
<p>![Screenshot 51](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-51.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>I went ahead and put N as mg (remember N is not ALWAYS mg). Otherwise, everything looks great. Ok. Now I will do this again. I will start at A and end at B, but I will take a different path:</p>
<p>![Screenshot 52](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-52.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>In this case, I push the block a distance L<sub>2</sub> past point B and then back to B. The work done by friction while the block is moving to the right is:</p>
<p>![Screenshot 53](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-53.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>But now to push it back to B, friction has to do more work. During this movement to the left, friction is now pushing to the right (because that is what friction does). So, the work while moving to the left is:</p>
<p>![Screenshot 54](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-54.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>This gives a total work due to friction going from A to B as:</p>
<p>![Screenshot 55](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-55.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…</a>)</p>
<p>Clearly this is not the same work done by friction as the first path. So the work done by friction DOES depend on the path - it can not be made into a potential energy.</p>
<p>Notice what would happen if you did this with gravity (turn the table vertical). For the second case, the part of the path that goes past point B would be extra negative work. However, when you come back down to point B, gravity is still pointing down. This would be positive work that would cancel the extra negative work.</p>
<p>Ok. I am done with this.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Tue, 11/04/2008 - 04:51</span>
Tue, 04 Nov 2008 09:51:40 +0000rallain107750 at https://scienceblogs.comBasics: Work Energy
https://scienceblogs.com/dotphysics/2008/10/20/basics-work-energy
<span>Basics: Work Energy</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>**Pre Reqs:** [What is a Force](<a href="http://scienceblogs.com/dotphysics/2008/09/basics-what-is-a-force.php">http://scienceblogs.com/dotphysics/2008/09/basics-what-is-a-force.php</a>)</p>
<p>[Previously, I talked about the momentum principle](<a href="http://scienceblogs.com/dotphysics/2008/10/basics-forces-and-the-momentum-principle.php">http://scienceblogs.com/dotphysics/2008/10/basics-forces-and-the-moment…</a>). Very useful and very fundamental idea. The other big (and useful) idea in introductory physics is the work-energy theorem. Really, with work-energy and momentum principle, you will be like a Jedi with a lightsaber and The Force - extremely powerful.</p>
<p> Well, what is work? What is energy? How are they related? In [another post, I talked about energy.](<a href="http://scienceblogs.com/dotphysics/2008/10/what-is-energy/">http://scienceblogs.com/dotphysics/2008/10/what-is-energy/</a>) I think it is interesting to look at how most textbooks define energy:</p>
<p>*Energy is the ability to do work*</p>
<p>This is really a stupid definition. Kind of circular logic, if you ask me. In the post I mentioned earlier, I claim there are two kinds of energy, particle energy and field energy. At low speeds (not near the speed of light), particle energy can be written as:</p>
<p>![Screenshot 53](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-53.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Where *m* is the mass of the particle, *c* is the speed of light. So, if you just look at a particle, that is it for the energy. Now, what about the "work" portion? Work is defined as:</p>
<p>![Screenshot 54](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-54.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Where *F* is the net force on the particle, ?r is the vector displacement of the particle. The "dot" in between F and ?r represents the "dot product" operation between vectors (also known as the scalar product). In a [previous post](<a href="http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-addition/">http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…</a>) I showed that you could multiply a scalar quantity by a vector quantity. Here I need to do "something" with two vectors. You can't multiply two vectors in the same sense that you multiply scalars. A general definition of the dot product for two vectors:</p>
<p>![Screenshot 55](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-55.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>That looks a little more messy than I wanted, but it can not be helped. Really, it is not that complicated. The dot product is simply the projection of one vector on the other. Let me explain in terms of work.</p>
<!--more--><p>
Suppose I pull a block 2 meters with a force of 10 Newtons as shown in this diagram:</p>
<p>![Screenshot 56](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-56.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Since both force and displacement are in the same direction, this would give a work of 20 Joules. However, I will do this the long way. Assume the x-axis is horizontal, then:</p>
<p>![Screenshot 57](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-57.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Note that I am calculating the work done by THAT force. There could be other forces acting on that block, but even if they aren't it doesn't change the work done. Now let me look at another example that is similar. Suppose I again push with a force of 10 Newtons, and I again move the block 2 meters:</p>
<p>![Screenshot 58](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-58.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>In this case, I push perpendicular to the direction the block moves (to do this, there would need to be other forces acting on the block). How much work would be done?</p>
