Review of Chapters 1-8

Interactive Transcript

spk_0 So a couple of things related to grade. The first one is at this point you can log into

spk_0 the website see all of your grades. The homework that's up there is accurate as of yesterday,

spk_0 which means you have a score and a grade and that's the worst you can do in the class.

spk_0 There's a couple ways you can improve that grade. There's three ways you can improve the

spk_0 grades that's on the website. There's no way to make it any worse than it is. Okay, so the first

spk_0 way to improve it is by doing well in the final. If you do better on the final than you've

spk_0 done on any of the midterm, I will replace your midterm, your any scores on midterm that are

spk_0 lower than your final score, I will swap out your midterm for your final. Okay, whether that

spk_0 be a single midterm or all of them. If you do better on the final than you do than any of the

spk_0 midterm, the final score will replace each of those midterm. Okay, so what that means is no

spk_0 matter what your grade is right now, if you ace the final you can ace the class. Okay, if you

spk_0 don't want to count on that happening, and it's probably not a wise strategy to just assume that

spk_0 you will ace the final. Two other ways to improve your grade. One is you have until Tuesday,

spk_0 December 16th, that's the day of the final at 9 a.m. to complete any homework that you haven't

spk_0 yet completed. Now you don't get full credit for turning in late homework. It gets reduced,

spk_0 and at this point you can get up to 50% of the original credit for submitting any homework

spk_0 that you didn't submit initially. So if you missed an assignment or something like that you might

spk_0 want to go back. You can still do that. They will add points to your homework score and that will add

spk_0 everybody points to your overall grade. Yeah homework 11 is fixed. Yeah, if you try to submit homework 11 late,

spk_0 it probably told you you were going to get negative percent. The reason is I have it set so that every hour

spk_0 it takes off like 10% or something like that. It's supposed to stop doing that once it gets to 50%.

spk_0 I somehow forgot to set that floor. I just kept going negative and that's fixed. If you encounter any

spk_0 sort of weird behavior in any of the scores that you see, let me know. One of the things that was

spk_0 pointed out in the discussion forum is that it would be really useful if you could see your individual grade

spk_0 on the class webpage, which is exactly what the webpage is supposed to be doing. It wasn't doing that.

spk_0 It was only showing your aggregate grade. So I fixed that. So if you want to go back and see how you

spk_0 scored an each midterm and just verify what's there to accurate, I'd encourage you to do that. If there

spk_0 is a difference from what's written on the webpage and what's on your exam that I handed back, it's not

spk_0 a problem. If you tell me that, now, if you wait until after I submit grades, it's a real problem.

spk_0 So I double check. I triple check to make sure that everything is accurate. But I put up the webpage

spk_0 so that you can also check that yourself since it's in your best interest to make sure that's accurate.

spk_0 Okay, so that's two ways you can improve your grade. The third one is with extra credit.

spk_0 Right, so I put up an extra credit opportunity. There's one extra credit opportunity that's been

spk_0 available all along and a few people have taken advantage of it. If you find mistakes in any of the

spk_0 lecture slides, you can go on to the webpage and submit the mistake. And for every mistake you find,

spk_0 I'll give you a half percent extra credit on your overall grade. So there's been a few people who

spk_0 caught some errors. They have to be significant. If I misspele a word, I'm not going to give you extra

spk_0 points for that. You're welcome to submit it because I do want to make the slides better, but it's

spk_0 generally physics mistakes or math mistakes that I'll give credit for. So you can do that.

spk_0 The other thing that I'll give extra credit for is when I hand out the solutions or when I print

spk_0 the solutions online, you know, I tell you what the right answer is for the multiple choice

spk_0 questions, but I don't go into any detail on that. So I started a discussion thread.

spk_0 You can see it here, multiple choice solutions. You can go in there and you can write up a solution,

spk_0 an explanation as to why this is the right answer. And I will give a half percent credit

spk_0 to the author of each solution. Right, so we've had four midterms. There's about 10 multiple

spk_0 choice questions per midterm. That's about 40 opportunities for extra credit right there.

