Continuing from yesterday's notes on acceleration, velocity, speed and vector quantities and scalar quantities and so on. The Dr. Pepper group has been acting up.
<br /><br />
<h1>Motion w/ constant acceleration</h1>
v(t) = integral of a * dt = at + c = at + v_0<br />
<br />In the above case, the "a" variable is a constant, so you have to evaluate it as such. This is why this is known as <u>constant</u> acceleration. When do you most often have constant acceleration? Free-falling is when you have constant acceleration, until you hit the terminal velocity. In this case, "c" is an initial velocity. <br /><br />
<br />
v(t) = at + v_0<br />
<br />
x(t) = integral of v(t) dt = integral of (at + v_0)dt = (1/2)at^2 + v_{0}t + c<br />
So when you take the integral of velocity, you get the position function. What is that value of "c"?<br /><br />
x(t) = (1/2)at^{2} + v_{0}t + x_{0}
<br /><br />
Using those two equations, find a single equation in terms of x, x_{0}, v, v_{0}, a. You can always start with acceleration and integrate to get the velocity, and then integrate again to get that above x(t) equation. Note that in this list of variables, there's no "t". So you want to substitute.<br /><br />

So how would you begin with that task? <br /><br />
t = (v - v_{0}) / a = (delta v)/a<br />
And the second equation becomes: x = (1/2)a(v - v_{0})^2 + v_{0}( (v - v_{0}) / a) + x_{0}
<br /><br /><br />
This notation is not quite LaTeX. Need to get the preview-latex module installed for visual emacs. Use this file with the written notebook notes.<br /><br /><br />

All of these equations are related. You can get all of these three equations by taking the integral of acceleration.<br /><br />
 Apparently Lockhart's final was "make a Rube Goldberg machine." 
 <br /><br />

<h1>For objects falling</h1>
For objects falling freely due to gravity:<br />
g = 9.81 m/s^2 = 32.2 ft/s^2
<br /><br />
The Tipler book keeps the negative sign with "g" into the equation itself rather than with the variable. The negative would indicate that something is falling down. You could also say that going towards the earth is positive, like when you have to wonder how far something falls and so you would want to call falling down as positive etc. We may "g" negative from time to time.<br /><br />

<h1>Interpreting graphs</h1><br />
<ul>
<li> x(t) also known as displacement <u>vs</u> time. What isi t about this graph? What can you find out besides deplacement? You can see that slope is velocity.
	<ul>
	<li>Slope is the velocity.
		<ul>
		<li>Velocity is increasing when graph is concave up.</li>
		<li>Velocity is decreasing when graph is concave down.</li>
		<li>Velocity is zero? Velocity is zero when the slope is zero, so it's the maxs/mins of the x(t) graph.</li>
		<li>Velocity is a maximum or minimum? At inflection points, which are <b>maxs/mins of v(t) graph</b></li>
		</ul>
	</li>
	<li>Acceleration
		<ul>
			<li>Acceleration is positive when graph is concave up.</li>
			<li>Acceleration is negative when graph is concave down.</li>
			<li>Acceleration is zero when it's (1) constant velocity, or at (2) an inflection point (the point where it changes concavity on the original graph). Or when the graph has a constant slope (i.e. is linear).<br />
		</ul>
	<li>
	</ul>
</li>
<li> Velocity versus time
	<ul>
		<li>Slope = acceleration</li>
		<li>Acceleration increasing for concave up</li>
		<li>Acceleration decreasing for concave down</li>
		<li>Velocity is constant when the slope is zero.</li>
		<li>What is the area under the curve of a velocity-versus-time graph? The area under the curve here is your <u>displacement</u>- change in position.
			integral of v(t)dt from a to b = x(t) from a to b
			</li>
		
		<li></li>
	</ul>
</li>
</ul>
Homework: "Interpreting graphs." Find average speed, average velocity, average acceleration, integration via finding the geometric area of the graphed shapes, etc.<br />
<br />
<br />
h
<h1>2007-09-04</h1>

An arrow is shot at 40 m/s and one second after shooting it the person who shots it hears it hit the target. The speed of sound is 340 m/s. How far away was the target?<br />
- There was also the interesting problem in the Tipler "Physics for scientists and engineers" book on page 44, #9, where you have to be running away from aliens at approximately 120 km/h and you have to touch somebody in the seat next to you etc.<br />
<br /><br />

And now for a question from the UT Austin calculus homework. We start off with v(t) = 6sin(t) - 5cos(t) and find s(t) if s(0)=1. 

<br /><br />
Did four or five problems from some UT homework concerning the derivative of velocity or integrals of accelerations in order to determine other values. 
<h1>2007-09-05</h1>
Physics homework due tomorrow.

<br /><br />
Still working with the Sally and Joe problem. Sally is walking and it takes it 60 seconds to go distance d. Joe takes 120 seconds to go distance d. How long for Sally if she walked twice as fast as she walked in the first place? What if Sally doubles her speed? The distance is two times the speed of the belt. The distance is also 2*V_{sally}.
<br /><br />
We are going to be doing some other problems out of the book. #31. Walk in the class room with a negative velocity and positive acceleration (increasing velocity). <b>Deceleration is not negative acceleration.</b>. Any time it becomes less negative, that means it is increasing. Think of an example where you have negative acceleration and negative velocity- like when you drop something off a cliff where g at all times is -32.2 ft/sec^{2}. Say you throw it up at 96.6 ft/sec. After one second it's going only 64.4 ft/sec upwards. And after two seconds it's going 32.2 ft/sec. And after three seconds it's going 0 and it's reached the top and then in the fourth second it will go downwards 32.2 and then in another second it will be going down at 64.4 ft/sec and another second it will be going down at 96.6 ft/sec. 
<br /><br />
You throw a ball in the air and it's in the air for ... Dakota throws a ball in the air and it lands 8 seconds later. Bryan didn't understand that he was supposed to catch it, so he throws it and 8 seconds later and it lands a ways away. Which ball went higher? They land at the same time, both in the air for eight seconds. Does larger initial velocity indicate that the ball will stay in the air longer? Yes, because it will take longer to decelerate. <u>If it takes the same time, then they went the same height</u>. Dakota threw the ball straight up, but Bryan threw it at an angle because it's going off to the side. The y-components are the same. If the time takes the same, then it takes the 4 seconds to go up and another 4 seconds to go down. <b>So they went the same height, since the time is the same.</b> The distance that the ball lands away is due to the <u>horizontal vector</u>, which Dakota had of value zero. 

<br /><br />
At the top of the ball being tossed up, the velocity is zero, but the acceleration is still -32.2ft/sec.<br /><br />

Is it possible for a body to have zero velocity and nonzero acceleration? Yeah, just like a ball at the top of being tossed. So that's instaneously zero of course.<br />
True or false? If the acceleration is zero, the body cannot be moving. False- because constant velocity means zero acceleration. The derivative of a constant is zero.<br />
True or false? If the acceleration is zero, the x-vs-t graph must be a straight line- not necessarily horizontal. Sure, because velocity is constant, so there's constant slope, so there's the x-vs-t graph is a line.<br />
<br />
<br />

State whether the acceleration is positive, negative or zero for the following functions x(t).<br />
Where is acceleration positive, negative or zero? Acceleration is positive where it's concave up, negative when concave down, 

<br /><br /><br />

There's a homework problem similar to this, the same concept, similar to Sally & Joe problem - if you double the speed, what happens to the result? This one is different. #45. Object projected upwards with initial velocity = v, attains height h.<br />
v_{0} up to H<br />
Another object had initial velocity of 2v_{0} and attained a height up to what? Will it go higher or not as high? It will go <u>higher</u>. How much higher? 4H? 3H? 2H? H? You can throw out H. So which of the other three choices will it be? <br />
s = (1/2)gt^{2} + V_{0}t + S_{0}, and in this scenario S_{0}=0. So in the first scenario:<br />
1) s = (1/2)gt^{2} + V_{0}t<br />
2) s = {1/2}gt^{2} + 2V_{0}t<br />
So get the derivative for #1. So then you set velocity to zero, which is at the top, and that's how you can get your time. And then you can solve for the 's' value again, so the key here is to notice what happens to V_{0}- if you double V_{0}, you're going to have 2V_{0} all squared, and that's 4V_{0}^2 so the answer is going to be 4H.<h1>2007-09-06 </h1>
<u>Today's nonsense</u>:<br />
- All "Megan"s missing<br />
- Two volleyball players missing<br />
- Dakota will be missing tomorrow<br />

<br /><br />

There was a question on the homework, #37. You start at 48.3 km/h and go up to 80.5 km/h in 3.7 seconds. So obviously they want the average acceleration in m/s^2, so average acceleration is going to be the change in velocity divided by the time. So that's 32.2 km/h per 3.7 seconds. 

<br /><br /><br />
<b>Test hint</b>: the Joe&Sally problem from the notes.<br />
<br />

<br /><br />
And now for some examples that you will need for the test: at the moment we have learned everything we need for the test, but now we're just trying some different approaches to solve basically the same problems.
<br /><br />

Ch. 2 #69- would have been a problem on homework but is obviously not so anymore. A stone is thrown vertically from a 200 m cliff.  So starting position = 200. During the last half second of its flight, it travels 45 meters. Find the initial velocity of the stone- somebody threw the stone vertically. So was the stone thrown vertically up? Or was it thrown vertically down? Take the fact that it travels 45 m / 0.5 sec, that's your <u>average velocity</u>. So average velocity is going to be 45 m / 0.5 sec or 90 m/sec. And we are still trying to find initial velocity. What's another way to find average velocity over a time period? .... Anyway, eventually you are able to get V_final, so now you're able to use something like (V_f)^2 = (V_0)^2 + 2a(change in x).<br /><br />
Your delta x (or delta s, whatever) is -200. Because it's S-S_0, and S when you hit the ground is 0, and S_0 was 200. 

<br /><br /><br />
So is s = (1/2)aT^2 + V_initial * T + S_0 always true? No, <u>only for constant acceleration</u>. See the <a href="2007-08-29.html">2007-08-29</a> lecture. Otherwise you'd have to integrate the given acceleration.<br /><br />

<br /><br />

Chapter 2 #71. A bus accelerates at a = 1.5 m/s^2 from rest for 12 seconds. S_0 = 0, V_0 = 0, and then afterwards it goes at a constant speed for 25 seconds, and then it slows to a stop at a = -1.5 m/s^2.<br /><br />
Acceleration is positive at first, so it's going to start with concave up, if you want to graph it. But you could graph velocity. <h1>2007-09-07</h1>
<br />
<u>Today's nonsense</u><br />
- Math club Monday morning will be bagging stuff to sell<br />
- WD40? Use it to clean your shower doors, there's nothing harmful about it. Or use it as a lubicrant for hinges. Apparently it's a product that displaces water, it's antirust, and on their 40th attempt, they came up with this formula, so WD40 - Water Displacement 40. You've got to see the uses. Anyway.<br />
- Katie being charged $6.66<br />
- Megan's change of $5.55<br />
- Phoebe and the movie "The Number 23" and being freaked out by her jacket with the number 23 on it.<br />
- Dropping a monkey from the air from a magnet when a dart-shooter is shot at it .. some sort of wiring system to set it all up?<br />

<br /><br />
<br />
You do not have to do #64. We will get through some homework problems and then give Dakota that phone call that he needs so much.

<br /><br />
<h2>Still working on that #71 bus problem</h2>
We are starting with the same #71 as from yesterday. A bus accelerates 1.5 m/s^2 from rest for 12 seconds, and then goes at constant speed for 25 seconds, and then slows to a stop with an acceleration of -1.5 m/s^2. What was the displacement (distance) traveled?<br /><br />
The total distance is distance 1 + distance 2 + distance 3 because the problem is broken into three parts. So the first distance is acceleration = (change in v)/(change in t), so the change in v is going to be a * (change in t). And we know that the change in time is 12 seconds and that it accelerates at 1.5 m/s^2. So that's 1.5 * 12, and that's our change in velocity. Since it started from 0 the final velocity is 18, and the average velocity over that time frame is going to be 18/2, and that's 9 m/s, so the first distance is 108 m.
<br /><br />
What about the second distance where it's going for a constant speed for 25 seconds? So it's going at 18 m/s at this point, for 25 seconds, so that's 450 m for the second distance.
<br /><br />
And for the third distance. It's decelerating to zero, and since the numbers are the exact same, you're going at 18 m/s and your acceleration is the same but opposite, so it too is 108 m, so then you just add all of these numbers together, and it's 666.<br /><br />
Part B on this problem: what is the average velocity? If you want the average velocity, do you find the final displacement minus the initial displacement, so the change in displacement, and then divide by the change in time? Approximately 13.6 m/s.

<br /><br /><br />
How about a rocket being fired vertically?
<br /><br />
<br />
<h2>The Perfume Problem</h2>
<br />
The rocket is fired vertically upward with an acceleration of 20m/s^2. After 25 seconds, the rocket shuts down. And then the rocket continues as a free particle until it reaches the ground. What is the maximum height? What is the total time in the air? What's the speed just before impact? The velocity at 25 seconds is 500 m/s. At 25 seconds the problem is like launching a rock from a cliff, which only has velocity. 

So basically we are going to find our S_{0} on our new start of the problem. In that time period, we have average velocity is 500+0 all over 2, and you were traveling for 25 seconds. So our S_{0} is the first part where we start where the acceleration goes to gravity instead of the rocket's propulsion system. So it's 62.50 m or- average velocity times the time. <h1>2007-09-10</h1>
<u>Today's nonsense</u><br />
- Homework due<br />
- Cheezits<br />
- <b>Test</b> on Thursday/Friday.<br />
- Test review tomorrow.<br />
- Homework due tomorrow.<br />
- 
<br />
<br /><br />
There will be past homework problems on the physics test, so make sure you study the problems that most people got wrong (like the #12 problem that was 66 + 1/3 or something.).<br />
<br />
<br />
Before we get started to the one-page homework due tomorrow, we're going to get back to the perfume rocket problem. The magnitude of velocity is speed. 
<br /><br />
#86 - Two balls are thrown into the air at a bus stop. The first ball has a velocity of 24 m/s and the second perfume ball has a velocity of 14 m/s. The perfume ball collides with the other ball when the perfume ball is at the highest point. In order to solve this you take the derivative of the position function for the perfume ball and then set the velocity function (the derivative) to zero in order to find the time. This problem will probably be on the test. (#86). There's also a problem where it says that "in the last second the ball traveled X seconds."

<br /><br />
<h1>2007-09-12 - Test review</h1>
<u>Today's nonsense</u><br />
- Spiller will not keep a pig pet<br />
- Goats and pigs<br />
- 

<br /><br /><br />

#A12 - An object that is dropped at the same velocity as it would be thrown horizontally, would hit the ground at the same time, if dropped or thrown from the same height.
<br /><br />
#A16 - For airbag 2, approximately how much distance does the dummy cover between the moment the car crashes and the moment the dummy first makes contact with the airbag? Remember with the airbag graphs, this is a graph of the velocity of the person, not of the cars or anything like that. You're looking at the area under the graph from the first part until it starts to change (which is sort of like at 0.01). This was an interesting question, but you will not have this question on the test.