<p>![Screenshot 59](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-59.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>So, forces perpendicular to the displacement do no work, only forces in the same (or opposite) direction do. So, what if there is a force not perpendicular, but also not in the same direction?</p>
<p>![Screenshot 60](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-60.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>I could calculate the work done by the force in the normal fashion (dot product) or I could say I just need the component of force in the direction of motion:</p>
<p>![Screenshot 61](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-61.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Or:</p>
<p>![Screenshot 62](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-62.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Either way, same thing. This is why sometimes you will hear people explain the dot product as a projection of one vector onto the other. It is only the components of the two vectors that are in the same direction that matter.</p>
<p>So, that is work. Now for the good part. The work energy theorem says:</p>
<p>*The work done on a particle is equal to it's change in energy*</p>
<p>![Screenshot 63](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-63.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Notice that it is the CHANGE in energy, not the energy. At low speeds, the mass-energy (mc<sup>2</sup>) doesn't really change, so I will typically just relate the work to the change in kinetic energy.</p>
<p>But wait!! What about potential energy? Yes, I know. But for a PARTICLE, there is no potential energy. You can have potential energy for a system. Instead of talking about potential energy, I will give a short example of work-energy. I will save potential for another day.</p>
<p>Here is an example I used [previously](<a href="http://scienceblogs.com/dotphysics/2008/10/basics-numerical-calculations/">http://scienceblogs.com/dotphysics/2008/10/basics-numerical-calculation…</a>). Suppose I throw a ball straight up with a speed of 10 m/s (and the ball has a mass of 0.5 kg, but that doesn't really matter). How high will it go? I did this problem both analytically and numerically using the momentum principle. The momentum principle deals with force and TIME - but in this case, I don't want the time. Work-energy deals with force and displacement. This will be perfect for this problem. So, while the ball is going up, here is the free body diagram:</p>
<p>![Screenshot 01](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-011.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>I can now write an expression for the work (the total work) done on this ball as it rises:</p>
<p>![Screenshot 64](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-64.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>In this case, the ball is moving up, so ?y is a positive number. The gravitational force is down but in the same direction (opposite direction - so that there is no cosine term). I can set the initial position of the ball to * y = 0 m*. As for the change in energy, the mass energy does not change so I just have change in kinetic energy.</p>
<p>![Screenshot 65](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-65.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Since I know the initial velocity, I can get a value for the final y:</p>
<p>![Screenshot 66](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-66.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>If you look back at the analytical solution using the momentum principle, this probably looks easier. It should, because it is. Remember:</p>
<ul><li>Work-Energy deals with force and change in position. This problem specifically was looking for the position - a perfect match.</li>
<li>Work-Energy does not give a vector answer. The kinetic energy has v<sup>2</sup> which is technically the square of the magnitude of velocity</li>
<li>The stuff I have done here deals with a PARTICLE. If you have something that can not be represented as a particle (like a car with internal combustion engine) then you will need to do something else.</li>
</ul></div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Mon, 10/20/2008 - 16:37</span>
Mon, 20 Oct 2008 20:37:41 +0000rallain107740 at https://scienceblogs.comSpring Motion and Numerical Calculations
https://scienceblogs.com/dotphysics/2008/10/16/spring-motion-and-numerical-calculations
<span>Spring Motion and Numerical Calculations</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>Maybe you know I like numerical calculations, well I do. I think they are swell. [VPython](<a href="http://vpython.org">http://vpython.org</a>) is my tool of choice. In the post [Basics: Numerical Calculations](<a href="http://scienceblogs.com/dotphysics/2008/10/basics-numerical-calculations.php">http://scienceblogs.com/dotphysics/2008/10/basics-numerical-calculation…</a>) I used vpython and excel to do something simple. I will do that again today (in that this problem could also be solved analytically). However, there is one big difference. This problem has a non-constant forces. Suppose I have a mass that is connected by a spring to a wall. This mass-spring is sitting on a table with no friction. </p>
<p>![Screenshot 27](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-272.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>There is a very interesting property of springs. The more you stretch them, the greater the force they exert (in the usual model of springs). This model works very well. </p>