spk_0 Now I'm not going to give them all the one person. I set a half a 3% per student. So what that

spk_0 is, is that's enough extra credit to get you bumped up half a grade. So if you're a borderline,

spk_0 you know, you're worried that you have a B minus right now, but you almost have a B.

spk_0 Go ahead and do the extra credit. Make sure you get enough to get up over that threshold.

spk_0 And then you'll know going into the final that you have that minimum grade that you want.

spk_0 Any questions?

spk_0 You do not have to take the final. If you're happy with the grade you have right now,

spk_0 you do not have to take the final.

spk_0 Same.

spk_0 No, I only award one extra credit per question. So for example, I posted an example

spk_0 solution. What I expect. So I took the first midterm, very first question, and I posted

spk_0 the solution to that. So I won't award any extra credit for that problem because I already

spk_0 answered it. But that leaves you like 39 other problems you can submit solution for.

spk_0 So that's also a valuable resource if you're studying. And you're looking at the solution guide,

spk_0 and you can't understand why, you know, choice B was the right answer.

spk_0 You can look there and someone hopefully will have explained it for you. If they haven't,

spk_0 then maybe you can do some digging, figure it out. You can contact me, or you can come to my

spk_0 office hours. I'll still have office hours between now and the exam. Figure out what the solution

spk_0 is, post it, and get the extra credit for it.

spk_0 I did. It's down there. You can see there's already a few messages in this thread.

spk_0 To go in there and see what the example looks like.

spk_0 To follow that. Back guideline. Okay, so you got your midterms back.

spk_0 Couple of people who came in late.

spk_0 Can you see anyone else? Not a midterms.

spk_0 Okay, so if you're paying attention, I'll go over a few of the,

spk_0 a few of the multiple choice. You run to the computer after class. You can type up some of these

spk_0 responses and get extra credit for them. Okay, I'm not going to go over all of them.

spk_0 I'll try to focus on the ones that people had a lot of problems with.

spk_0 Number two.

spk_0 Ask about how fast a clock will run if we're on the moon.

spk_0 And it's a pendulum clock. So what we know about pendulum is that they're period

spk_0 or the frequency at which they run. I can get a frequency of square root of g over l.

spk_0 So on the moon, the acceleration due to gravity is less. Six times less.

spk_0 So the rate at which they're going to run is the square root of six times slower.

spk_0 That wasn't one of the options. None of the above.

spk_0 Assuming that your ears pop when the pressure increases by a certain amount in a pool,

spk_0 the trick is you have to figure out how far down the pressure is increased by this amount.

spk_0 Use path scales law to do that. I'm not going to work through that, but if you write down path scales

spk_0 law on your test, you can reference that in your textbook and figure out how you would do that.

spk_0 Problem four was a little tricky. Not tricky, just a little bit involved because there were two parts to it.

spk_0 We're asked about water coming out of a faucet. We have a given flow rate.

spk_0 After it's fallen 20 centimeters, we want to know the radius of the stream of water.

spk_0 So we have flow of water. So we should think Bernoulli's equation.

spk_0 We can pick two points. So the, yeah, top of the flow at the output of the faucet in 20 centimeters below,

spk_0 would be two logical points to pick. Bernoulli's equation lets us relate the energy pre-unit volume at point

spk_0 one to the energy pre-unit volume at point two. At both of these points, the fluid is outside

spk_0 of the pipe. So it's open to the air, which means the pressure both points is the same as the

spk_0 atmospheric, the low's terms cancel. There's a difference in height between the two points,

spk_0 and therefore there has to be a difference in velocity. So you're given the velocity at one point,

spk_0 you can find the velocity at the other. If you know V2, the velocity of the water down here 20

spk_0 centimeters below the output of the pipe, you can say the flow rate,

spk_0 which is area times velocity, has to be the same as the flow out of the pipe, which you're given.

spk_0 So you know what that equals to, 10 to the minus 4 meters cube per second.

spk_0 So you can solve for the area of that flow of water. Once you know the area, you can find the radius.

spk_0 Question 5 involves a standing wave on a string, a free end at x equals 100, a fixed end at x equals 0.