<br /><br />

#B1CD. A two stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s^2. At 10 sec after launch, the first stage of the rocket (now without fuel) is released. The second stage now has an acceleration of 6 m/s^2. What will be the maximum height attained by the first stage after separation? What will be the distance between the first and second stages 2 sec after separation? You find the height of the first stage (in this case, 200 m) because the average velocity is 20 m/s, and acceleration was 4 m/s for 10 seconds, so the average (0+40 / 2) = 20 m/s for the average velocity, and then you multiply that by the 10 seconds in order to get how far it traveled in the first stage. When you release the first stage, the rocket does not just fall down, there's still velocity, so it's going to go up and it no longer has acceleration, so the first stage will fall while the second stage keeps going up, and the second stage now has an acceleration of 6 m/s^2 according to the setup of the problem. What will be the maximum height attained by the first stage after separation -> you can start at the 200 m point. The first stage (after that 10 seconds), you have T=0 at that point and so s=-4.905T^2 + 40T + 200----- and the 40 is because the velocity is 40 m/s. So how will you find maximum height? Take the derivative, to get velocity, find where velocity equals zero, and then find time, and plug that time back into the position function, and that will tell you the maximum height. You could use the quadratic equation, and then take the average of the two zeroes in order to get the maximum height, or you could take the derivative (as previously explained). So the highest position is something like 281.549 meters. So what will be the distance between the first and second stages two seconds after separation? Stage2 at this moment has an s=(1/2)(6 m/s^2)T^2 + 40T + 200. <u>If they tell you the acceleration, then g is already out of there.</u>. So now you find the distance between T=2 for both of those 's' or position functions.

<br /><br />

#B7. Initial velocity is 20 m/s upwards. Be careful, if they say that an object is thrown downward with a speed of 20, the velocity would be -20, so be careful. But this one is indeed +20. The maximum height? Well, s=(1/2)gT^2 + 20T + 0. And so: V=(-9.81)T + 20. Maximum height occurs when V=0 so you solve for T and then plug it into the position function in order to determine the height at that time. We're talking about free fall and neglecting air resistance.<br /><br />

#B6. A car starts from rest heading due east. It first accelerates at 3.0 m/s^2 for 5 sec and then continues without further acceleration for 20.0 sec. It then brakes for 8.0 sec in coming to rest. What is the car's velocity after the first 5.0 seconds? What is the car's acceleration over the last 8.0 sec interval? What is the total displacement? How fast is it going after 5 seconds? It's going 15 m, because 3*5 = 15. To get how far it travels? And then continues without acceleration for 20 seconds. So here, it's velocity is going to be 15 m/s but it is without acceleration for the average velocity is the velocity because the velocity stays the same the whole time. In the first part, the average velocity was something like 7.5 m/s. The third part of this: it brakes for three seconds and comes to a rest, for 8 seconds, and so we don't know the acceleration. So the question. Part A = 15 m/s. What's the car's acceleration over the last 8 seconds? Alright, how fast was it going? It was going 15 and it ended at 0, so acceleration = change in velocity all over change in time, so we're looking at -15/8, and then of course we have our units: m/s^2. Because, and things that you can do to remember, you know that acceleration is the derivative of velocity, well, we hope you know, and v' has to be dv/dt, which is your change in velocity over your change in time, so all of those concepts help each other, and then part C, what's the total displacement. Now, all of these are moving forward, so how far does it go during this first little part? So the total displacement, this first little section had 5 * 7.5 m/s, and the second part was a constant 15 for 20 seconds so that's 300 m, and then the third part was from 15 to 0 so its average velocity was 7.5.  So get the total displacement. The average, now we back up, it's going different velocity, so displacement will be velocity * time, so if you're going 15 m/s for 20 sec, you multiply, and these are different velocities, so you find the average velocity over that time period, and take the average velocity and multiply by that last 8 seconds, and when the time is changing at a constant rate and you want to know how far it traveled you can take the average by adding them together and dividing by two.


<br /><br /><br />
<u>Problems Spiller "likes":</u><br />
Traveling so much in the last second. If you're going on a trip, and the first half of your trip you travel at this speed, what would you have to travel at the second part to get a certain average. You drop something, you throw something, they land at the same time, how tall was the ...<h1>2007-09-17- Test given back and g lab</h1>

L = length<br />
T = time<br />
What is the A in x = Acos(BT).<br />
When you take the cosine, you take the cosine of the scalar so there's no units, so B will always have the reciprocal units of T, so if T is in seconds then B would be 1/seconds. So this has no units in the parameter ultimately. So whatever B is, T is the reciprocal units. A and X are going to have the same units. So whatever it is that X is representing (in this case, position), then A would have those units as well, which would be L for length.<br /><br />

v = C_{1}e^{-C_{2}T}. <br />
So V is going to have the same units as C_{1} in this case as well.
<br /><br />
<br />
g/cm^3 to kg/m^3 ?? So you do 1 * 100 * 100 * 100 / 1000.

<br /><br /><br />

A car making a 200 km journey travles 20 km/hr for the first 100 km. How fast must it go during the second 100 km to average 40 km/hr? 200/40 = 5 hr. So that's impossible.

<h2>Introduction</h2>
To determine the acceleration due to the force of gravity, taking the derivative of the position function provides the velocity function and the second derivative of the position function provides the acceleration, which should be experimentally confirmed to be -9.81 m/s^2 for constant acceleration. 

<h2>Methods</h2>
The TI-84+ SE calculators were wired to the rangers which allow for motion detection. The bouncy ball was dropped directly below the ranger as in the following diagram:<br />
<img src="2007-09-17.png"><br />
It is important to note that the information that was collected with the detector was the distance per millisecond that the ball was starting from the detector. 

<h2>Data</h2>
<center><u>Plotted ideal position function</u><br /><img src="2007-09-17-plot.png"></center>
<center><u>Experimental position data</u><br />(How do you plot points in gnuplot?)</center>

<blockquote>

</blockquote>


<h2>Analysis</h2>
First we need to trace and find the maximum of the first parabola in order to start the construction of our equation. The maximum is (1.299, 0.310562), so: y_{1} = (x-1.299)^{2} + 0.310562. <br /><br />
The equation that matches the first bounce is -3.550(x-1.299)^{2} + 0.310562.<br />
The equation that matches the second bounce is -5(x-1.79896)^{2} + 0.277808<br />
The equation that matches the third bounce is -1(x-2.2986712)^{2} + 0.17<br />

<h2>Conclusion</h2><h1>2007-09-20 - Vectors</h1>
<u>Today's nonsense</u><br />
- Dakota's watchlessness and anxiety<br />
- Bryan's new position as the "math person" of the math club<br />
- Dakota and Bryan running at each other, example<br />

<br /><br /><br />
This is chapter 3. Vectors represent quantities that have <b>magnitude</b> and <b>direction</b>. So mangitude can be stuff like size, it has a value, or speed. The direction makes the value either positive or negative. Component form of vectors, well, we can talk about how we can arrange it so that the vector starts at the origin, and the magnitude is just the length, and the direction would be the angle. We're going to try to answer all of that. So what are some quantities that are vectors?<br /><br />

Velocity, acceleration, displacement, force, momentum, --- quantities with magnitude but with no direction are going to be <b>scalars</b>.<br /><br />

Scalars: speed, distance.

<br /><br /><br />

Lockhart overdoes the visual addition of vectors.<br />
<h2>Graphically</h2>
Graphically they are represented by <b>arrows</b> (rays, but rays go on forever, and that's not right for vectors), or "line segments with arrowheads on one end." The arrows indicate direction and whose length is proportional to its magnitude.

<br /><br /><br />

Two vectors can be added graphically by placing the tail of one to the head of the other. The resultant, will be the connection from the beginning of the first one to the head of the second one that has been visually added on (this could be done algebraically with the Pythagorean theorem or distance formula, of course). There's also the <u>parallelogram method</u> where the diagnol of the parallelogram would be the, yeah.
<br /><br />
So let me ask you. Now, vector c = vector a + vector b, so you need to realize, or understand what this statement says, this does not apply that C = A + B, because the magnitudes are different, so when you do not mention that it is a vector, it defaults to magnitude. 
<br />
<b>The Triangle Rule</b>: The sum of any two legs is always more than the third leg. <br /><br />

<u>The Parallelogram Method</u><br />
The parallelogram method is another way to look at adding vectors. Place the two vectors tail to tail. The diagonal of the parallelogram formed by VectorA and VectorB. equals VectorC. So take VectorA, and remember VectorB, and this says that you put them tail to tail, and so here's VectorA and here's VectorB, and then make a parallelogram out of them, so that VectorC is the vector splitting B & A so that it meets from the intersection of vectors A&B and then goes to the imaginary intersection of the two other sides.

<br /><br /><br />

Two vectors are equal if they have the <b>same magnitude and same direction</b>. This means they have the same length and are parallel. Moving a vector so that it remains parallel to itself does not change it. So that's why we can do this graphical technique of adding vectors.

<br /><br /><br />

Tho subtract vectors graphically we add the opposite vector that you have to subtract. So VectorA - VectorB = VectorA + ( - VectorB). <u>Opposite vectors have the same magnitude but opposite direction</u>.

<br /><br />

An equivalent way of subtracting VectorB from VectorA is to draw them tail-to-tail and then draw a VectorC from VectorB to VectorA. In this scenario, VectorC must be added to VectorB to find VectorA). So VectorA - VectorB = VectorC so VectorA=VectorB + VectorC. 

<br /><br />

Vectors can easily be added algebraically, rather than graphically, by using the components of vectors, their projected horizontal and vertical lengths. <h1>2007-09-21</h1>
<u>Today's nonsense</u><br />
- Luisa's wrong, Spiller right<br />
- Katie's still getting lost on which class she's in<br />
- Dakota is still watchless<br />


<h2>Unit vectors</h2>
<br /><br />

Unit vectors have a magnitude of <b>1</b>- because it's a unit vector, like a unit circle. Unit vectors point in their corresponding directions. We have the <b>i unit vector</b> which points <u>horizontally</u> and the <b>j unit vector</b> which points in the <u>vertical directions</u>. Then there's also the <b>k unit vector</b> which points in the <u>z direction</u>. Each begins at the origin. Remember, you can move around vectors as much as you want, so unit vectors are with respect to the origin and the origin is as you define it, so you're not necessarily on an x-y plane. 

<br /><br />
ex 1) You walk 3 km west, then 4 km headed 60 degrees north of east, and then 6 km north 40 degrees west (N-40-degrees-W). 60 degrees north of east is the same thing as 30 degrees east of north, and remember that 90 is the key here. So in this scenario, what is the resultant displacement? So that's: -3i + 0j (not pointing up or down) + then you figure out a triangle via SOHCAHTOA because the "60 degrees north of east" is in the first quadrant so it's going to be (cos60 = adjacent/hypothesis) -> (x=4cos60) -> so that's going to be what you do as !THAT THING! (which is 2) * i. So far: -3i + 0j + 2i + "same but different"- this time you find the y-value of that triangle, and so sin60=y/4, so then y=4sin60,, so you get: -3i + 0j + 2i + 3.464j .... and now what about that final part where you go 6 km north and 40 degrees west? So the angle from the x-axis to the vector is going to be 50 degrees in this scenario. So what's the component form of this vector? -3.857i + 4.596j. So to get your resultant, you add up all of the i's and j's and you get your ultimate displacement vector (call it the 'd' vector). So VectorD = -4.856i + 8.06j 
<br />
-- But we are not done yet. We are trying to find the resultant displacement, not the resultant displacement vector. So if I want to find how far, we're going to do the magnitude, and so what quadrant is this displacement vector? It's in quadrant 2, so it's to the top-left somewhere, So the magnitude of that vector is going to be the hypotanuse of the right triangle, so if I want to find the magnitude of the vector, I'm going to do a^2 + b^2 = c^2, so the absolute value of VectorD = 9.410 km. So now how do you get the direction? You have to find the angle that you want, the angle of the resultant displacement vector- so it's going to be the arctan of 8.06/4.357 or whatever. So that turns out to be 58.927 degrees. So now we have the resultant magnitude and the resultant direction.<br /><br />
How do we report this answer? 58.927 degrees North of West. Or you can say: West 58.927 degrees North. Or you can say: North 31.7310 degrees West. But we're not going to get into that. Typically you do North/South first, and they said that you started with North and then either go East or West. So in that case we would have to subtract. So our final answer is: 9.410 km 58.927 degrees North of West.
<br /><br />
You took the arctan because you were trying to figure out the angle, in order to get the final direction of the resultant direction, so that we may report our answer properly. So whenever you are looking for the angle and you know some sides, then you take the arc{trigfunc}.
<br /><br /><br />

<h2>Position, Velocity, and Acceleration</h2>
The <u>position vector</u> of a particle is a vector drawn from the origin of a reference frame to the xy position of the particle. <br />
VectorR = x(VectorI)+y(VectorJ)<br />
So we can talk about an <b>average velocity vector</b>:
<br />
How would you find the average velocity vector? VectorAVG = ( change in VectorR ) / (change in time)<br />
<br />

So now we can get now to instantaneous velocity vectors, which is when there's no delta-t and delta-t would be approaching zero? <br />
<b>Instantaneous velocity vector</b><br />
VectorV = lim (as delta-t -> 0) of ( delta VectorR)/(delta-t) = d(VectorR) / dt<br />
The derivative of the position vector, is the velocity vector. So you would take the derivative of the two components, separately, so next we are going to write:<br />
<br /><br />

<b>Average acceleration vector</b><br />
VectorA_avg = (change in VectorV)/(change in t)<br />
<b>Instantaneous acceleration vector</b><br />
VectorA = lim of ( delta-VectorV / delta-t ) as delta-t -> 0 == dVectorV / dt
<br /><br /><br />
The magnitude of any vector, VectorV = a(VectorI) + b(VectorJ)<br /> So this is just rehashing the idea of using the Pythagorean theorem. <br />
So, the magnitude of that vector is going to be abs(VectorV) = sqrt(a^2 + b^2)<br />
Likewise, for a vector that might be VectorV = a(VectorI) + b(VectorJ) + c(VectorK)<br />
abs(VectorV) = sqrt(a^2 + b^2 + c^2)<br />

<br /><br /><br />
So let's ask a couple theoretical questions or two.  The magnitude of the displacement of a particle is, what the distance the object traveled? How does displacement compare to distance? <b>Displacement is less than or equal to distance</b>. So there are some simple questions here. So let's take a look at #5 on the homework or on the same pages. It says: a man walks around a circular arc from x=5,y=0 to the final position x=0,y=5, so what's the overall displacement? You just use half the circle. So part of his displacement is the distance, then plus his angle, so you use the pythagorean theorem to find the distance. You know the magnitude is going to be the sqrt(5^2 + 5^2) so since it's displacement you also need to include an angle. So abs(VectorDisplacement) = 5sqrt(2). So now we need an angle. The angle is 45, because it's isoscoles triangle. So it's 45 degrees North of West. Or: North 45 degrees West. Or 45 Degrees N of W. 
<br /><br />
Quick mention of prime factoring<br />
<br /><br />
VectorC = VectorA + VectorB. <br />
VectorC = aVectorI + bVectorJ <br />
VectorA = dVectorI + eVectorJ<br />
VectorB = fVectorI + gVectorJ<br />
One of the coefficients on VectorA or VectorB could have been negative.<h1>2007-09-24 </h1>
<u>Today's nonsense</u><br />
- Meg's still missing<br />

<br /><br /><br />
Today we have a new assignment in two days. "So! Let us begin, shall we?" Homework: pg 76-77, #31-34,36-40. 

<br /><br />

ex 1) A particle's position coordinates (so we can even use the same terminology if you wish) (x,y) are the following coordinates: (3,7) at t=0; (7 m, 12 m) at t=4 sec; and finally (13 m, 17 m) at t= 10 sec.<br /><br />
(A) Find the V_avg from t=0 to t=4.sec<br />
(B) Find the V_avg from t=0 to t=10 sec<br />

<br />
So we have to look back to <a href="2007-09-21.html">yesterday's notes</a> to find that VectorA_avg = (change in VectorV)/(change in t). On #32 they ask for the average velocity, so maybe they just want a number here and not the vector? So the average velocity is going to be change in position over change in time. So in this case we're not going to be using specific vectors ... right now for (A) they are just looking for a specific number. On the test this will be further specified. So this is going to be "4i + 5j" / 4 ==> i + (4/5)j. So the magnitude is sqrt(41)/4 = 1.601. 