<p>![Screenshot 28](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-282.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>This is known as Hooke's law. I have written it as a scalar for simplicity. The "k" is called the spring constant. It is a measure of how "stiff" the spring is. The value "s" is the amount the spring is stretched. Typically, there is a minus sign in front of the ks to indicate that the force is in the opposite direction that the spring is stretched. Really, in a scalar equation this is rather silly to include (but everyone does anyway).</p>
<p>**Question: What will the motion of the mass be like if I pull it back and then let go?**</p>
<p>Although this can be determined analytically, I am going to first calculate this with vpython. I will try to show all the details so that you can reproduce this also. If you have not already installed [vpython](<a href="http://vpython.org">http://vpython.org</a>), do that now (don't cost nothing).</p>
<!--more--><p>
In the IDLE editor, enter the following:</p>
<p>![Screenshot 30](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-301.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<ul><li>The first line (it's two underscores, then future, then two more underscores) imports a better division than the standard python division.</li>
<li>The next line imports all the vpython stuff. This lets you use functions like "sphere(), and vector()".</li>
<li>The last import makes it so it is easy to create a graph.</li>
</ul><p><br /><br /></p>
<p>Now you should save and run your program just to make sure everything is ok. If it is fine, you will get a yellow sphere. </p>
<p>Ok, I trust it worked. Here is the plan for the program. </p>
<ul><li>Set up constants and set up stuff for the graph</li>
<li>Make a loop</li>
<li>In the loop, update the force (force of the spring depends on position)</li>
<li>Use the force to update the momentum</li>
<li>Update the position using the change in time and the momentum divided by the mass</li>
<li>Update the graph</li>
<li>Do the loop again....and again....</li>
</ul><p><br /><br /><br />
Note that I am only representing the ball, not the floor or wall or spring. Also, I am going to set up a weird spring. This spring is attached at the origin and as a zero natural length. This means that any displacement of the mass from the origin will result in a force proportional to that displacement. This is not realistic, but easy to calculate and it gets the point across. </p>
<p>So, let me get started with the setup part:</p>
<p>![Screenshot 31](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-311.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<ul><li>I calculated the initial momentum as the mass times the vector(0,0,0) which is STILL the zero vector. However, this way I could easily go back and put the initial velocity as something else.</li>
<li>ks is the spring constant</li>
<li>I set the time interval to be small. If you have this too large, things don't work too well. It IS possible to fix this by changing the "recipe" a little, but I want it to look simple.</li>
<li>posgraph sets up the graph that I will make</li>
</ul><p><br /><br /><br />
If you want, you can run the program (it won't do anything) to see if you have any syntax errors.</p>
<p>It works? Ok, here is the rest of the program.</p>
<p>![Screenshot 32](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-32.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<ul><li>First thing, I changed the position of the mass to pos=(.5, 0, 0). If I didn't do this, it would never move because there would be no force to cause a change in momentum. Of course, it would work if I left it at the origin and gave it an initial momentum - try that.</li>
<li>The fnet vector is just the opposite of the position from the origin of the mass (times some constant). If you want, you can make your mass go in any direction</li>
<li>posgraph.plot puts a point on the graph. It plots the point at pos=(horizontal value, vertical value). In this case, I want time on the horizontal axis and just the x-component of the position of the mass.</li>
</ul><p><br /><br /><br />
Now run your program. It will happen fast because computers are fast (there is a way to slow down the motion, but I am too impatient for that). Hopefully, your program will produce a graph that looks like this:</p>
<p>![Screenshot 34](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-341.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>This is probably more awesome than you realize. What function does this graph look like? If you had to guess? Does it kind of look like the cosine function? Why is that awesome? </p>
<p>**What is cosine**</p>
<p>The cosine function is simply the ratio of the sides a and c in the diagram below of a right triangle.</p>
<p>![Screenshot 35](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-351.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Cosine is can also be explained as the projection of a radius of a circle onto one of the axes. </p>
<p>![Screenshot 36](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-361.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>As the red line moves around the circle, the length of the green line changes. The ratio of the green line to the red line is the cosine of the angle of the red line. So, the sine and cosine functions are associated with triangles and circle (and ? - sorry, I just wanted to give a shout out to ?). Well, who cares. Obviously I care. I care because here in the program, do you see a circle? No. Do you see a triangle? No. Do you see ?? No. But, the program clearly produces a cosine function. I think that is awesome.</p>
<p>Well, maybe the program is wrong. No, it's not wrong. You can easily set up an experiment and get similar data. Also, I can do this problem analytically and get the same answer.</p>
<p>**Analytical Solution**</p>
<p>For this solution, I will already assume that all motion takes place in the x-direction. So, I can write Newton's second law as:</p>