spk_0 So what that tells us is that there is a node here and an anti-node here,

spk_0 and you're asked to find where there can be an anti-node. So the fundamental mode for standing

spk_0 wave on the string is going to look like this. If an anti-node at the right or a node at the left,

spk_0 those come from our boundary conditions. There are higher harmonics that we can have as well.

spk_0 So you can quite draw that one scale.

spk_0 Here's the second harmonic, and it has an anti-node, one third of the way over at x equals 33 centimeters.

spk_0 That was the correct answer. I can make arguments why it's impossible to have an anti-node at 050 or 67.

spk_0 But you can see clearly that there definitely can be one at 33, so I'll leave it at that.

spk_0 Question number 6.

spk_0 Yeah.

spk_0 No, I said where can there be an anti-node? And clearly there can be an anti-node at this point

spk_0 if it's the second harmonic. It doesn't mean there will be an anti-node there,

spk_0 but there are some points like here where there cannot be an anti-node because there has to be a

spk_0 node because it's six. Likewise, I can argue that there has to be a node, but there cannot be an

spk_0 anti-node at 50 or 67. But so now I'll just, I've showed that there can be one at 33, so I'll leave it at that.

spk_0 Question 6 about the functional form for a wave on a string. Part A and B, your options A and B

spk_0 are not dimensionally consistent. You just look at this term inside of parentheses.

spk_0 B times Z, this is a length for unit time, that's a velocity, and this is a length distance.

spk_0 So the units on this are distance squared over time, the dimensions that are distance squared over time,

spk_0 the dimensions of this are time. These two terms are not dimensionally consistent. We can't subtract two

spk_0 quantities at a different dimension. So choice A or B physically are meaningless.

spk_0 So that leads us with choice C or D, and the difference between them is the sign, that affects whether the wave is going to the

spk_0 right or to the left, whether it's positive or negative. We're told we want a wave going in the positive Z direction.

spk_0 So choice C, as if the time increases, the position has to increase as well for this argument to stay the same.

spk_0 That is to say if we follow a point on the wave, it will move to the right.

spk_0 If the wave is described by choice C, it will move to the left, if it's described by choice B.

spk_0 If the choice C is the right answer there.

spk_0 Number 7, we increase the volume on our stereo by 13 dB, how much does the sound change?

spk_0 Increasing by 13 dB on a logarithmic scale, which is the scale that loudness is measured in.

spk_0 The dB is the logarithmic unit. It means that the intensity increases by 10 to the 13 over 10,

spk_0 or 10 to the 1.3, which is a factor of 20.

spk_0 So the first thing you had to realize is that this dB is not a direct measure of intensity, so it's not 13 quatt's per centimeter squared.

spk_0 It's not a factor increase, so it's not 13 times.

spk_0 And then if you work out using the relationship between sound level and intensity, you could solve for 20.

spk_0 Question 8, a bagpipe player has a pipe that's 30 centimeters long, open it one end closed at the other.

spk_0 So it's like this string, and then it will have an anti-node on one side and a node on the other.

spk_0 We're going to ask about the frequency of the fundamental.

spk_0 It's fundamental standing wave.

spk_0 One that has a wave form that looks like this.

spk_0 This is one quarter of a wavelength, going from zero to a max.

spk_0 I'm going to go back to zero, into a min, and back to zero.

spk_0 That would be one full wavelength of the sound.

spk_0 If you recognize that, you can see that the length is a quarter wavelength.

spk_0 The wavelength is 4L or 1.2 meters.

spk_0 So if you're given the wavelength and the speed of sound, you can find the frequency.

spk_0 Frequency is the speed of sound divided by the wavelength, and you're given those values.

spk_0 You're given the velocity of sound in the problem, and you can solve for the wavelength and plug it in.

spk_0 You get 287 hertz.

spk_0 And then in question 9, you're asked how that sound would change if the bagpipe player is walking

spk_0 towards you. So when you have a problem that involves sound and motion, you should think

spk_0 of Doppler Shift. So here we have a Doppler Shift, because the sound is coming towards us.