<br /><br /><br />

So this next one ... oh my, I don't want to do this one, it's a thought problem. It's one of those where when the ball is at the maximum height what's the acceleration, things that are just testing to see if you know what's going on. 

<br /><br />

2) At T=0, a bug located at the origin that has a velocity of 24 m/sec at theta = 30 degrees. At T= 3 sec the bug's at x=60 m and y = 100 m with velocity 30 m/sec at theta = 60 degrees. So we have to find two things: (A) the average velocity and (B) the average acceleration (because it changed velocities). Find the velocity and acceleration vectors. So what's the average velocity vector? So that's the change in VectorR over change in T. So that's 60i + 100j (because we start at 100) and all divided by 3, so we can write that as 20i + (100/3)j so the magnitude of this is going to be sqrt(400 + (100/3)^2)--- but we're just finding the vector so we stop at 20i+(100/3)j.<br /><br />
Now we're going to find the average acceleration vector. So we do VectorA_avg = change in VectorV / change in T. So the starting velocity vector (V_1) is what and what's the ending velocity vector (V_2)? The V_1 is 24 m/s at 30, so you have to find your i and j-- you know that your i=12sqrt(3) and j=12, so you're going to be using a <b>30-60-90 triangle</b>. So the V_1 is going to be 20.785i + 12j. The 30-60-90 triangle: the short leg is (1/2)hypotanuse, and the other one is .... right.  So what about V_2 = 15i + 25.981j (where i and j are your unit vectors). The longer side is the sqrt(3) * hypotanuse, and the short side is (1/2) * hypotanuse, in a 30-60-90 triangle.<br />
- So now we can do the delta to get the average acceleration vector. VectorA_avg = -5.785i + 13.981j all over 3 = -1.928i + 4.660j. 

<br /><br /><br />
In cal 2, it's like the second and third test of next semester, where we have vectors, and one will not even have calculus in it, and that's next semester. 
<br /><br />
There are a couple problems. <b>Mary and Robert</b>. We might do #40 together tomorrow. So they are going to rendevouz on Lake Michigan, but we don't have time to make one up today like this, and we can maybe xerox the pages. <h1>2007-09-25 </h1>

At 2:00 AM an enemy ship is located moving with velocity 10 mph northeast. A gunner located 60 miles N40degreesW of the ship is notified and will prepare to launch a torpedo. It will take the gunner 30 minutes from the time of the ship's sighting to fire a torpedo with a speed of 25 mph. In what direction should the gunner fire, and when, to the nearest minute, will the hit occur?

<br />

A diagram would be a good start. See the handout with this problem and the work done in class. <h1>2007-09-26</h1>

How do I find the equation in terms of x and y given a position vector and velocity vector? Given VectorPosition = (something1)i + (something2)j then x = something1, and y=something2. Remember, even if the position vector has some T^2 values in there, it could easily be linear. This is sort of like parametric equations. Vectors are easier to think of than parametrics, so that's why we teach them before. In precal, you're taught parametric equations, and you are never taught to relate vectors and parametrics together.

<br /><br />

<h1>2007-09-27 Basic relativity and projectiles</h1>
<u>Today's nonsense</u><br />
- Only six people in class today (golf (Alli?), sick (Luisa), homebound (Amanda), homebound (Meg))<br />
- No wind-up toy car (!)<br />
- Lots of people walking around with playdoh (which we are told not to eat)<br />
- Spiller knows what a 'tiddle' is (not a tilde) but not an apostrophe? What's with this?<br />
- Should we make a bananna gun/launcher?<br />
<br />
<br />
#47 from the handout<br />
<br />
ex 6) A car is traveling east at 60 km/h. It rounds a curve, and 5 seconds later it is traveling north at 60 km/h. Find the average acceleration. Find the magnitude and direction of the average acceleration. So: find the vector, then the magnitude and direction. Note that this is at a constant speed- it's also changing direction. The average acceleration is going to be <b>the change in velocity-vector of the change in time</b>. So what are the two velocity vectors? 

<br /><br /><br />

<h1>Projectile motion</h1><br />
Don't tell PETA about this, but we're going to try to shoot monkeys with a bananna gun. "Alright, you want monkey brains with your eggs this morning so you're shooting the bananna at the monkey ..." When you shoot the bananna, at the same time you fire, the monkey is going to drop from the tree. <h1>2007-09-28 - More projectile motion</h1>

Homework: pg 78, #50-56, 58-60, 62, 63, <u>66</u>, 67-70. #66 is bonus. Due Monday. 
<br /><br /><br />
An object is shot with an initial velocity of 84 m/s at 30 degrees to the horizontal (or same thing: "from the horizontal"). Initial velocity horizontal = 84cos30, and initial velocity vertical = 84sin30. 

<br /><br /><br />

Remember that we are now working with velocity and acceleration in terms of vectors, so at its maximum height, the horizontal velocity is going to be whatever the initial velocity was, and it's not going to be zero like the 'j' component of the velocity vector. So we're talking two-dimensional now. Before it was only one-dimensional.<h1>2007-10-09 - New six weeks</h1>
Will be playing with terminal velocity later this six weeks in a lab. Dropping coffee filters- drop downwards to get maximum air resistance, and wouldn't they flip over because we're dropping them cup up? We want them to reach terminal velocity so that they just sort of float down, and we don't want them to be flat, because then we'd get that swinging motion of a flat paper in the wind. 

<br /><br />
Speed of a rocket as a function of time with nonconstant acceleration. Your graph needs to be able to explain itself, so label the axes, label your tick marks, etc. What is the standard form of the exponential equation? <h1>2007-10-11</h1>
Just working on the "analyze the data" packet from two days ago. And plotting on semilog paper. 
<br /><br />
We will be playing with log graph paper today. Two days ago we did a linear regressional analysis. The data that we plotted was time-versus-ln(speed) where P was the speed. Our ultimate goal- we know that the equation is P= P_{0}e^{kt} which should be P = ___ e ^ { ___ T). So what is k and what is P_{0}? So the slope of the graph was your k value. <h1>2007-10-12</h1>
<u>Today's nonsense</u><br />
- Brook and the coffee stain on the ceiling due to "coffee vapor"<br />
- "Eachother" and "a lot" v. alot<br />
PV = nRT = Boyle's law<br />

- Homework<br />
<br /><br />
Lots of notes and information today on Sir Isaac Newton and some of his contributions to the wonderful world of physics. We're on chapter 4 now. Isaac Newton is also responsible for the calculus stuff. He's the guy that causes headaches every now and then. Newton presented three fundamental laws of physics. 

<br /><br />
<b>Newton's first law</b> the law of inertia. An object at rest, stays at rest, unless acted on by an external force. As well, an object in motion continues in motion at constant velocity unless acted on by an external force.
<br /><br />
<b>Force</b>-- an external influence on an object that causes it to accelerate relative to an <u>inertial reference frame</u>. Do you see how this can be never ending? <br />
<b>Reference frames</b>-- a set of coordinate systems at rest relative to each other  Should "each other" be one word?<br /><br />
<b>Inertail reference frame</b>-- a reference frame in which the law of inertia holds exactly. So suppose you saw you were outside, and a dog went past you at 60 mph? That dog's velocity is zero if its frame of reference is the car that it's riding in. As the car slow down, the dog's velocity remains zero.

<br /><br />
<b>Mass</b> - an intrinsic property of an object that measures it's resistance to acceleration. If it's moving, it's going to continue moving. So if you take two objects and give them the same force, and one takes off faster, what can you tell me about the mass of the two objects? The one that accelerates less has the more mass. So, it's a measure of the object's inertia. So how easy is it to change it's inertia? You can't change mass, that's an intrinsic property, it's the same no matter where you are. Weight is different of course. 

<br /><br />

The <b>ratio of two masses</b>  is defined quantitatively by applying the same force to each and comparing their accelerations. How would you measure the mass of something on the moon? Well, you would have a standard mass, and then you'd have to apply equal forces to each one, and find the accelerations which is the change in velocity over the change in time. <br /><br />

m2 / m1 = a1 / a2

<br /><br /><br />

The mass of the standard object (so, if you have a standard object, its mass is 1 kilogram) is 1 kg. The force required to produce an acceleration of 1 m/s^2 on the standard object is defined to be <b>1 newton (N)</b>. 
<br /><br />

<h2>Newton's Second Law</h2>

<br /><br /><br />
The acceleration of an object is in the direction of the net external force acting on it. You can have forces from different directions, but we add up all of those force-vectors, and the resultant force is your net force. So this is all of the i's and the j's and vector components and vector manipulation and so on. It is proportional to the net external force and is inversely proportional to the mass of the object: 
<br /><br />
VectorA = VectorForce  /   (mass of the object on which the force is applied)
<br /><br />
OR: VectorF = mass * VectorAcceleration<br />
<br />

<br />

Then: the sum of the forces is equal to the net force. 

<br /><br /><br />
Physics homework due Tuesday.<br /><br />
-- Read about the dog that can travel at 60 mph<br /><br />
<h1>2007-10-15</h1>
Homework due tomorrow.<br />
<u>Today's nonsense</u><br />
- Katie and Aly absent on Friday from calculus.<br />
- Brandon went to go see the pig. They let the pig out. The pig would not go back in the pin. So, a girl there that has her own pig, got her whip out, and started hitting the pig, and so the pig bruises and marks easily, and so there was a mark over the pig, and Brandon was very distraught over it, and he said, "I can't do this, I can't hurt the pig like that," and so now how is he going to ever train the pig? The pigs are stubborn. They squeel often. Apparently they mark easily, but it goes away quickly. How is he going to send that pig to the slaughter house? The story is going to have to be that the pig is going to the uncle's farm. We are going to have to take pictures of the pig and send them to Brandon every so often and show that he's still alive and not slaughtered .. right.<br />
- What happens to the dogs in the dog races that chase rabbits and can no longer chase? Can you just shoot them? Isn't that against the law?<br />
- Dakota sounds like he's on the redneck side of his family. "I've shot several dogs," - Dakota.<br />
- "Why not" and knots in boy scouts and apparently in Girl Scouts you don't learn knots? What do the boy scouts do with all the knots if the girl scouts don't need them? Maybe the girls weren't paying attention? 
What's a slug? (Unit of measure)<br /> Not the mollusk.<br />
<br /><br />
Today we are doing some examples similar to said homework. The lab is going to be due later, as soon as we can get to it, and if we did the lab today then there'd be homework griping tomorrow and so to avoid political disaster ... So, when we last left off. Related rates. That's what they are doing in AB. They're so simple. Yeah, you look at them again now this year, and oh, wait, you don't think they are simple? They will be on the next test. Find an AB person to help you, get some tutoring. Are you ready for some examples?
<br /><br />
1) A single force of 20N (kilogram meter per second squared) acts on a particle of mass m_1. The particle starts from rest and travels in a straight line a distance of 24 m in 4 sec. What's the mass of the object? Use the position = 0.5aT^2 equation.

<br /><br />
2) A certain force, F_1, causes an acceleration of 5 m/s^2 when it acts on an object of mass m, sliding on a frictionless surface. Find the acceleration of the same object if ... if two equal forces of F_1 act at 90 degrees on the same object. What's the magnitude of F_1? Anyway, it seems that what you have to do is combine the force vectors (because they are at 90 degree angles) and then there's the mangitude of the combined forces, which you use later after you discover F_1 = 5m, and then you show that m = F_1 / 5, and you can plug that into the the new acceleration that you are trying to find.
<br /><br />
"For every rule there is an exception, and that exception is Dakota." - Spiller<br />
"Your whole room is a distraction." - Dakota's response.
<br /><br />
2B) Two equal forces of F_1 act at 60 degrees on the same object. So we need our two forces, and add them together. If you want to keep it simple, what I would do is have the first vector going to the right (F_1i + 0j) and the other one would be going up at an angle of 60 degrees. If these two forces ar at 60 degree angles, then ... is acceleration going to be greater or less than, and why do you think? The closer the two forces are, then it's going to be twice the force, and so double the acceleration, the closer they are the more effective they are. Your intuition was indeed correct. 

<br /><br />
3) A 10 kg object is subjected to forces F_1 = ( 3i + 5j )  and F_2 = (5i - 11j) N. The object is at rest, located at (-3,2) m at time t=0. So, what is the object's acceleration? Find the overall net force (F_{A+B}) which is 8i - 6j. Isn't acceleration equal to F/m? So in this case, the mass is 10, and so that's Force / 10, so that's .8i - .6j. What's the velocity at time T=4 sec? Velocity is the integral of acceleration, so integrate acceleration, and you'd get (0.8T + C)i - (-.6T - C)j. The object is at rest, so these two C values are zero, since you were told that, and 

<br /><br />
<h1>2007-10-16 </h1>
<u>Today's nonsense</u><br />
- Fire alarms going off (can't hear them in Gardner's room).<br />
- Phoebe's overuse of exclamation points in email<br />
- Homework due today.<br />
<br /><br />
If you have different forces, you have to combine the forces. 
<br /><br />
From the homework: 11) A tugboat tows a ship with constant force F and in 10 seconds it increases in velocity at 4 km/h. Change in velocity over change in time is acceleration. Essentially, here, you combine the vectors and figure out how the second force compares to the first by knowing the mass and using that with F_1 and (F_1+F_2) data.
<br /><br />
Worked on the semilog graph paper. <h1>2007-10-17</h1>
<u>Today's nonsense</u><br />
- Donuts! (Thanks Brook.)<br />
- PSAT testing for sophomores and juniors today.<br />
- Homework due tomorrow: pg 105 #....<br />
- Spiller wants the administration to start requiring me to write down in the pass-book every time I use an electrical outlet.<br />
- The autotech guys don't even know what's wrong with the car.<br />
- 
<br /><br />
<br />
The SI unit of mass is the kilogram. <br />
SI unit of force is the Newton or kg-m/sec^2<br />
US customary system unit of force is the pound<br />
US unit of mass is the <b>slug</b> which is the mass of an object that weighs 32.2 lb<br />
<br />
<u>Weight</u> - the force of gravity on a object causing it to accelerate. Your weight is going to change from planet to planet ... during your career? Whatever. Your mass is constant. What's the relationship between weight and mass? <br />
Weight is a force. So, force=ma, so ... according to the book, g = 10 m/s^2. <br /><br />
1 pound (lb) = 4.448222 Newton.<br /><br />
Hence weight is a force.

<br /><br /><br/>

ex 1) What is the weight in pounds of 1 kg? 

<br /><br />

2) The net force acting on a 130 lb student is 25 lb. What is her acceleration?<br />
<br />
ex 3) What force is needed to give an acceleration of 3 ft/sec^2 to a 5 lb block? The force is 0.4658 pounds (lbs).