<p>![Screenshot 37](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-371.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Where F<sub>x</sub> is the total force (there is only one in this case anyway). Please forgive me, but I am going to drop the "in the x-direction" notation since EVERYTHING is only in the x-direction. The force from the spring is:</p>
<p>![Screenshot 38](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-381.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Remember that this spring has zero natural length, so the x position IS the "stretch". Putting this together, I get:</p>
<p>![Screenshot 39](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-391.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Here I wrote acceleration as the second time derivative of position. If that is something completely foreign to you, don't worry - you will see this later maybe. Anyway, what I have here is a differential equation. How do you solve a differential equation? There are lots of strategies, but I find the best one is to guess. Yes, just guess a solution and see if it works. First, let me re-write the differential equation:</p>
<p>![Screenshot 40](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-40.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>If you look at this equation, it says "take the derivative with respect to time twice and get something times the original function" (really, it says that, you might have to listen closely). Once function that does that is ....cosine. So, let me try the function:</p>
<p>![Screenshot 41](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-41.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Where A and ? are constants. Let me take the first derivative:</p>
<p> ![Screenshot 42](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-42.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>And now the second derivative:</p>
<p>![Screenshot 43](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-43.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>So this means that:</p>
<p>![Screenshot 44](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-44.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Well, does this agree with the numerical solution? One easy way to compare is the period. The period is:</p>
<p>![Screenshot 46](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-46.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Using the values from the original situation:</p>
<p>![Screenshot 47](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-47.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Is this what the numerical calculation gives also? I will add the following inside the while loop:</p>
<p>![Screenshot 48](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-48.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>This should print the time if the position of the mass is near where it started. The output is:</p>
<p>![Screenshot 49](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-49.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>So from this, the period is right around 1.256 seconds. They agree. I think this is an excellent example of how numerical calculations are really the same thing as analytical calculations. Ok, they are not the same, but they do the same thing.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Thu, 10/16/2008 - 16:57</span>
Thu, 16 Oct 2008 20:57:42 +0000rallain107738 at https://scienceblogs.comBasics: Numerical Calculations
https://scienceblogs.com/dotphysics/2008/10/14/basics-numerical-calculations
<span>Basics: Numerical Calculations</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>**Pre Reqs:** [Kinematics](<a href="http://scienceblogs.com/dotphysics/2008/09/basics-kinematics.php">http://scienceblogs.com/dotphysics/2008/09/basics-kinematics.php</a>), [Momentum Principle](<a href="http://scienceblogs.com/dotphysics/2008/10/basics-forces-and-the-momentum-principle.php">http://scienceblogs.com/dotphysics/2008/10/basics-forces-and-the-moment…</a>)</p>
<p>What are "numerical calculations"? Why are they in the "basics"? I will give you really brief answer and then a more detailed answer. Numerical calculations (also called many other things - like computational physics) takes a problem and breaks into a WHOLE bunch of smaller easier problems. This is great for computers ([or a whole bunch of 8th graders](<a href="http://scienceblogs.com/dotphysics/2008/09/computational-physics-and-a-group-of-1000-8th-graders/">http://scienceblogs.com/dotphysics/2008/09/computational-physics-and-a-…</a>)) because computers don't mind doing lots of little problems. Why are they "basic"? Well, most text would say they are not basic. I disagree. I think this is a legitimate method for solving problems. In particular, this is a great way of solving problems that can not be solved analytically (meaning solving one hard problem).</p>
<p>**Numerical Calculations are Theoretical Calculations**</p>
<p>Let me just get this out of the way. Numerical calculations and analytical calculations are really in the same "class". Often people will lump numerical in with "computational experiment" but that is a really bad thing to do. Some others will claim that there are three different "paths" to discover stuff in science: theory, experiment, and simulations. Simulations are the same thing as numerical calculations which are the same as theory. ([I wrote a letter about this in the American Journal of Physics](<a href="http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000076000009000797000001&idtype=cvips&gifs=yes">http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJ…</a>)) </p>
<p>**Example Problem**</p>
<p>Let me start with a problem that can be solved analytically. Suppose I have a ball of mass 0.5 kg and I throw this straight up with a speed of 10 m/s. How high will it go?</p>
<!--more--><p>