spk_0 We'd expect the frequency to go up. So one of these two positive shifts.

spk_0 And if they're walking at one meter per second, and the speed of sound is 344 meters per second,

spk_0 that fractional shift is about 0.3%.

spk_0 You can work that out using the Doppler formula that was given in the last problem.

spk_0 But that simple argument shows you that it increases by about 0.3%.

spk_0 Any questions then on the multiple choice?

spk_0 Yeah, that's your question.

spk_0 That's not a question.

spk_0 Well, A and B are dimensionally inconsistent.

spk_0 Right? C is a wave that's moving in the positive Z direction.

spk_0 D is a wave moving in the negative Z direction.

spk_0 All right. The first of the two work it out problems.

spk_0 This problem really had, there's three parts to the problem, but really there's two logical parts.

spk_0 The first two parts relate to the forces that act on this floating cup.

spk_0 And then the third part relates to the oscillatory motion that you'd have if it were left

spk_0 creed above back and forth in the water.

spk_0 So in part A, we're asked how far below the water line this cup will sit when it's floating.

spk_0 And what that means is the sit there when it's floating means it's an equilibrium.

spk_0 Equilibrium means the forces balance.

spk_0 And when you deal with forces, the basic problem solving strategy is always the same.

spk_0 Draw a free body diagram, add up the forces, the net force, you said equal to the math

spk_0 times acceleration. That's Newton's second law.

spk_0 Then you solve whatever quantity you're interested in.

spk_0 So in this case, it's an equilibrium.

spk_0 So we know the net force has equal zero.

spk_0 That net force comes from the sum of the forces on it.

spk_0 There's the weight pulling it down.

spk_0 And then because it's in the water, there's a buoyant force pushing it up.

spk_0 That allows it to float.

spk_0 So you have to recognize that the buoyant force is equal to rho Vg.

spk_0 And that the volume that's displaced is related to how far below the water it floats.

spk_0 Once you've done that, you can solve for x at 2.4 centimeters.

spk_0 The way this was graded, you got one point for recognizing that this was an equilibrium

spk_0 and writing this equilibrium constraint.

spk_0 You got 2.4 correctly recognizing that the buoyant force is given by this expression here.

spk_0 One point, if you identified this more generic term of the expression,

spk_0 one point for identifying the proper weight of the cuff that opposed that buoyant force,

spk_0 one point for solving for x and one point for getting the correct numerical answer

spk_0 with the appropriate unit.

spk_0 So pi d squared over 4 is the area of a circle of diameter d.

spk_0 The area of the cylinder, the area of that cylinder times the type, x,

spk_0 is the volume of the cylinder that's underwater.

spk_0 That book gives us the volume of this weight.

spk_0 Yes, I do.

spk_0 If the floating cup is pushed down an additional centimeter at part B,

spk_0 what is the net force on it?

spk_0 Basically the same equation, not the same equation, but the net force is the left side of this

spk_0 equilibrium expression.

spk_0 So we just have to take the left side of the expression that we had up here

spk_0 and evaluate it not at x but at whatever we got here plus one centimeter.

spk_0 And that will give us our answer.

spk_0 So instead of saying it's 2.4 centimeters below the surface, we'll say it's 3.4,

spk_0 we'll plug that in to get about half a Newton pointing up.

spk_0 So 3 points for correctly writing the expression for the net force.

spk_0 The buoyant force pointing up, the weight pointing down.

spk_0 1.4

spk_0 1.4 simplifying this into this form,

spk_0 which you can then evaluate for 2.2 to get the final answer.

spk_0 And then finally part C will be the frequency of oscillation after it's released.

spk_0 So this is pushed down.

spk_0 It's not an equilibrium.

spk_0 At least once we let go, it's not an equilibrium, it's going to shoot back up.

spk_0 And that's going to cause it to sort of bounce back and forth and oscillate.

spk_0 So the key here is we know how much force it takes to push it down 1 centimeter.

spk_0 So it can treat it like a spring.

spk_0 With a spring constant, that's given by the necessary force to push it 1 centimeter divided by 1 centimeter.