<br /><br />

<h2>Newton's Third Law</h2>
Forces always occur in equal and opposite pairs. If object A exerts a force ofn an object B, an equal but opposite force is exerted by object B on object A. 
<br /><br />

ex) A 5.6 kg object hangs at rest from a rope-like substance attached to a beam. <br />
A) Draw a diagram showing all forces and reaction forces acting on the object (and indicate each reaction force). So you have an object hanging, so here's the string, here's the object, The force in the rope is tension. <br />
B) So, in the string and the ceiling, you have a tension going down, and an equal tension pulling up. <h1>2007-10-18 - </h1>
<u>Today's nonsense</u><br />
- Homework due. pg. 105 #18,20-29,31<br />
- You will be working on a lab writeup tomorrow while Spiller is sick. He's going to get blood taken, which takes, what, he gets there early, so he'll be out by 8:15, and he doesn't want to drive or pass out and hit his head, so he's not going anywhere. Phoebe understands. Mrs. Spiller is subbing- it keeps the money in the family.<br />
- Need to start keeping track of time that is spent slacking off.<br />
<br /><br />

"At Wall-Mart, for a dollar, they have these, this is for the Halloween thing, they were like the collapsable pumpkins, which had wires in them for storage and all, and they had a little basket, should Math Club get some of them as prizes? I don't know because we'd have to buy enough for everybody if we were to get enough for everybody and we don't know how many there are. At the last Bingo game we could say "Okay now we're playing for all of the Halloween stuff." No, they're like pumpkins but a basket that's collapsable, so it springs up so that you can collapse it, it's like a dollar each. I'm curious, I want to hear what Luisa has to say. No, it just has a little face like it was a jackolantern. Hand-sewn quilt, Mrs. Gilson and the quilting guild .... Are we ready? Are we waiting for the toaster? I have to have my keys. This whole NHS thing and them just changing days ... I mean, we could do Wednesday, what's Wednesday mornings, meetings, and then if we have a faculty meeting we will not have one that week. I'm surprised calculus two hasn't griped about this. Physics 2, and then the degree symbol, it's very creative, you just don't get it. ... If you're going to refuse to wear the ID when you have it, we should just write you up. If they didn't give us such cheap IDs, maybe we would wear it. I've had the same ID for ... so you have another one that is broken? Do you have another one? Is it broken? May I see it? If this is an angle, then what is this? " 

<br /><br />
<center><img src="2007-10-18.png"><br /><h3>A WRECKED ANGLE</h3></center>
<br /><br />

Imagine a spring sitting there doing nothing. But what if we stretched it or compressed it? The amount of force required to compress it, well, each spring has a certain constant that applies to it. 
<br /><br />
<h2>Hooke's law</h2>

When a spring is compressed or extended by a small amount, <b>delta x</b>, the force required to do so is proportional to "delta x":<br />
<br />
F<sub>_x</sub> = -k(delta x) where k is the unique force constant<br />
The value k is different for each spring.

<br /><br />
ex 1) A 110 kg baseball player hangs on the rim following a dunk, when he comes to rest, the rim is bent down 15 cm. Assume the rim has spring-like behavior.and claculate its force constant, k. <br />
<img src="2007-10-18_2.png"><br />

"So they wiped out all of those stores? The movie place by the gas station? They took down all of that. It's supposed to be a CVS. Aren't we getting one in Buda too? No, Walgreens. Somebody keeps talking about a Supertarget coming into Kyle, by Home Depot, and an HEB Plus? This is a big time. It's called competition. And we're going to have a hospital. We should ask Alex, his father is an economic developer. You may begin!"

<br /><br />
He's just hanging there on the rim. So we know that 110 kg is his mass. And of course, gravity we're going with 9.81 m/sec^2. So his weight comes out to be 1079.1 Newtons. So that's the force that's coming over here. So Weight = 1079.1 N. What are the units on k? Force does not equal distance. This is kind of like with the gas constant in chemistry. It has to have units to make the equation come out all the same. Newton is a kg * m per sec^2. 

<br /><br />
Quite often when we are talking about force, we're talking about the magnitude, so to say the positivity/negativity of the force means that you're just looking at the direction as well.

<br /><br />
Let's talk about that monkey that we were talking about a few days ago. The monkey and that bananna. <br />
2) A 4 kg bunch of bannananas is suspended motionless. from a spring balance whose force constant is k = 300 N/m. By how much is the srping stretched? <br />

<br />
<center><img src="2007-10-18_3.png"></center>
<br /><br />

<center><img src="2007-10-18_4.png"></center>
<br />

Tomorrow we will do lab writeups. There will be no particular style that is mandated, but particular things need to be included. This will be the "one free ride" lab writeup.

<br /><br />
What are we doing with this data? Now, every time you do a problem with objects falling in the air, we always say to neglect air resistance, the idea with the coffee filters is that they will reach terminal velocity very quickly, so one coffee filter versus four, which one will hit terminal velocity first? The one. That's the kind of thing that you will point out in your final conclusions in your lab, or things that you could do differentfly if you were to do it again, the brilliant idea of doing the four first and then taking one out, that's a really good idea, but Dakota's group found that doing them from 1 to 4 is kind of bad, so ...  

<br /><br />
You don't need to write this down, it will be on the handout: mass * acceleration = mass * acceleration due to gravity if there is no wind resistance. Obviously, and this is given on the paper, you don't need to write this, unless there's a drag force, which would mean that m * a = m * g - F_drag. This drag force is  F_drag = bv^n. Do you know what the b stands for? The b value is a velocity. 

<br /><br />
When you drop something, it gets faster and faster theoretically, but if you were to drop a feather or coffee filter, it did not get faster and faster, it got faster at first, and then it just floated down because it hit terminal velocity, because of air resistance. So the terminal velocity is dependent on the shape of the filter etc. So hopefully all of the filters were the same shape and as we stretched them out they were not the same. Our goal is to find the "N" value. <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
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		<TD WIDTH=20%>
			<P><BR>
			</P>
		</TD>
		<TD WIDTH=20%>
			<P>1</P>
		</TD>
		<TD WIDTH=20%>
			<P>2</P>
		</TD>
		<TD WIDTH=20%>
			<P>3</P>
		</TD>
		<TD WIDTH=20%>
			<P>4</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>1</P>
		</TD>
		<TD WIDTH=20%>
			<P>5.65</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.87</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.78</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.16</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>2</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.75</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.31</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.50</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.59</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>3</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.89</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.72</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.72</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.81</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>4</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.78</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.03</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.82</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.75</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>5</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.63</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.15</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.72</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.80</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>6</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.56</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.22</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.25</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.31</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>7</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.81</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.18</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.75</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.56</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>8</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.68</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.28</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.83</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.91</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>9</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.89</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.40</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.93</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.16</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>10</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.65</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.22</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.68</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.56</P>
		</TD>
	</TR>
</TABLE>
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</HTML><h1>Coffee filter experiments</h1>

<h2>Hypothesis</h2>
The coffee filters will reach their terminal velocity quickly when dropped and to figure this out we will be solving the drag equation for the exponent (n).<br />
<center><img src="2007-10-19_2_ytime_xvel.png"><br />Time-Velocity Graph</center>
<br /><br />
We will be testing the following equations: <br />
<center>
M * a = M * g - F<sub>_drag</sub><br />
F<sub>_drag</sub> = bv<sup>^n</sup><br />
ln v<sub>_f</sub> = (1/n) ln m   + ( (1/n)(ln g - ln b) )<br />
</center>
<br /><br />

<h2>Design of the experiment</h2>
To discover exponential equations, we needed an experiment that would show us an exponentially increasing variable with another increasing controlled variable. In this case we decided to drop coffee filters from the balcony of the school's hallway. We were going to drop one, two, three, and four coffee filters at a time for 10 trials per each amount, recording the amount of time it takes to fall the specific distance. So this required two people, one person to drop and the other to play with the stopwatch and come up with numbers.
<br />
<br />
<center><img src="2007-10-19.png"></center>
<br /><br />
<h2>Data and observations</h2>

	<h3>Data</h3>







<TABLE WIDTH=100% BORDER=1 BORDERCOLOR="#000000" CELLPADDING=4 CELLSPACING=0>
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			<P><BR>
			</P>
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		<TD WIDTH=20%>
			<P>1</P>
		</TD>
		<TD WIDTH=20%>
			<P>2</P>
		</TD>
		<TD WIDTH=20%>
			<P>3</P>
		</TD>
		<TD WIDTH=20%>
			<P>4</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>1</P>
		</TD>
		<TD WIDTH=20%>
			<P>5.65</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.87</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.78</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.16</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>2</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.75</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.31</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.50</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.59</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>3</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.89</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.72</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.72</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.81</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>4</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.78</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.03</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.82</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.75</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>5</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.63</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.15</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.72</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.80</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>6</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.56</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.22</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.25</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.31</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>7</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.81</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.18</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.75</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.56</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>8</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.68</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.28</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.83</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.91</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>9</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.89</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.40</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.93</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.16</P>
		</TD>
	</TR>
	<TR VALIGN=TOP>
		<TD WIDTH=20%>
			<P>10</P>
		</TD>
		<TD WIDTH=20%>
			<P>4.65</P>
		</TD>
		<TD WIDTH=20%>
			<P>3.22</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.68</P>
		</TD>
		<TD WIDTH=20%>
			<P>2.56</P>
		</TD>
	</TR>
</TABLE>









	
	<h3>Observations</h3>
	The coffee filters quickly crumpled up, especially when throwing them in a shoe onto the balcony.
<h2>Calculations</h2>
<h2>Conclusions</h2>
	The hypothesis was correct and <b>here's the exact value</b> that we calculated. 

<br /><br />
We also found that it would have been more effective if we started dropping four at a time, so that there would be no structural impairment to the filters and they would have dropped at the same rate. There's a calculation that we can do to show that the effects of the structure influenced the amount of time that it took to fall. Each time that we dropped a single coffee filter, out of the set of 40, there was a certain amount of "structural degradation" that we could label with the variable <b>delta C</b> and multiplied over each trial for each filter, by the near the ends of the rounds, the filters would be falling at a comparatively similar rate that would make it match the actual exponential curve that we came up with in the calculations section.
<h2>Discussion</h2>


<br /><br /><br />
<center><img src="2007-10-19_3.png"></center>
<h1>2007-10-29</h1>
<u>Today's nonsense</u><br />
- No Dakota.<br />
- It's a setup- they're all Twix.<br />
- Balto the movie?<br />
- Oil content of the cheez-it bag (Katie). Is it intimidating? <br />
- Spiller likes to mock me typing.<br />
- Coach Burnett came in yesterday and said that Spiller had messages on the phone because of the red light, and Burnett went through and explained it, and how come when you call in to the school and try to dial the extension it doesn't let you? Maybe to prevent interruption of the class. But what about on the weekends? <br />
- 
<br /><br />
There will be a take-home physics quiz because the grades on the test were pathetic. 

<br /><br />
<h1>Free-body diagrams</h1>
Be neat. Free-body diagram- a diagram that shows all the forces acting on a system. 
<br /><br />
ex 1) disguised, you enter a dogsled race. You're disguised and dressed up as a dog. You're catching on, you can do this. Excuse the excitement. You pull on a rope attached to the sled with a force of 150 N at 25 degrees w/ the horizontal. So you will have to realize that the larger the angle is with the horizontal, that you're wasting a lot of force, and the only force causing it to move is the horizontal force, and the vertical force is trying to lift the sled. The mass of the sled is 80 kg and there is negligble friction. A) Draw a free-body diagram for this. 
<center><img src="2007-10-29_1.png"></center>
<br />
B) Find the acceleration of the sled. The y-acceleration is going to be zero, unless you're traveling over a frozen lake that's not so frozen, or you've just gone over a hill. You can use the forces and the masses (as in the diagram) to solve for the horizontal acceleration. In this case we do not have to deal with vertical acceleration, so there will be no 'triangle for acceleration'. Find a_x. So we can use Tcos25 = m(a<sub>_x</sub>). So when you pull with a force of 150 N, the force is going to have that tension. <br />
<br />
a_x = 1.699 m/sec^2<br />
<br />
C) Find the normal force, F<sub>_n</sub>, exherted by the surface on the sled. The rope is pulling up on the sled as well- the rope is not horizontal. We said that the sum of the forces in the y-direction was zero for the rope. We have 150sin25 + F<sub>_n</sub> - 80(9.81) =0 and somebody will realize that "g" is going to be 9.81 and that's the acceleration due to gravity and we are not doing -9.81 because we are taking into account that the weight is going down, and that's why we said -80(9.81) in that line of equations from just a few moments ago. F<sub>_n</sub> = 721.407 N. Now, do you understand, well, actually, you might not understand, but I may as well ask the question, and make you think about it, did you figure out what you did wrong, okay, Brooke is still writing if you're still wondering, I shouldn't have to single out anybody, and we're still on the problem, D, so, it could be rounding up but that would be rounding but, so [pause].<br />
<br />
D) What is the greatest tension that can be applied to the rope without lifting the sled off the surface? Now if I was to pull with a force great enough, I could just pull it off the ground, so the question is going to be what is going to be the greatest tension that can be applied without lifting it off the ground, and the greater the tension in the rope, the more vertical force as well as. So we want the vertical to always be zero. <br /><br />

F<sub>_n</sub> > = 0 <br />
You want your normal force to be greater than or equal to zero. So let's look at the worse-case scenario. Take your sum of y-forces, solve for F_n, move your stuff to the other side, and you'll have F<sub>_n</sub> = mg - Tsin25 and then you plug in numbers, setting F<sub>_n</sub> to zero = (80)(9.81) - Tsin25. Then solve and get T = 1856.995 N. Anything greater than that and you'll be lifting the sled off the ground.

<br /><br /><br />
2) You unload a truck by sliding its cargo down, so it is going to have a ramp, down a ramp, with roller, assume the ramp is frictionless. The ramp is inclined at an angle theta. For a box of mass m find the acceleration of the box as it slides down the ramp and the normal force exerted by the ramp on the box. The normal force is perpendicular. So in this case the normal force is off to an angle to the typical surface that we associate with. 
<center><img src="2007-10-29_2.png"></center>
The normal force is straight in the y-direction, but the weight is at an angle in the problem (but universally, the weight is as it usually is). So from this we are going to get our horizontal and vertical components. So let's look at the sum of the forces now. Shall we? We shall. So, what about our, horizontal forces? To find F_x it's going to be SOHCAHTOA and it's going to be sine. So we're looking for the one that's opposite, we have a parallel line below our dotted line that parallels the ramp. So in this case the sum of F_x is going to be wsin(theta) and then any other horizontal forces. It's just like as if we had a vector in the fourth quadrant. So, what's the net force in the y direction? It's mass times the a_y acceleration, but is this thing accelerating in the y-direction? So F_y should be zero.<br /><br />
Since F_y = 0, we know the mass is m, we know g, we know theta, so what is the normal force? <br />
F<sub>_n</sub> = (mg)cos(theta) <br />
And what's the acceleration? a<sub>_x</sub> = gsin(theta) because the masses just cancel themselves out via division. 
<br />
<br />
Where would friction come into play? In the sum of F_x and so on we would have other stuff to do.

<br /><br />
<h1>2007-10-30</h1>
<u>Today's nonsense</u><br />
- Homework due today and tomorrow, and calculus homework.<br />
- RSS feed for the calculus/physics websites?<br />

<br /><br />

ex 3) A picture weighing 8 N is supported by two wires at angles of 60 degrees and 30 degrees to the ceiling. Find the tension in each wire.<br />
<center><img src="2007-10-30_2.png"></center>
Let's look at what's going on- you have to know alternate interior angles where if a line is cut by .... So now we have to do our balances of forces. We're still using right as positive and left as negative. So the 30 degree component will be positive and the other one will be negative. <br /><br />
Sum of F<sub>_x</sub> = T<sub>_2</sub>cos(30) - T<sub>_1</sub>cos(60) ---- this second part is negative because it's in the opposite direction of positive x direction.<br />
Sum of F<sub>_y</sub> = T<sub>_2</sub>sin(30) + T<sub>_1</sub>sin(60) - mg ... and so since the picture is not moving both of these summed forces are going to be nothing. <br />
So therefore the F=ma and a=0. Find the tension in each wire. You have two equations and two unknowns. Everything else you know, you know m, you know g, so we can take one equation and solve for, or we can try to do simultaneously, couldn't we just do where we put them next to each other and subtract? But it would be difficult. So let's take the first equation and solve for, what, T_1 or T_2? T_2 first.
<br /><br />

So what is T<sub>_1</sub>? T<sub>_1</sub> = ( T<sub>_2</sub>cos(30) ) / cos(60) ... so now we will substitute this in to the other equation and solve for T<sub>_2</sub>.<br />
T<sub>_2</sub>sin(30) + T<sub>_2</sub>cos(30)sin(60)/sin(60) = mg. <br />
But now what? Oh my. So we can plug in values right now. The only benefit to not plugging in values is getting you used to the degrees because in some problems you will be asked to do a totally general equation.