**Analytical Solution:**</p>
<p>First, let me draw a free body diagram for the ball while it is in the air.</p>
<p>![Screenshot 01](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-01.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Yes, this is correct. Even though the ball is moving up, or if it is at the highest point, or if it is on the way down, there is only one force on it (neglecting air resistance, we can add that later). Remember your training. What is touching the ball? Nothing. What long range forces are on the ball? Just the gravitational force of the Earth. If you were in class with Aristotle, he would probably come up with something like *"If it is moving up, it still has some of the force of the hand on the ball."* Don't listen to him. He thinks he is a Jedi.</p>
<p>Ok, so how do you find out how high this will go? I am not going to do this the obvious way. The obvious way is to use the kinematics equations and assume that the acceleration is 9.8 m/s<sup>2</sup> in the downward direction. Not going to do that. I am going to start with the momentum principle. I will write it as:</p>
<p>![Screenshot 02](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-02.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Since the only motion and only force are in the y-direction (I am calling up as the positive y direction) I can write the following scalar equation:</p>
<p>![Screenshot 03](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-03.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Notice that I went ahead and put the gravitational force as -mg where g is the local gravitational field (that is what I like to call it) and has a value of 9.8 Newtons/kg. So, what do I do with this equation? Well, there is something I know. I know the initial momentum (it was shot up at 10 m/s) AND I know the final momentum at the highest point is 0 kgm/s (because at the highest point, it stops for an instant). I can use this to find the time it take to get to the highest point.</p>
<p>![Screenshot 04](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-04.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>First, this is clearly going to be a positive value (good for time). I am calling the initial velocity and momentum by putting a subscript (1). Yes, I should have put units on the zero final momentum, but I am lazy. Ok. So, I have the TIME it takes to get to the highest point, but that is not what the question asked. Here I can use a small "trick". If the force is constant, the momentum (and thus velocity) change at a constant rate. Then:</p>
<p>![Screenshot 07](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-071.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>If I choose the ground to be the origin so that the initial y is zero, then solving for the final y (y<sub>2</sub>)</p>
<p>![Screenshot 08](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-081.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Now putting in what I found for the change in time from above:</p>
<p>![Screenshot 09](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-09.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Now, if I want to put in some numbers, I can:</p>
<p>![Screenshot 10](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-10.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>I know what you are going to say: "you are doing it wrong. That is way more difficult than it needs to be. Just use the kinematic equation." Well, there is nothing wrong with this method. I like it, it starts from more fundamental ideas than the kinematic equation you are referring to (I know the one).</p>
<p>**Numerical Solution**</p>
<p>What if I wanted to do this numerically? Well, the key is to break this into a whole bunch of little steps. This is such an easy example that anything can be used. I am going to show how to do this in a spreadsheet - though typically I like to use [vpython](<a href="http://www.vpython.org">http://www.vpython.org</a>). I am not going to give a real intro to spreadsheets, but if you are not familiar with them maybe this will help.</p>
<p>First, I am going to set up my constants and variables:</p>
<p>![Screenshot 11](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-11.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>A couple of notes on this first step. </p>
<ul><li>Computers are stupid. I am sorry, but it is true. They really don't know much. They don't know the units. I wrote them down on the side, but the computer is just going to crunch the numbers. It is my (the human) responsibility to make sure the units are correct. I wrote them to the side, just for us humans.</li>
<li>The momentum is actually calculated as 10*mass (it is good to get in a practice of calculating everything that you can). </li>
<li>There are only 3 variables. Time, momentum (in the y-direction), and y position. You might need more for more complicated problems.</li>
<li>What is "dt"? dt is the "time step" this is how long in time each step is. You will see.</li>
</ul><p><br /><br /></p>
<p>Now, I will calculate the times. To do this, I will enter "0" for the first time cell (in my example, the first time cell is A7). The next time (cell A8) will be the time before plus "dt". Here is the expression I entered in Excel.</p>
<p>![Screenshot 12](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-12.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Instead of entering "0.1" for cell A8, I entered "=A7+B$4". If you are not familiar with spreadsheets, the $ sign in front of the 4 means "don't change this". Now I will copy this formula to the cells below A8. In Excel you can do this by highlighting cell A8 and dragging the small box at the lower left hand of the cell:</p>
<p>![Screenshot 14](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-14.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>How far down do you copy? It really doesn't matter, you can change this later.</p>