spk_0 Once we know that spring constant, omega is square root of k over m.

spk_0 The frequency that we're trying to find is omega over 2 pi.

spk_0 So we can evaluate all those things, you get 3.2 hertz.

spk_0 So when you're doing problems with oscillation, the key is always to be able to find omega.

spk_0 Because omega is related to forces and the motion.

spk_0 And in this problem, you got one point for recognizing the frequency you wanted with omega over 2 pi.

spk_0 2 points for figuring out what omega is, omega's square root of k over m.

spk_0 One, two points for figuring out what k is, k is the effective spring constant for this system.

spk_0 And then one point for evaluating and getting the final answer.

spk_0 Any questions and problem pen?

spk_0 Okay, then problem 11, this is definitely the one that people struggled with the most.

spk_0 Maybe because the last thing we covered, or the fact that we only spent a single day talking about Doppler shift.

spk_0 I will say if you struggled on this, you probably want to brush up on the Doppler shift before the final exam.

spk_0 So I gave you the formula for the Doppler shift.

spk_0 But I didn't explicitly explain what any of these terms were.

spk_0 So really all the points that you got were for figuring out what the terms were.

spk_0 I didn't give any points for writing down formulas because they are given you the necessary formula.

spk_0 So it's just a matter of figuring out what's going on in the problem.

spk_0 We've got two police cars, car 1 and car 2.

spk_0 And then we've got us, we're the listener over here.

spk_0 These things are moving faster than we are.

spk_0 We know that because they pass us.

spk_0 So they must be moving faster.

spk_0 We're told that we hear a frequency of 680 hertz.

spk_0 So the frequency that the listener hears is 680 hertz.

spk_0 In part A, we're told that once one of these police cars has passed us.

spk_0 So the situation looks like this.

spk_0 Where car 2 is beyond us and car 1 is still approaching.

spk_0 We're right here in the middle.

spk_0 When one of the cars has passed us, we hear a beep frequency of 2 hertz.

spk_0 So a beep frequency means there's two different frequencies that differ by 2 hertz.

spk_0 Now, as cars, as a source of sound approaches us, we hear it's Doppler frequency gets up shifted.

spk_0 As it's receding from us, it gets down shifted.

spk_0 So if car 1 is approaching us, we know that that one, we hear 680 hertz.

spk_0 That was given in the problem.

spk_0 The frequency 1 is 680 hertz.

spk_0 While that was given in the problem, very few people actually identified that in part A

spk_0 as the frequency we'd hear from the cars is behind us.

spk_0 For car 2, we know it's frequency as to be less than that.

spk_0 As it gets down shifted since it's receding from us.

spk_0 And it differs by 2 hertz.

spk_0 So frequency 2 must be frequency 1 minus 2 hertz.

spk_0 Or 678.

spk_0 Okay, so you get one point for recognizing that the approaching cars are frequency of 680 hertz.

spk_0 I consider that an easy part of the problem.

spk_0 So I only awarded one point for that.

spk_0 Two points for recognizing that the other frequency that you hear is down shifted.

spk_0 One point for recognizing is down shifted by 2 hertz.

spk_0 And then one more point for putting all that together and getting 678 hertz.

spk_0 So the other part.

spk_0 Part B asks, what's the frequency that one of the

spk_0 policemen would hear in his own car?

spk_0 So if what we hear is the frequency of the listener, what we're after here is the frequency of the source.

spk_0 That's what the policemen would hear in his own car.

spk_0 Okay, so a couple of ways to go about this.

spk_0 One is we can consider this situation over here.

spk_0 Where this car is behind us.

spk_0 This car is in front of us.

spk_0 They both are producing the same sound as heard by the policemen in the cars.

spk_0 But to us, one sounds like it's at 680 hertz and one sounds like it's at 678.

spk_0 So what we can do is we can take this equation.

spk_0 We can plug in 680 to the sound we hear and the policemen is approaching us.

spk_0 The policemen is approaching us.

spk_0 It has a negative source velocity.

spk_0 And we can plug in 678 when the policemen is going away from us

spk_0 that's a positive source velocity.