<br /><br /><br />
4) So we have a box. A 32 kg box hangs motionless from a vertical rope. It is then accelerated upward to a speed of 6 m/sec in 0.8 seconds. What was the tension in the rope?

<center><img src="2007-10-30_3.png"></center>
Sum of F<sub>_x</sub> = 0. <br />
Sum of F<sub>_y</sub> = T - mg = ma<sub>_y</sub><br />
What is the acceleration in the y direction? Change in velocity / change in time. So that's ( 6 m/sec ) / 0.8 sec = 7.5 m/sec^2. So this type of problem is one of those where, say, an AP test question free-response where it's part of the free-response, you could lose credit by not showing your work.<br />
<br />
T = ma<sub>_y</sub> + mg<br />
Remember, you are just looking at the forces, and that's the weight and the tension. So now you plug and chug. 553.6 N. 

<br /><br />

5) This one's pretty cool. Yeah. You could do this one easily. Next time you fly on a plane. All you need is a protactor with a string hanging down, and you just have to see as the airplane accelerates that the string will not hang straight down, so you can calculate the acceleration ...... As your plane speeds down the runway, you notice your suspended yo-yo's string makes a 22 degree angle with the vertical. 
<br />
A) What is the acceleration of the plane?<br />
B) If the mass of the yo-yo is 40 g, what is the tension in the string?<br />

<center><img src="2007-10-30_4.png"></center>
Sum of F<sub>_x</sub> = T<sub>_1</sub>sin(22) = ma<sub>_x</sub><br />
Sum of F<sub>_y</sub> = T<sub>_1</sub>cos(22) - mg = 0<br />
So now we can figure out T<sub>_1</sub> = mg / cos(22) = 0.423 N.<br />
a<sub>_x</sub> = T<sub>_1</sub>sin(22) / m = 3.963 m/sec^2.<br />
<br />

<br /><br /><br />
Nick pointed out this <a href="http://www.fearofphysics.com/Probs/mech037.html">cool physics-problem helper</a> relevant to this set of notes.<br />
<h1>2007-10-31</h1>
<u>Today's nonsense</u><br />
- Slang physics ("convo")<br />

<br /><br />
Bryan and Dakota went mountain climbing, they tied each other with ropes, and one slipped and fell, and so one person is hanging from the rope, and now you have to get the tension, it's "just another typical problem," and that's how you should look at it. It's the same problem all over again. Need to include free body diagrams. 

<br /><br />
Here's the story: this is example six. Bryan, tied by a long rope to Dakota, slips off an icy edge. Before setting his climbing ax, Dakota slides along the frictionless ice. What is the acceleration of each and the tension in the rope? Dakota will be m<sub>_1</sub> and a<sub>_1</sub>. <br />
<center><img src="2007-10-31.png"></center><br />
Bryan has no i and Dakota has no j. <br />
<br />

Sum of F<sub>_x</sub> = T<sub>_1</sub> = m<sub>_1</sub>a<sub>_1</sub><br />
Sum of F<sub>_y</sub> = T - m<sub>_2</sub>g = m<sub>_2</sub>a<sub>_2</sub><br />


<br />
The physics problems become more difficult when you try to rationalize what's going on, instead of just following straight the laws of forces and balancing the forces. Overthinking. Just realize. Well. What actually makes a difference? If both of you are attached, then both must accelerate at the same rate. Unless the rope was an elastic band. And then that's just like a hypothetical related rate with a ladder that's on fire etc.<br />
<br />
So: a<sub>_1</sub> = -a<sub>_-2</sub> = a (instead of a<sub>_1</sub> or a<sub>_2</sub>-- so Dakota has a positive a and Bryan has a negative a).<br />
The tensions: those tensions are the exact same. T<sub>_1</sub> = T<sub>_2</sub> = T because it's all the same rope.<br />
<br />
m<sub>_1</sub>a = m<sub>_2</sub><font color="blue">(-a)</font> + m<sub>_2</sub>g -- just solving for T in both cases. T<sub>_1</sub> = m<sub>_1</sub>a etc. <font color=red>(be careful)</font><br />
Our goal is to solve for "a." So you want to move the (-a) to the other side. And then factor out the a.<br />
m<sub>_1</sub>a + m<sub>_2</sub>a = m<sub>_2</sub>g<br />
a = m<sub>_2</sub>g / (m<sub>_1</sub> + m<sub>_2</sub>)<br />

<br /><br />
<br />

While working in space, you push on a box of mass m<sub>_1</sub> with force F. The box is in direct contact with a second box of mass m<sub>_2</sub> <h1>2007-11-01</h1>
<u>Today's nonsense</u><br />
- <font color="red">Dakota rule</font>- something about not having to bring cookies? Who came up with this nonsense?<br />
<br /><br />
Today we are doing two examples, and then we will get the homework that's due Monday, and tomorrow we may go over the quiz that was straight off of the review and the test and even then some of you failed it entirely so we're going to look at it.

<br /><br />
The problem is this: In this case we need two free-body diagrams. <br />
<center><img src="2007-11-01_2.png"></center><br />
The total mass of the system is m<sub>_1</sub> + m<sub>_2</sub>. The other part of this question is: what is the force of one block on the other. <br />
a) What is the acceleration of the boxes?<br />
F / (m<sub>_1</sub> + m<sub>_2</sub>) = a<br />

<br />
Let's look at the different free-body diagrams and see how this all works. So let's do the free-body diagram around m<sub>_1</sub>.<br />
Sum of the forces = F<sub>_hand</sub> - F<sub>_(the force of block 2 on block 1)</sub> = and that net force is going to equal = m<sub>_(1)</sub>a<sub>_(1)</sub>.
<br />
Sum of the forces on the second block = Sum of forces = F<sub>_(1,2)</sub> = m<sub>_(2)</sub>a<sub>_(2)</sub>.<br />
So do you realize that a<sub>_1</sub>=a<sub>_2</sub> therefore it all equals a?<br />

<br />
VectorF<sub>_(2,1)</sub> = -VectorF<sub>_(1,2)</sub><br />
F<sub>_(2,1)</sub> = F<sub>(1,2)</sub> (read this as "Force of 1 on 2")<br />

<br />
So what are we going to do with this information? So can we do substitution? F - the force between the blocks = m1a, but the force between the blocks = m2a, so can we say <br />
F - m<sub>_(2)</sub>a = m<sub>_(1)</sub>a<br />
So we can solve for "a" by moving over m<sub>_(2)</sub>a and so on and we get the same exact conclusion as F / (m<sub>_1</sub> + m<sub>_2</sub>) = a ... just like before.<br />
This just confirms the answer for part A.<br />

<br /><br />
b) What is the magnitude of the force exerted by one box on the other?<br />
F<sub>_(1,2)</sub> = m<sub>_(2)</sub>a = m<sub>_(2)</sub>F / ( m<sub>_(1)</sub> + m<sub>_(2)</sub>)

<br /><br />

#72 in the book or example #8. <br />

ex 8) #72 pg 110 ch. 4. They give you a diagram, so if you draw the diagram, ... <br />
<center><img src="2007-11-01_3.png"></center>
Find a and T. <br />
The acceleration's magnitude of block1 is going to be the same as block2. If one moves, the other will move. The tension in the rope at either locations will be the same. They can be called T_1 or T_2 but they're both the same thing. So the free body diagram, how do we begin? <br />
<center><img src="2007-11-01_4.png"></center><br />
Sum of F = net force = what causes acceleration = and this will only be on the x-axis = so we do not have to concern ourselves with the vertical = <br />
Sum of F = T - you have congruent triangles in the diagram, you flip it and turn it <br />
Sum of F = T<sub>_(1)</sub> - m<sub>_(1)</sub> * g * sin(theta) = m<sub>_(1)</sub> * a<sub>_(1)</sub> <br />
-- sine(theta) because it's sine-theta times the adjacent to tell us the hypotanuse


<br /><br />
For m<sub>_(2)</sub>? <br />
Sum of F = T<sub>_(2)</sub> - m<sub>_(2)</sub> * g = m<sub>_(2)</sub> * a<sub>_(2)</sub><br />

<br />


If T<sub>_(1)</sub> = T<sub>_(2)</sub> then accelerations are equal and opposite (a<sub>_(1)</sub> = -a<sub>_(2)</sub>)<br />
If you say a<sub>_(1)</sub> = a<sub>_(2)</sub> then T<sub>_(1)</sub> = -T<sub>_(2)</sub><br />
<br />

m<sub>_(1)</sub> * a<sub>_(1)</sub> + m<sub>_(1)</sub> * g * sin(theta) = m<sub>_(2)</sub>*g + m<sub>_(2)</sub> * a<sub>_(2)</sub><br />
m<sub>_(1)</sub> * a + m<sub>_(1)</sub> * g * sin(theta) = m<sub>_(2)</sub> * g - m<sub>_(2)</sub> * a<br />


<br /><br /><br />
Law of sines<br />
The sine of an angle divided by the side opposite to that angle, is equal to the side of another angle ... <br />
sin(A)/side A = sin(B) / sideB = sin(C) / sideC ... etc. <br /><br />
Law of Cosines = pythagorean theorem with some extras.<br />
<h1>2007-11-02</h1>
<u>Today's nonsense</u><br />

<br /><br />

Reviewing the quiz.<br />
1. What force is needed to give an acceleration of 13.8 ft/sec^2 to a 23 lb block? What is that 23 lb block? The 23 lb represents weight. Weight is mass * acceleration due to gravity. So we know that mass is 23/32.2 and that's in slugs. Force is mass * acceleration. We have our acceleration, so multiply the mass by the acceleration and move on to the next problem.

<br /><br />

2. A single force of 62 N acts on a particle of mass m. The particle starts from rest and travels in a straight line in a distance of 45 m in 5 seconds. Find m. So S = (1/2)aT^2 + V_o * t + S_o ... and 45 = (1/2)a(5)^2 and a=3.6 ... so 62 = m(3.6) and m=62/3.6 ... 

<br /><br />

3. A certain force F_1 causes an acceleration of 6 m/sec^2 when it acts on an object of mass m sliding on a frictionless surface. Find the acceleration of the same object if the two equal forces of F_1 act at 110 degrees on the same object. So F_A = F_1 * i + 0 * j. So then <br />
F_B = -0.342 F i + .940 F_1 * j. <br /> 
Be aware of your frame of reference. Now add these two force vectors together to get = 0.658 * F_1 * i + 0.940 * F_1 * j. Take those two numbers, square them both, add them, take the square root, and the magnitude ||F|| = 1.417 * F_1. Adding vectors. That shouldn't have been difficult. What does F_1 do? F_1 takes some mass and moves it with an acceleration of six, and now they want me to tell what this new acceleration is going to be, so it's mass times acceleration, 1.147 F_1 = ( F_1 / 6 ) * a ... therefore a = 6(1.147).


<br /><br />

4. An 8 kg object is subjected to two forces F_1 = -9i + 11j Newtons and F_2 = 11i + 27j Newtons. The object's initial velocity is v = 3i - 8j m/sec and is located at (4, -5) at time t = 0 seconds.<br />
A) What is the magnitude and direction of the object's acceleration?<br />
B) What is the object's velocity vector at time T=4 seconds?<br />
C) What is the object's position vector at time T=6 seconds?<br />

<br />
VectorF = 2i + 38j.<br />
VectorA = F/m = (1/4)i + (38/8 ==> 19/4)j<br />
Magnitude and direction? Do (1/4)^2 + (19/4)^2 and take the square root, and you get a = 4.757 m/sec^2, and then this is the first quadrant because it's a positive i and a positive j, so all you had to do was take the arctan, and the arctan of 19 ... and the arctan of 19 came out to be 86.987 degrees. 

<br /><br />
What's the velocity vector? You integrate acceleration to get velocity. What's the integral of 1/4 ? It's (1/4)T + constant ... so VectorV = ((1/4)T + C_1)i + ((19/4)T + C_2)j. What was the initial velocity for this problem? So you can rewrite the velocity vector as ((1/4)T + 3)i + ((19/4)T - 8)j. <br />
<br />
Position- you integrate your velocity vector and then plug in the given position values that the problem started out with. 

<br /><br />

5) You pull a sled, using a rope, along a frictionless ice surface with a force of 230 Newtons at an angle of 35 degrees to the horizontal. The mass of the sled is 118 kg.

<br /><br /><br />
<h2>Working on the homework for Monday</h2><h1>2007-11-05 </h1>
<u>Today's nonsense</u><br />
- Meg's baby pool<br />
- Test review today. Homework due.<br />
<br /><br /><br />

<h1>2007-11-14 - Friction</h1>
<u>Today's nonsense</u><br />
- We need to pay for a few tickets to send Spiller up to Florida over Thanksgiving.<br />
- Homework due tomorrow.<br />
- "I have pretty bad aim. I can't guarantee I'll miss." - Katie ... kind of like "my battery is out of cameras"<br />

<br /><br />

<h2>Friction</h2>
When you apply a horizontal force to an object resting on the floor. It either moves or does not move. What if it doesn't move? Then the friction is greater. There are two types of friction: static and kinetic (friction of movement). The object may not move because the force of <u>static friction</u>, Vector f<sub>_(s)</sub>, exerted by the floor on the object, opposes the applied force. Until you apply enough force to overcome the statitic friction, there will be no movement. Once you get the object moving, so if you have no net acceleration and no force, the .... <br /><br />

f<sub>_(s,max)</sub> = μ<sub>_(s)</sub> * F<sub>_(n)</sub>. <br /><br />
f<sub>_(s)</sub> < = μ<sub>_(s)</sub> * F<sub>_(n)</sub>.<br />
Where Mu is called the coefficient of static friction. 

<br /><br /> Say it takes 80 Newtons to move a box from a floor, if you apply 60 Newtons, then the static friction is one value, then 40 Newtons, well, the static friction will be that value. 

<br /><br />
If you push the object hard enough, it will slide across the floor. To keep up the object sliding with constant velocity, you must exert a force on the object that is equal in magnitude and opposite in direction to the force of <u>kinetic friction</u>, Vector f<sub>_(K)</sub> (also called sliding friction), exerted by the floor<br /><br />

f<sub>_(k)</sub> = μ<sub>_(k)</sub> * F<sub>_(n)</sub><br />

<br />
where μ<sub>_(k)</sub> is called the coefficient of kinetic friction. 

<br /><br />
<center><img src="2007-11-14.png"></center>

<br /><br />
ex 1) A bartender slides a mug of mass 0.45 kg horizontally along the bar with initial speed 3.5 m/s The mug comes to rest near the customer fter sliding 2.8 m. Find the coefficient of kinetic friction. <br />
<br />

HWDWSTP? FBD.
<br /><br />
<center><img src="2007-11-14_2.png"></center>

<br />

Calculate a<sub>_(x)</sub><br />
V<sup>^(2)</sup> = V<sub>_(0)</sub><sup>^(2)</sup> + 2a<sub>_(x)</sub>(delta x)<br />
0 = (3.5)<sup>^(2)</sup> + 2a<sub>_(x)</sub> * (2.8)<br />
a<sub>_(x)</sub> = -2.1875 m/sec^2. <br /><br />

μ<sub>_(k)</sub> = (-a<sub>_(x)</sub>) / g = 2.1875/9.81 and it's unitless, it's just a coefficient.