<p>Now, what about y? What is the first value? I already said that I would like the origin to be on the ground. This would mean the first value is 0. What about the value after that? I can use the following formula:</p>
<p>![Screenshot 15](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-15.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Great, but I don't know what the average velocity is. That is ok. I know what the initial velocity is, and that is good enough. </p>
<p>**This is the secret of numerical calculations**</p>
<p>It is not technically correct to use the initial velocity for the average velocity. However, if my time interval is short, it won't really matter much. So, I will enter the following for the second value of y:</p>
<p>![Screenshot 16](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-161.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Notice that instead of v<sub>y</sub>, I put p<sub>y</sub>/m - this will matter in the future (I am being prepared). Also, notice that I called the momentum C7, but dt as B$4 and mass as B$2. I am going to want to change the momentum, but not the time step nor the mass.</p>
<p>Finally, for the momentum. I will use the momentum principle:</p>
<p>![Screenshot 17](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-17.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Here I am referring to p<sub>1</sub> and p<sub>2</sub>. These refer to the momentum for the first and second calculations, You can see that generically, the following is true:</p>
<p>![Screenshot 18](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-18.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Where *n* can be any integer. Here is the formula entered in Excel for the momentum:</p>
<p>![Screenshot 19](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-19.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Copying down some, you should have something that looks like this:</p>
<p>![Screenshot 20](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-201.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Here you can already see your answer. What is the highest the ball goes? Well, you can look at the y positions and see that 5.61 is the highest. You also know this is the highest because it occurs when the momentum (and thus velocity) is around zero. I can also plot the position as a function of time:</p>
<p>![Screenshot 21](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-211.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>I know what you are thinking. I am a cheater and I am wrong. When I solved this problem analytically, I found the ball went 5.1 meters. With this method, the highest point is 5.256 meters (according to Excel). This problem is easily fixed. What if I go and change dt from 0.1 to 0.01? Try it. The first thing you will find is that you don't have enough data. You will need to copy your calculations down around 100 cells. Now looking for the highest point, I find:</p>
<p>![Screenshot 22](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-222.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Now I get on the order of 5.15 meters. What do you think would happen if I made dt = 0.001? Well, I would definitely need more Excel cells to find the answer.</p>
<p>I know this is long, but I want to show you another way of doing this numerical calculation.</p>
<p>**VPython**</p>
<p>It is useful to realize that with a spreadsheet, each row was a different time. Spreadsheets are nice in that people typically feel more comfortable with them than with a programming language.</p>
<p>What is [VPython?](<a href="http://www.vpython.org">http://www.vpython.org</a>) VPython is a set of modules for python (the programming language) that does a lot of the hard work in 3-D graphics for you. In this case, you wouldn't REALLY need vpython because you don't really need a visual representation of the ball going up. Oh - vpython (and python) are FREE and run on Mac OS X, Windows, and Linux.</p>
<p>Let me go ahead and show you the vpython program</p>
<p>![Screenshot 25](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-252.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<ul><li>Notice the vector nature of stuff. VPython likes vectors and so should you. Even though this problem does not require vectors, it is good to get in the habit.</li>
<li>Do not be afraid of the "sphere" function. It is part of the visual module and it will make a nice and pretty sphere.</li>
<li>There are more sophisticated ways of getting the highest point, I could even make a graph, but printing out all the values is the clearest way.</li>
</ul><p><br /><br /></p>
<p>Here is part of the output that was printed:</p>
<p>![Screenshot 26](<a href="http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-262.jpg">http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…</a>)</p>
<p>Here also the highest point is 5.15 meters. One of the cool things about a program vs. a spreadsheet is that it is very easy to decrease the time steps. What if I re-run the program, but change dt to 0.001? The first thing you will notice is that the program takes longer to run (because it is printing out a lot more stuff). If you remove the "print" statements, it goes blazingly fast still. Looking at the output, I get the max height as 5.11 meters.</p>
<p>Ok. That is good enough for now. I still need to show you some more things with numerical calculations, but this is a good start. The next thing I need to show is a case where it is not so easy to do an analytical calculation.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/rallain" lang="" about="https://scienceblogs.com/author/rallain" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">rallain</a></span>
<span>Tue, 10/14/2008 - 04:29</span>
Tue, 14 Oct 2008 08:29:22 +0000rallain107737 at https://scienceblogs.com