<br /><br /><br />

2) A block rests on an iclined plane surface. The angle of inclination is increased until it reaches a critical angle, θ<sub>_(c)</sub>, after which the block begins to slide. Find the coefficient of static friction, μ<sub>_(s)</sub>. We are going to put this on a tilted frame of reference, because we want our normal force of friction to be straight up. 

<br />
<center><img src="2007-11-14_3.png"></center>
<br /><br />

Sum of F_y = F_n - mgcos(theta) = ma_y = 0<br />
Sum of F_x = mgsin(theta) - f_s = ma_x = 0 (because the static friction is keeping it in place). <br />
What's the largest that theta can be such that the acceleration is zero, and it's not moving, so do you know how to start that problem? You start with the soon to be infamous f_s, soon to be infamous because we haven't used it enough to, .... then we did just like we did on the last example. The static friction is equivalent to what, based on what this says? mgsin(theta) = Normal force * coefficient of static friction. <br />

<br />
Coefficient of static friction = tan(theta)


<br /><br /><br /><br />
μμμμμμμμμμμμμμμμμμμμ = Mu

<h1>2007-11-15</h1>
<u>Today's nonsense</u><br />
- Lil' Jibble the Hip-Hop Funky-Fresh Integration Turtle<br />

<br /><br /><br />

ex 3) Two children are pulled on a sled over snow. The sled, initially at rest, is pulled by a rope that makes a 40 degree angle with the horizontal. The children's combined mass is 45 kg and the sled's mass is 5 kg Mu_s = 0.2 and Mu_k = 0.15.Find the frictional force exerted by the ground on the sled and the acceleration of the children and sled, starting from rest, if the tension in the rope is (a) 100 N and (b) 140 N. 

<br /><br />

Is the frictional force static or kinetic? <br />
<center><img src="2007-11-15.png"></center>

<br /><br />
<h1>2007-11-16</h1>
<u>Today's nonsense</u><br />
- Smarties<br />
- Physics homework due Tuesday. pg. 139 #14-16, 18-20, 25, 26, 29, 34, 35, 38<br />
- 3564. "I don't think you could have possibly put this in a worse order." - Nick.
<br /><br /><br />

Examples.<h1>2007-11-19</h1>
<u>Today's nonsense</u><br />
- Evacuation tomorrow. Homework tomorrow.<br />

<br /><br /><br />

See the handout from Friday that we started on. "I wouldn't want any electrical currents flowing around my body while I sleep." - Dakota

<br /><br />

<h1>2007-11-26</h1>
<u>Today's nonsense</u><br />
- Seniors have "passes" to come in late on Wednesday.<br />

<br /><br /><br />

Unofficially starting a review for the test. Test is Thursday/Friday.<br />

<br /><br />
Homework #14 - Rock band, Dead Weight, the band starts off with the sound of an automobile accident, and the lead singer slides out from the back with some extreme guitar moves that go BAM BAM BAM. So, lead singer, Sharika, goes to the floor and slides on her knees, with an initial speed of 3 m/s (V_o) and after sliding for d = 2 m, and she rests in between flashpots. What's the coefficient of kinetic friction between Sharika and the stage?<br /><br />
Free body diagram. Maybe. Absolutely. <br />
<center><img src="2007-11-26.png"></center><br />
F_n - mg = 0 <br />
f_k = mu sub k * F_n =mu sub k * mg
<br />
<br />
Sum of F_y = F_n - mg = 0 <br />
Sum of F_x = -f_k = ma<br />

<br />
Your f_k = -ma, so <br />
-ma = mu_k * mg <br />
-a = mu_k * g<br />
mu_k = -a / g<br />
Now, the acceleration, they had an initial velocity and final velocity, so we can do our final velocity squared = initial velocity squared + 2ad<br />
Final velocity = 0, so:<br />
0 = 3^2 + 2a(2)<br />
So that's -9/4 = a<br />
So: mu_k = -(-9/4)/9.81 = 0.229<br />

<br /><br /><br />

#15) We have a 5 kg block that is held against the wall with a horizontal force of 100 Newtons. <br />
<br />
<center><img src="2007-11-26_2.png"></center>
<br />
Sum of F_x = F_n - F = 0 (because it's holding it against the wall and the acceleration is therefore zero)<br />
Sum of F_y = f_s - mg = 0 <br />
<br />
Part A) Find the friction force, and that just equals mg so f_s = mg = 49.05 N<br />
Part B) What's the minimum horizontal force to prevent the block from falling if the coefficient of friction is 0.4?<br />
mu_s = 0.4<br />
f_s = mu_s * F_n<br />
F_n = f_s / mu_s = 49.05 / 0.4 = 122.625 N<br />

<br /><br /><br />

#16) We have an angle.. We did this problem as far as angle and the mu_s as relationship between mu_s and the angle. You can probably find this in your notes but we'll rederive it. <br />
<br />
On a snowy day with the temperature near the freezing point, the coefficient of static friction between the car's tires and the road is 0.8. And it says what's the maximum incline that the four-wheel drive vehicle can climb with zero acceleration? <br >
<br />
<center><img src="2007-11-26_3.png"></center><br />
Sum of F_x = f_s - mgsin(theta) = 0<br />
Sum of F_y = F_n - mgcos(theta) = 0<br />
We are trying to solve for theta. <br />
f_s = mu_s * F_n<br />
Say that f_s = mu_s * F_n = mu_s * mgcos(theta) = mgsin(theta)<br />
mu_s = mgsin(theta) / mgcos(theta) = sin(theta)/cos(theta)<br />
mu_s = tan(theta)<br />
theta = arctan(mu_s)<br />
theta = 4.574 degrees<br />

<br /><br /><br />

#18) A 3 kg resting on a horizontal shelf is connected to a 2 kg block by a light string.  What is the mimum coefficient of static friction (mu_s) so that the two objects remain at rest?<br />
<br />
mu_s = ? so that a = 0 <br />
<center><img src="2007-11-26_4.png"></center><br />

f_s = T = 2g = mu_s * F_n = mu_s * 3g so that mu_s = 2/3<br />
<br />
Suppose the actual mu_s is less than the one they actually gave us, so that means that we'd actually be moving and we'd have a coefficient of kinetic friction. Find the time for the 2kg mass to fall two meters. <br />
mu_k = 0.3<br />
So the whole issue with falling 2 meters, we don't care about that yet, we have to find the acceleration, and once you find the acceleration you can do the whole velocity squared = velocity squared initial + 2ad and then we'd do our sort of s = 1/2 at^2 + initial velocity * t plus initial position or something.<br /><br />
Sum of F_x = T - f_k = ma = 3a<br />
Sum of F_y = F_n - 3g = 0 because in the y-direction it is not moving<br />
<br />
Then on the other side:<br />
Sum of F_y = T - 2g = 2(-a)<br />
<br />
a_1 = -a_2 = a<br />
So solve for a. Basically you should realize that you have to solve for T. <br />
T = T <br />
3a + f_k = 2g - 2a<br />
3a + mu_k * F_n = 3a + mu_k * 3g = 2g - 2a<br />
5a + mu_k * 3g = 2g<br />
5a = 2g - 0.3(3g)<br />
a = 1.1g/5 = 2.1582 m/sec^2 <br />

How long is it going to take that block to fall 2 meters? <br />
S = (1/2)aT^2 + V_o * T + S_o<br />
2 m = (1/2)(2.1582)T^2 + 0T + 0 meters<br />
4 = 2.1582 * T^2<br />
4/2.1582 = T^2<br />
T = sqrt(4/2.1582) = 1.361 seconds<br />

<br /><br /><br />

19) <br />
<center><img src="2007-11-26_5.png"></center><br />

Sum of F_x = -f_k = ma<br />
Sum of F_y = F_n - mg = 0<br />
f_k = ma<br />
mu_k = -ma / mg ???<br />
f_k = mu_k * F_n<br />
-ma = mu_k * mg<br />
mu_k = -ma/mg<br />
mu_k = -a/g<br />
That answer is not good enough. You know your velocity, and you need to find your a in terms of v and d. And in this scenario V_f = 0<br />
0^2 = V_o squared + 2ad<br />
0 = V^2 + 2ad<br />
<h1>2007-12-03 - Circular Motion</h1>
<u>Today's nonsense</u><br />
- Homework due Wednesday. pg. 142 #44-53 (omit 46c)<br />
- Rehash of the "<b>Moving on</b>" <b>rule</b>: 10% has to get it in physics, and 10% doesn't have to get it in the calculus class.<br />
- <b>Item spotting rule</b>: If you're in the class when the new item is put up on the wall, then you can't call it out unless you call it out at the end of the period, after everybody had a chance to call it. Furthermore, if you have already been in the class for one period (not a "visit"), then you are disqualified from making a claim on the new item.<br />
- "There's no point in physics, but we treat it all as if there was."<br />
- The story of Sir Cumference, and he was making pies.<br />
- Learning: "you can't force it" <-- there's supposed to be something funny here.<br />
- <font color="red">WE DO NOT LEAVE ANYBODY BEHIND IF THEY ARE TRYING.</font> <br />
<br /><br /><br />

a = v<sup>^(2)</sup> / r<br />
v = 2 * pi * r / T in meters/second<br />
Angular velocity would be in degrees/second.<br />

T is the period , time required for one complete revolution.<br />
Distance traveled during one period is 2 * pi * r.<br />
<br />

<br />
<b>Centripetal acceleration</b> - acceleration of a particle moving with constant speed in a circle. It is directed radially inward toward the center of the circle.<br />
<b>Centripetal force</b> - a net force in the direction of the acceleration to produce it. It is always directed inward, toward the center of the circle of motion. The centripetal force may be due to a string, spring, or other contact force.<br />

<br /><br />

Circular motion. The test grades are making more and more sense. Well there are many things you can try, try eating some starbusts or drinking some water, there are some things to work. Wednesday. It's online. The very homework that was half the test, yes. What goes around, .... and Brooke said it. That's omit 46c, so, oh, kind of like a circle. It all fits together. So you have the notes and now the notes are very simple. Let's talk about this acceleration and where this acceleration is when you are swinging something, what, pardon? Because someone said "I was printing the examples" and so I am going to go over it, and we're going to go over angular velocity if you don't know angular velocity and all that, velocity squared over the radius is your acceleration, and notice that this is not a magical formula, and 2 * pi * r is your circumference, so your <b>period</b> is 2 pi * r / V, and the period is the time required to complete one revolution.<br /><br />

Centripetal acceleration, that's what's going on when you're talking about it, so a car going around a circular ramp, we could talk about tops, or grades that don't exceed the number 73, who knows? It's directed radially inward toward the center of the circle, the force and acceleration are directed inwards. Whatever direction of the force, that's the same direction as acceleration. At any given time, the velocity is going to be tangental, and the acceleration is going to be inwards. 

<br /><br />
<center><img src="2007-12-03.png"></center><br />

If vectors are perpendicular, they are <b>orthogonal</b>. The next calculus test is over AB material, like related rates. Ooh yes. And we are going to have some UT homeworks. It's back to back to back homework assignments. 
<br /><br />

Centripetal force is a <u>net</u> force. What can cause this? Angular velocity.<br /><br />

<center><img src="2007-12-03_2.png"></center>

<br /><br />

ex 1) A satellite moves at a constant speed in a circular orbit about the center of the earth and near the surface of the earth. If its acceleration is 9.81 m/s^2, find (a) its speed and (b) the time for one complete revolution.<br />
Radius of the earth = r<sub>_(e)</sub> = <b>6378</b> km. <br />
a = 9.81 = v<sup>^2</sup> / r <br />
v = sqrt(a * r)<br />
v = 7910 m/sec = 7.910 km/sec<br />

<br /><br />
(b) The time for one complete revolution. V = 2 * pi * r / T.<br />
T = 5066.265 seconds = 1.407 hr

<br /><br /><br />

ex 2) You swing a pail of water in a vertical circle of radius r. The speed of the pail is v<sub>_(t)</sub> at the top of the circle. (a) Find the force exerted on the water by the pail at the top of the circle. (b) Find the minimum value of v<sub>_(t)</sub> for the water to remain in the pail. (c) Find the force exerted by the pail on the water at the bottom of the circle, where the pail's speed is v<sub>_(b)</sub>. 
<br /><br />

<center><img src="2007-12-03_3.png"></center>
Tangental acceleration later?<br /><br />

<h1>2007-12-04</h1>
<u>Today's nonsense</u><br />
- We have homework due tomorrow.<br />
- Katie and Dakota and Alyssa are alergic to pine trees. Dakota and Katie had a speech impediment. No correlation. "You're the derivative of an alergic."<br />
- Please note that it is <b>K</b>andy with a K. So if you're Ms. Dees and you call and you ask for candy, know that you're going to be asking for the candy in the back of the room all in the boxes.<br />
<br /><br /><br />

ex 2) You swing a pail of water in a vertical circle of radius r. The speed of the pail is v<sub>_(t)</sub> at the top of the circle. (a) Find the force exerted on the water by the pail at the top of the circle. (b) Find the minimum value of v<sub>_(t)</sub> for the water to remain in the pail. (c) Find the force exerted by the pail on the water at the bottom of the circle, where the pail's speed is v<sub>_(b)</sub>. 
<br /><br />

<center><img src="2007-12-03_3.png"></center>
Tangental acceleration later?<br /><br />

Sum of F<sub>_(y)</sub>= -F - mg = ma<sub>_(y)</sub> = m (-1)v<sup>^(2)</sup> / r<br />
F = ma<sub>_(y)</sub> + mg<br />
F = m(-1)v<sup>^(2)</sup>/r - mg<br />
F = m(-v<sup>^(2)</sup> / r   -  g)<br />
a = (-v<sup>^(2)</sup> / r   -  g)<br />
<br /><br />

(b) Find the minimum value of v<sub>_(t)</sub> for the water to remain in the pail. What's the minimum velocity at which you can swing this thing so that the water stays in the bucket? Your minimum velocity is going to be when your force is as at an absolute minimum, and what's the smallest that a force can get? <br />
minimum where F=0<br />
(-v<sup>^(2)</sup> / r   -  g) = 0<br />
-v<sup>^(2)</sup> = 9.81r<br />
v = sqrt(9.81r)<br />
The smaller the radius, the smaller the velocity (because of the square root).<br />

<br /><br />
 (c) Find the force exerted by the pail on the water at the bottom of the circle, where the pail's speed is v<sub>_(b)</sub>. <br />
<center><img src="2007-12-04_1.png"></center>
Sum of F<sub>_(y)</sub> = F - mg = ma<sub>_(y)</sub> = mv<sup>^(2)</sup> / r ... and that's a positive v^2 because the acceleration is upwards, towards the center of the circle.<br />
We're looking for force. F = mg + mv<sup>^(2)</sup> / r <br />
Put it into the terms that you already know.

<br /><br /><br />

ex 3) A tether ball of mass m is suspended from a rope of length L and travels at constant speed v in a horizontal circle of radius r. The rope makes an angle theta given by sin(theta) = r/L, as shown in Figure 5-28. Find (a) the tension in the rope, and (b) the speed of the ball. <br /><br />

The angle of the rope to the supporting beam of the rope, hanging over the circle, is theta. The "horizontal circle" is a disk on a plane. The rope has a tension. Down is mg. <br />
Sum of F<sub>_(x)</sub> = Tsin(theta) = ma<sub>_(x)</sub> = mv<sup>^(2)</sup> / r <br />
Should that be negative Tsin(theta)? Either direction you can call negative. We can call "to the left" as negative, or we can think of the tether ball on the other side of the pole. If we call T negative, then the a<sub>_(x)</sub> would have to be negative.<br />
<br />

Sum of F<sub>_(y)</sub> = Tcos(theta) - mg = ma<sub>_(y)</sub> = 0<br />
<br />
Find the tension in the rope, solve for T. <br />
mgtan(theta) = (1/r)mv<sup>^(2)</sup><br />
V = sqrt(rgtan(theta)<br />

<br /><br /><br />

ex 6) A pilot of mass m = 56 kg comes out of a vertical dive in a circular arc of radius 420 m. Her speed at the bottom of the circle is 200 km/h. (a) What's the magnitude of he(r) acceleration? So draw a half circle, with at the bottom point we have our free-body diagram, and there's mg going down, and F going up, and so ... <br /><br />

Sum of F<sub>_(y)</sub> = F - mg = ma<sub>_(y)</sub> = mv<sup>^(2)</sup> / r<br />
a<sub>_(y)</sub> = v<sup>^(2)</sup> / r = (55.556 m/s^2) / 420 m = 7.349 m/sec^2<br />
<br />
(b) What's the net force acting on her at the bottom of the circle?<br />
F<sub>_(net)</sub> = ma = 7.349 * 56 kg = 411.523 N<br />
<br />

(c) What's the force exerted by the airplane seat on her?<br />
We're looking for the original F from the free-body diagram. Keep in mind that the sum of forces gives us the net force which equals F - mg. So we're looking for F. We have mg, we have F<sub>_(net)</sub>. <br />
F = 960.883 N = F - mg. 

<br /><br /><br />

We have an object on a string, and we're swinging it over head but the rope is at an angle, ... <br />
ex 7) <br />
m = 326 g<br />
d<sub>_(string)</sub> = 1.4 m<br />
theta = 35 deg with the horizontal<br />
v = ?<br />
<br />
a = v<sup>^(2)</sup>/r and v = 2 * pi * r / T.
<br /><br />
cos(theta) = r / d<br />
r = d cos(theta)<br />
r = 1.4cos(theta)<br />
<br />
v = 2 * pi * 1.4cos(theta) / T<br />
Sum of F<sub>_(y)</sub> = Tsin(theta) - mg =  0<br />
Sum of F<sub>_(x)</sub> = Tcos(theta) = ma<sub>_(x)</sub> = mv<sup>^(2)</sup> / r <br />
T = mg/sin(theta) = (1/r)(1/cos(theta))mv<sup>^(2)</sup><br />
v<sup>^(2)</sup> = g(1.4cos(theta))cos(theta)/sin(theta)<br />
v = sqrt (g(1.4cos(theta))cos(theta)/sin(theta))<br />


<br /><br /><br />

ex 5) A curve of radius 30 m is banked at an angle theta. Find theta for which a car can round the curve at 40 km/h even if the road is frictionless. <h1>2007-12-05 - More examples</h1>
<u>Today's nonsense</u><br />
- Homework due Friday. pg 146 #91-95,97-99,101,110<br />
- Classwide church-based support-the-needy gift plans<br />
<center><img src="lego_2007-12-05.jpg"></center>
<br /><br /><br />

ex 4) In a skid test, a recent-model BMW 530i was able to travel in a circle of radius 45.7 m in 15.2 sec without skidding. (a) What was its average speed? (b) Assuming v to be constant, what was the centripetal acceleration? (c) Again assuming v to be constant, what is the minimum value for the coefficient of static friction?<br /><br />

 a = v^(2) / r<br />
 v = 2 * pi * r / T<br />
2 * pi * r = distance<br />
(a)        v = 18.891 m/sec.<br />
<br />

(b)         a<sub>_(c)</sub> = v<sup>^(2)</sup> / r = 7.809 m/sec^2<br />
<br />

(c)          Do a freebody diagram.<br />
Static friction is the centripetal force here (f<sub>_(n)</sub>), and it moves inward here. Imagine that your car is out of control on a curve-turn, and so the friction is pointing inwards towards the circle, while you're still moving outwards from the circle. <br />
<center><img src="2007-12-05.png"></center>

Sum of F<sub>_(y)</sub> = F<sub>_(n)</sub> - mg = 0<br />
Sum of F<sub>_(radial direction)</sub> = -f<sub>_(s)</sub> = ma<sub>_(radial)</sub><br />

Coefficient of static friction<br />
mu<sub>_(s)</sub> * mg = mv<sup>^(2)</sup> / r<br />
mu<sub>_(s)</sub> = 0.796<br />

<br />


<br /><br /><br />

ex 5) A curve of radius 30 m is banked at an angle theta. Find theta for which a car can round the curve at 40 km/h even if the road is frictionless.
<br />
<br />

<h1>2007-12-06 - More examples</h1>
<u>Today's nonsense</u><br />
- Homework due tomorrow.<br />
- Averytoss contest
<br /><br /><br />

"If I get hurt, I will not hold anyone responsible." - Avery.
<br /><br />

<center><img src="2007-12-06/DA.png"></center>
Sum of F<sub>_(y)</sub> = Fcos(theta) - mg = 0<br />
Sum of F<sub>_(r)</sub> = Fsin(theta) = ma<sub>_(r)</sub> = mv<sup>^(2)</sup> / r<br />
F = mg/cos(theta) = mv<sup>^(2)</sup> / rsin(theta)<br />
tan(theta) = v<sup>^(2)</sup> / rg<br />

<br /><br />
T = 1.675 seconds per revolution<br />
If you want to find the force, you need to know the mass.<br />
<br />
Velocity, we said, is 2 pi * r / T. <br />
tan(theta) = (2pi * r)<sup>^(2)</sup> / rgT<sup>^(2)</sup><br />
r = gT<sup>^(2)</sup>tan(theta) / (4 pi<sup>^(2)</sup>)<br />
r = (9.81)(1.675)<sup>^(2)</sup> tan(42.967) / (4 pi<sup>^(2)</sup>)<br />
r = 0.637 m <br />
<h1>2007-12-10</h1>
<u>Today's nonsense</u><br />

<br /><br /><br />

Objects falling into gravity experience a drag force of bv<sup>^(n)</sup>. The value b depends on the shape, the density of the air, etc. Approximately 1 at low speeds and 2 at high speeds. 

<br /><br />

Using down as positive,<br />
Sum of F<sub>_(y)</sub> = mg - bV<sup>^(n)</sup> = ma<sub>_(y)</sub><br />
sum of forces for free falling objects. That's weight, and then holding it back a bit is the drag force. So, when you reach terminal velocity, .... <br />

<br />
When terminal velocity, V<sub>_(t)</sub>, is reached, you have a<sub>_(y)</sub> = 0 and: <br />
mg = bv<sub>_(t)</sub><sup>^(n)</sup><br />
or: V<sub>_(t)</sub> = ( mg / b)<sup>^(1/n)</sup>

<br /><br />

The <u>larger</u> the b, the <u>lower</u> the terminal speed.<br />
A parachute is designed to have a <u>maximize</u> b. A submarine (as well as a car, or a boat) is designed to minimize b. 

<br /><br />

73) 

<br /><br />

r = 150 m<br />
theta = 10 deg<br />
m = 800 kg<br />
V = 85 km/h w/o skidding = 23.611 m/sec<br />
<br />
<br />

Find: (a) normal force, (b) frictional force, and (c) mu<sub>_(s)</sub>.
<br /><br />

Sum of F_y = F_n * cos(theta) - mg - f_s * sin(theta)<br />
Sum of F_x = f_s * cos(theta) + F_n * sin(theta) = ma<sub>_(r)</sub> = -mv<sup>^(2)</sup> * (1/r)<br />

<br />
F<sub>_(n)</sub> = (mg + f_s * sin(theta)) * sec(theta)<br />
f<sub>_(s)</sub> = (F<sub>_(n)</sub>cos(theta) - mg ) / sin(theta)<br />
f<sub>_(s)</sub> = -F<sub>_(n)</sub> * sin(theta)  * sec(theta)<br />
F<sub>_(n)</sub> = ( ( mv<sup>^(2)</sup> / rcos(theta) ) + ( mg / sin(theta) ) ) / ( cot(theta) + tan(theta) )<br />
F<sub>_(n)</sub> = ( ( mv<sup>^(2)</sup> / rcos(theta) ) + ( mg / sin(theta) ) ) / ( cot(theta) + tan(theta) ) either multiply by (sin * cos) / (sin * cos) or just plug in numbers now<br />
But make sure your values are in the right form. The radius is in meters, but the velocity is in km/h, so we need velocity in meters per second.<br />

<br /><br /><h1>2008-01-08</h1>
<u>Today's nonsense</u><br />
- No Meg?<br />
- Thursday: we get to throw some calculus into W=f*d, and what about when you have a varying force? And that's where you will throw in integration.<br />
- Dot product, cross production in cal 2<br />
- Rube Goldberg machine<br />
- Find the car commercial (unAmerican) <br />


<br /><br />
Chapter 6. In physics, work is done when a force acts on an object that moves through a distance, and there is a component of the force along the line of motion. So, if your force is perpendicular to your line of motion, then there's no work occuring there. So, if you're holding the sky up so it doesn't fall, are you doing any work? The force that you are applying is counteracting the gravitational force, so the net force is zero and there's no work.

<br /><br />

<b>Energy</b> is transferred between two systems when work is done by one system on another. Energy is Newton meters, which are known as <b>joules</b>. 

<br /><br />

Work: motion in one dimension with constant force.<br />
W = Fcos(theta) * (delta x) = F<sub>_(x)</sub> * (delta x)<br />
That's, of course, in the direction called x. Your F<sub>_(x)</sub> is just a little trig. This is kind of something you should "common sense it". 

<br /><br />
When there are several forces doing work, <br />
W<sub>_(total)</sub> = F<sub>_(net,x)</sub> * (delta x) but if you're not worrying about this then you generically say W = F * d <br />
<br />
<b>Work is scalar</b> and positive or negative. The SI unit for work and energy are the same unit -- the <b>joule</b> (symbol: J). <br />
1 J = 1 Newton * meter = 1 N * m = 1 J<br />
J/m = N<br />
J/N = m<br />

<br />
electron volt, <b>eV</b>, is a common unit in atomic/nuclear physics. <br />
1 eV = 1.6 * 10<sup>^(-19)</sup> J<br />
Other common units and prefixes that you know of are:<br />
keV = 1000 eV<br />
MeV = 1000000 eV<br />

<br /><br />
ex 1) A force of 12 N is exerted on a box at an angle of 20 degrees with the table top. How much work is done by the force as the box moves along the table a distance of 3 m? <br />
F<sub>_(x)</sub> = <br />
F<sub>_(y)</sub> = <br />
W = F * cos(20 deg) * (change in x = 3 m)  = 33.82
<br /><br />
<br />
<h1>2008-01-09 - Kinetic energy</h1>
<u>Today's nonsense</u><br />
- Meg's back.<br />

<br /><br /><br />

F<sub>_(x)</sub> = ma<sub>_(x)</sub><br />
W<sub>_(total)</sub> = F<sub>_(x)</sub> * (delta x) = ma<sub>_(x)</sub> * (delta x)<br />
V<sub>_(f)</sub><sup>^(2)</sup> = V<sub>_(o)</sub><sup>^(2)</sup> + 2 * a * (delta x)<br />
W = (1/2) * m * (V<sub>_(f)</sub><sup>^(2)</sup> - V<sub>_(o)</sub><sup>^(2)</sup>) in other words, (1/2) * m * the difference between the two squares.<br />
Distribute. W = (1/2)mV<sub>_(f)</sub><sup>^(2)</sup> - (1/2)mV<sub>_(o)</sub><sup>^(2)</sup><br />
<br /><br />

Kinetic energy = K<br />
Potential energy = P<br />
K = (1/2) m * v<sup>^(2)</sup><br />
P = mgh<br />

<br /><br />
Notice that the equation for K looks like W somewhat. So, <br />
W = K<sub>_(f)</sub> - K<sub>_(o or initial)</sub> = (change in Kinetic energy or just "delta K")<br />
<br /><br />

1. inertia<br />
2. <br />
3. equal and opposite reactions<br />

<br /><br /><br />

<h2>The Work-Kinetic Energy Theorem</h2>
The total work done on a particle is equal to the <u>change</u> in its kinetic energy: <br />
W<sub>_(total)</sub> = delta K = (1/2)mV<sub>_(f)</sub><sup>^(2)</sup> - (1/2)mV<sub>_(o)</sub><sup>^(2)</sup><br />

<br /><br /><br />

ex 2) A 50 kg girl is running at 3.5 m/sec. What is her kinetic energy? 306.25 J.<br />
K = (1/2)mV<sup>^(2)</sup>
<br /><br />

Applied force is in the direction of motion, we're going to say work is positive. The gravitational force is opposite, so in the case of gravity <u>work is negative</u>. But otherwise work is positive.
<br /><br />

<h1>2008-01-10 - Examples</h1>
<u>Today's nonsense</u><br />
- No Meg. Okay, nevermind.<br />
- The magic of the grapefruit demonstrated again. Grapefruit is magical because you do better on your work and homework with grapefruit. Go home and try it. <br />
<br /><br />

ex 3) Your professor enters a dog in a sled race. m = 80 kg. The force is at 90 degrees.<br />
<center><img src="2008-01-10.png"></center>
m = 80 kg, F = 180 N. delta x = 5 m<br />
a) W<sub>_(x)</sub> = F<sub>_(x)</sub> * (delta x)<br />
W = 845.723 J<br />
<br /><br /><br />

See the handout from yesterday.

<br /><br /><br />

<h1>Work done by a variable force</h1>
For a constant force with variable displacement (delta x = x<sub>_(2)</sub> - x<sub>_(1)</sub>). Suppose we have a graph of Force in the x direction versus x. Well, if it's a constant force, so whatever that force function is -- it's a horizontal line -- from x1 to x2, then you can ask the area under the curve. So ... the area under the curve of force is work, because work is F of x times delta x.<br />
<br />

The area under the curve of Force represents work.<br />
<br />
<br />

For variable forces:<br />
W = integral from x2 to x1 of F<sub>_(x)</sub> dx<br />

<br /><br />

<h1>Work and energy in three dimensions</h1>
<h2>Work-kinetic Energy Theorem in 3D</h2><br />
<br />
W<sub>_(total)</sub> = Integral from s2 to s1 of F<sub>_(s)</sub> * ds = (1/2) mV<sub>_(2)</sub><sup>^(2)</sup> - (1/2)mV<sub>_(1)</sub><sup>^(1)</sup><br />
<h1>2008-01-14</h1>
<u>Today's nonsense</u><br />
- Video on last Friday.
<br /><br /><br />
Vector dot product. A X B = cross product. A dot B = dot product.<br />

<br /><br />
There are three unit vectors that you should be familiar with: i, j, k. And if you're Professor Lewin you can pronounce it as "i roof". A <b>unit vector</b> is a vector with magnitude (of 1) and direction. This is because you are going to be using a unit of measure. <br />
<br />

ex) Find a unit vector parallel to Vector R = 3i - 4j<br />
What we are looking for is the exact same slope. The magnitude of VectoR is |VectorR| and is 5. The vector that we are looking for is going to be (1/5) times what we already have so that the magnitude of R is converted to one.
<br /><br />
<br />

The unit vector parallel to VectorA is the reciprocal of the magnitude of VectorA times VectorA or in other words VectorA * (1/(||VectorA||)). <br />

<br /><br />

So what about this dot product? How do we calculate a dot product? 

<br /><br />

<h2>The Dot Product/ The Scalar Product</h2>
Given VectorA = a<sub>_(1)</sub>i + a<sub>_(2)</sub>j + a<sub>_(3)</sub>k and VectorB = b<sub>_(1)</sub>i + b<sub>_(2)</sub>j + b<sub>_(3)</sub>k,<br />
the dot product of vectors A and B, written as VectorA dot VectorB, is given by:<br />
VectorA dot VectorB = a<sub>_(1)</sub> * b<sub>_(1)</sub> + a<sub>_(2)</sub> * b<sub>_(2)</sub> + a<sub>_(3)</sub> * b<sub>_(3)</sub><br />
and is defined by: <br />
VectorA dot Vector B = ||VectorA|| ||Vector B|| cos(theta)<br />
where theta is the angle between the two vectors.<br />
<br /><br />

<h2>Properties of the Dot Product</h2>
#1.) If VectorA and VectorB are perpendicular, then VectorA dot VectorB = 0 (because the angle between them if they are perpendicular is 90 degrees, and the cosine of 90 degrees is zero, so ...)<br />
#2.) If VectorA and VectorB are paralell, then VectorA dot VectorB = ||VectorA|| ||VectorB||<br />
#3.) If VectorA dot VectorB = 0, then VectorA=0 or VectorB=0 or VectorA and VectorB are perpendicular.<br />
The only way that a vector can be zero is if all of the components are zero. There's no way that you can add up to zero because of the Pythagorean theorem (you can't square to zero). <br />
#4.) VectorA dot VectorA = (VectorA)<sup>^(2)</sup><br />
#5.) VectorA dot VectorB = VectorB dot VectorA (commutative)<br />

<br /><br />

The component of VectorA in the direction of VectorB or the component of VectorA along VectorB. <br />
Go back to "the unit vector parallel to VectorA is the reciprocal of the magnitude of VectorA all times VectorA."<br />
It's just the dot product divided by the magnitude of the vector (that you want to project on to).<br />
VectorA dot VectorB all over ||the vector that you want to project over||<br />

<br /><br />

<h1>2008-01-15</h1>
<u>Today's nonsense</u><br />
- Meg is back. Again.<br />
- Homework due tomorrow.<br />
- Med schools that don't need their incoming students to take the MCAT? Quality of incoming medical students is not so high.<br />
- Mona Lisa may have been da Vinci's mother.<br />
- Converting to fluorescent light bulbs.<br />
<br /><br /><br />

ex 1) a) Find the angle between the vectors <br />
VectorA = (3 m)i + (2 m)j <br />
VectorB = (4 m)i - (3 m)j<br />
<br />

VectorA dot VectorB = ||VectorA|| ||VectorB|| cos (theta)<br />
cos theta = ( VectorA dot VectorB ) / ( ||VectorA|| ||VectorB|| )<br />
12 m^2 - 6 m^2<br />
so cos theta = 6 m^2 all over ( sqrt(13) * 5 ) <br />
arccos ( 6 / (5 sqrt(13) ) ) = theta<br />
theta = 70.560 degrees<br />

b) Find the component of VectorA in the direction of VectorB.<br />
<br />

The component of VectorA in the direction of VectorB or the component of VectorA along VectorB. <br />
Go back to "the unit vector parallel to VectorA is the reciprocal of the magnitude of VectorA all times VectorA."<br />
It's just the dot product divided by the magnitude of the vector (that you want to project on to).<br />
VectorA dot VectorB all over ||the vector that you want to project over||<br />

<br /><br />

6 / 5 is the component of VecotrA in the direction of VectorB.<br />

<br /><br /><br />

ext) a) Find VectorA dot VectorB for VectorA = (3m)i + (4m)j and VectorB = (2m)i + (8m)j<br />
b) Find A, B and the angle between them.<br />
<br />

A dot B = 38 m^2<br />
b) The dot product is not zero, so therefore they are not parallel vectors. A = || VectorA || = 5 m. <br />
B = sqrt(68) <br />
cos theta = 38 / (5 sqrt(68) )<br />
theta = 22.854 degrees<br />

<br /><br /><br />

ex 2) A particle has a displacement of (change in S) = 2m i - 5m j <br />
constant force VectorF = 3N i + 4 N j<br />
Find a) work done by force.<br />
Take the integral. W = integral of VectorF dot dS. ... so what do we get when we take the vector product of those two? <br />
VectorF dot dS = -14 J<br />
So the integral is going to be VectorF * (delta S)<br />

b) component of the force in direction of the displacement <br />
VectorF dot (delta S) all over the magnitude of delta VectorS.  == -14 / (sqrt(29)) = - 2.600 Newtons. <br />

<h1>2008-01-16</h1>
<u>Today's nonsense</u><br />
- Homework due today.<br />
- Spiller wants to write a book, "The Tao of Physics"<br />
<br /><br /><br />

The longer the ramp, the smaller the force needed to push an object up that ramp.

<br /><br />

Why wouldn't you use an infinitely long ramp? Because there's time. And that's leading up into <b>power</b>.

<br /><br />
Work = mg sin (theta) * distance. Work does not bring in new forces, it's all what we've been doing with the net forces.

<br /><br />

<h1>2008-01-17</h1>
<u>Today's nonsense</u><br />
- Homework due Tuesday. # 31-39. 41,42, 44, 45, 46<br />
- AirMac.<br />
- Measles/mumps vaccination may cause autism? Double check this.<br />
<br /><br /><br />

<h2>Power</h2>

Power is the reate at which a force does work. <br />
<br />

Consider a particle moving with velocity VectorV in a short time interval, dt, its displacement is d(Vector s) which equals (VelocityV) dt<br />
The work done by a force VectorF on the particle in this time is <br />
dW = VectorF dot d(Vector s) = (VectorF) dot (Vector v)dt<br />

<br /><br />

Power is then P = dW/dT = VectorF dot VectorV with SI units of J/sec and then Joules per second is known as a watt.<br />
Power is work per time. <br />

<br /><br /><br />

In the U.S. customary system, the unit of energy is the foot-pound. <small>Joule is Newton-meter in the metric system</small>. The unit of power is a foot-pound / sec. And then there's horsepower (hp).<br />

<br />

1 hp = 550 ft lb/s = 746 W<br />
<br />
<br />

For a net force F<sub>_(x)</sub> acting on a particle in one dimension. <br />
P = F<sub>_(x)</sub>V<sub>_(x)</sub> and <br />
F<sub>_(x)</sub> = ma<sub>_(x)</sub><br />
So, P<sub>_(x)</sub> = ma<sub>_(x)</sub> V<sub>_(x)</sub> or<br />
a<sub>_(x)</sub> = P / (mV<sub>_(x)</sub>)<br />

<br /><br /><br />

How does the acceleration vary with respect to the speed? Inversely. <br />
If the power is constant, then the faster you go, the slower (the less quickly) you accelerate. <br />
<br />

Alternatively, it takes more power to give the same acceleration to a car moving 60 mph than than one moving 30 mph.<br />

Acceleration and power are directly related. Acceleration and velocity are inversely related. <br />
<br />

<br />

a<sub>_(x)</sub> = V<sub>_(x)</sub> / dt <br />
So, P = ma<sub>_(x)</sub>V<sub>_(x)</sub> = mV<sub>_(x)</sub> (dV<sub>_(x)</sub> / dT) = which is the derivative of (1/2)mV<sup>^(2)</sup> which is the equation for kinetic energy (or this is known as dK/dT)<br />
<br />

Kinetic energy = the integral of power with respect to time. <br /><br />

If power is constant, then instead of dk=dP/dt, if we're talking about a constant power, we don't need a differential, and it's just going to be P(delta T) = (delta K).<br />
<h1>2008-01-18</h1>
<u>Today's nonsense</u><br />
- Train accident last night?<br />
- 
<br /><br /><br />
A small motor is used to operate a lift that raises a load of bricks weighing 800 N to a height of 10 m in 20 s. What is the minimum power the motor must produce? <br /><br />
 
P = maV = <br />
P = 800 (1/2)<br />
400 W<br />


A new Cadillac can accelerate from 0 to 96 km/h in 6.5 sec. How quickly would you expect it to be able to accelerate from 80 km/h to 112 km/h?<br /><br />

Keeping power a constant. <br />
P (delta T) = delta K<br />
P = (delta K) / (delta T)<br />
P = (1/2)m(96)^2 - (1/2)m0^2  all over (6.5)<br />
P = 4608/6.5 = 708.923<br />
708.923 = ((1/2)m(112)^2 - (1/2)m(80)^2 ) / x
<br /><br /><br />


A car accelerates from 0 to 40 km/h in T seconds. If the power output of the car is constant, how long does it take for the car to accelerate from 40 km/h to 80 km/h?<br /><br />

(1/2)(40)^2 / X = ( (1/2)(80)^2 - (1/2)(40)^2 ) / Y<br />
T = (1/3)T<sub>_(2)</sub><br />
T<sub>_(2)</sub> = 3T<br />


A truck of mass m is accelerated from rest at T=0 with constant power P along a level road. (a) Find the speed of the truck as a function of time. (b) Show that if x=0 at time T=0, the position function x(T) is given by:<br /><br />

P = (delta K) / (delta T) = (1/2)mv^2 / T<br />
V = sqrt(2PT / m )<br />
<br />
v = dxt<br />
dx = vdt<br />
dx = sqrt(2P / m ) * T<sup>^(1/2)</sup>dt<br />
(2/3)(2P/m)<sup>^(-1/2)</sup> * something
The purpose of the demo is to compare the theoretical (5/2 the radius of the loop, include a derivation) release height with the actual release height, where we use trial and error to experimentally determine an optimal height to complete a loop-de-loop. The main reason for the difference in holding it higher than the theoretical is due to friction, the construction of the car, the coefficients of friction, and so on. The free-body diagram shows kinetic friction, opposing complimentary forces, and gravity. The angle of the track determines the final velocity of the Hot Wheel car, though given a frictionless surface you would want to have a steeper angle because the steeper it is the less the normal force which is associated with the mu in the coefficient of friction.

<br /><br />

Record the radius of the loop. And determine the height from which you released it (try the 5/2 * radius height first). And calculate the angle.

<br /><br />

<h1>The Physics of Hot Wheels</h1>

<h1>2008-01-24</h1>
<u>Today's nonsense</u><br />
- "We should start a Grapefruit Club. Whoever has a grapefruit today doesn't have to take the calculus test. How about that?" - Alyssa.<br />
- Shooting scare yesterday. <br />
- The Elm Grove Math/Sci is tonight. Be there at 5:45, done at 7. No physics homework tonight. Assignment due Monday.<br />
- Church Hill High School. Saturday, Febuary 9th. Math, calculator and numbersense people. Magic Time Machine (hasn't passed its health codes? Same as Salt-lik.). Just eat the cooked food, never the raw stuff, like lettuce.<br />
- Read Ch. 6 in the physics book. Read pages 162-168. You have conservative forces and nonconservative forces, like Professor Lewin's briefcase example about no work being done by picking up and setting down a briefcase all day.<br />
- The class has decided to move the test Friday.<br />

<br /><br /><br />

<b>Conservative force</b> - a force is conservative if the total work it does on a particle is zero when the particle moves around any closed path (you end up back where you started) returning to its initial position. The work done by a conservative force on a particle is independent of the path taken and the particle moves from one point to another.

<br /><br />

<b>Potential energy</b> is mgh.<br />
W = integral of VectorF dot d(vector)s = - (delta u)<br />
or delta u = u<sub>_(2)</sub> - u<sub>_(1)</sub> = - integral from s<sub>_(1)</sub> to s<sub>_(2)</sub> of VectorF dot d(vector)s = - (delta u)<br />
<br />

<br /><br />

Gravitational potential energy<br />
u = integral of mg dy = mgy + u<sub>_(o)</sub><br />
u = u<sub>_(o)</sub> + mgy<br />

<br /><br /><br />

<h1>2008-01-28</h1>
<u>Today's nonsense</u><br />
- Brandon won first place in the pig weight competition.
<br /><br /><br />
<h1>2008-01-30</h1>
<u>Today's nonsense</u><br />
- Callis has number sense tricks.<br />

<br /><br /><br />
The angle between two vectors is the dot product divided by the magnitude of the two vectors multiplied together.<br /><br />

For constant force, <br />
F = ma<br />
W = F * d = F * delta x<br />
Work can also be the change in energy (potential or kinetic), even with constant velocity (making Force zero), so there's two ways to approach problems involving work.<br />
W = change in kinetic energy for non-vertical situations<br />
W = integral of F<sub>_(x)</sub> * dx<br />
The slope of a Work-time graph is going to be power.<br />
dW/dt = Power<br />
Joules = Newton meters.<br />


<br /><br />
1 hp = 550 ft-lbs = 746 Watts
<br /><br />

<h2>2008-03-24</h2>

You have different types of record albums. So, do you remember how you might have a record album that looks like this? An LP. Then some might have a little record with a bigger hole in the middle. And then some might be a bit different.

<center><img src="2008-03-24.png"></center>

What is the angular velocity? Aha. ω is the symbol for angular velocity (omega).

<br /><br />

What is 1 rev equivalent to?<br />
1 rev = 2pi rad = 360 deg<br />
So, angular velocity would be radians per second.<br />
ω = radians / sec. So if 14 sec<sup>^(-1)</sup> is the rev, then divide that by 2pi to figure out how many revolutions.

ω = dθ/dT<br />
α = dω/dT = d<sup>2</sup>θ/dT<sup>2</sup> in rad/sec<sup>2</sup>.<br />

<br /><br />

The linear velocity on a disk is tangent to the circular path of the particle and V<sub>t</sub> = dS/dt = tangential velocity = r * dθ/dT = r * ω.<br />
Recall that a<sub>t</sub> = dV<sub>t</sub>/dt = r dω/dT (this is all substitution) ------ so a<sub>t</sub> = r α<br />
a<sub>c</sub> = V<sub>t</sub><sup>2</sup> / r = (rω)<sup>^(2)</sup> / r = r ω<sup>^(2)</sup><br />

<br /><br />

If angular acceleration is constant, recall:<br />
v = v_o + at --------> w = w_o + at<br />
x = x_o + v_c * t + (1/2)at^2 ---------> theta = theta_o + w_o * t + (1/2) alpha * t^2<br />
v^2 = v_o^2 + 2a(x-x_o) --------> w^2 = w_o^2 + 2a(theta - theta_o)<br />


<br /><br /><br />

<h2>2008-03-25</h1>
<u>Today's nonsense</u><br />
- Bingo tonight.<br />
- Continue working on yesterday's example problems today.<br />
- Homework due tomorrow. #1-15 on the handout.<br />
<br /><br /><br />

<h1>2008-03-31</h1>
<u>Today's nonsense</u><br />
- On Saturday, the bus didn't get to Del Valley at the right time, so Hays missed the number sense UIL test.<br />
- wench, winch, what?<br />

<br /><br /><br />

<h2>Rotational Kinetic Energy</h2><br />
K<sub>_rot</sub> = (1/2) I w^2<br />

<br /><br />

ex 1) A flywheel used for storing energy consists of a uniform disk of mass 150000 kg and radius 2.2 m that rotates at 3000 rev/min about its center of mass. Find its kinetic energy. <br />
<br />

1) Find the I. What is the moment of inertia for a uniform disc? I = (1/2)MR^2. We should be in radians/sec, kilograms, and meters.<br />
So: K_rot = (1/4)(150000)(2.2)^2 [ 100 pi ]^2<br />
So you need the angular velocity in radians per second, so that's 3000 * 2 pi / 60 to get rev/sec. <br />
K_rot = 1.791 * 10^10 Joules<br />

<br /><br /><br />

ex 2) A winch has mass M and radius R. A cable wound around the winch suspends a load of mass M. The cable has length and density lambda with total mass M<sub>_(c)</sub> = L * lambda. <br />
<center><img src="2008-03-31.png"></center>

The load begins to fall, unwinding cable as it goes. How fast is the load moving after it has fallen a distance d?<br /><br />

E_i = initial energy = zero = E_f = final energy = zero = u_f + k_f<br />
So for every gain in kinetic, we are going to have a loss in kinetic.<br />
u_f = -mgd because it will be falling a distance d<br /><br />

The cable is not massless, and so this is going down, so this makes the problem more difficult. As the cable falls, you are losing potential energy. Let's look at the cable that has fallen. Has the whole cable fallen?