spk_0 Install for both the frequency and the velocity.

spk_0 So you could do all that.

spk_0 Or I would also accept 679 based on the fact that it's halfway in between 680 and 678.

spk_0 And there's a lot of mathematical theory that goes into the justification of that argument

spk_0 involving things called first-order Taylor expansion, which I'm not expecting you to know.

spk_0 But I would still give you full credit for just writing down the pathway in between.

spk_0 Even though it's not very mathematically rigorous, it turns out to be accurate.

spk_0 Okay, so each siren is at 679 hertz.

spk_0 Part C, what speed are the police cars driving at?

spk_0 Well, here we know the speed they're driving at is V sub S.

spk_0 Speed of the source of the sound.

spk_0 We know the speed of the listener is 20 meters per second.

spk_0 We know the speed of sound is 340 meters per second.

spk_0 We know the frequency of the source is whatever we got in part B.

spk_0 Whether you got that right or not, it's irrelevant.

spk_0 How that should be 679.

spk_0 The frequency we hear would be 680 if the cars approaching us.

spk_0 And if the cars approaching us, we need to recognize that our velocity is negative.

spk_0 Because we're moving away from the car.

spk_0 So if you do all those things, you can evaluate what the speed of the car is by solving that expression.

spk_0 So one point, just for recognizing that the V in this formula is 340,

spk_0 which was actually stated in the problem.

spk_0 But if you wrote down this formula and you started plugging in numbers,

spk_0 you should get one point, it's plugging in 340 for V,

spk_0 one point for plugging in minus 20 meters per second as the speed of the listener.

spk_0 And that's assuming that you're using the Doppler shift formula

spk_0 or the situation described initially where the cars are approaching you.

spk_0 And you're moving away from them.

spk_0 One point for recognizing that the source frequency is not 680 hertz.

spk_0 Not what we hear, but rather whatever you got for part B.

spk_0 One point for recognizing that the frequency that we hear is 680 hertz.

spk_0 And a lot of people got those mixed up and said that the source frequency was 680.

spk_0 It's not the frequency that we hear.

spk_0 So a point for that, and then a point for solving

spk_0 for the final velocity of 20.5 meters per second.

spk_0 And you can just check that is faster than we're driving, which is what we expected.

spk_0 They're driving faster than us, that's why they passed us.

spk_0 Part D, what would you hear if you were driving at the same speed but in the opposite direction?

spk_0 Well, all we have to do is take this expression here and change the sign for V sub L.

spk_0 From it was minus 20 meters per second, we change it to plus.

spk_0 And we get 765 hertz.

spk_0 If we're driving at the police cars.

spk_0 Since we're driving at them, our relative separation is decreasing much more rapidly.

spk_0 Therefore, we get a much larger fast-forward shift.

spk_0 So one point for recognizing that you needed to change the sign and make that plus 20 meters per

spk_0 second. Two points for using the proper speed of the police cars, which is what you

spk_0 got in part C. And then one point for recognizing that the frequency is to be upshifted.

spk_0 And one point for getting the final answer.

spk_0 The question from problem 11.

spk_0 Okay.

spk_0 Then let's move on and do some review.

spk_0 Do you want me to work out a few problems in great detail?

spk_0 There are a lot of problems in little detail.

spk_0 How many people would rather see fewer problems in more detail?

spk_0 One, two, three.

spk_0 How many would rather see more problems in less detail?

spk_0 Okay, that seems to be the choice.

spk_0 So we started with vectors.

spk_0 Vectors was chapter one.

spk_0 Vectors are quantities that have a magnitude and a direction.

spk_0 We represent them by arrows.

spk_0 Arrows are things that have a length and a direction.

spk_0 And we can relate the length of an arrow to the magnitude of its components along the x and y

spk_0 axis using the Pygoryan theorem.

spk_0 You can find the angle of that.

spk_0 We usually measure angles to respect to the positive x-axis from trigonometry instead of the x and y

spk_0 components and the angles the arc tangent.

spk_0 And likewise, if we're given the components of a vector, we can find its magnitude

spk_0 using the Pygoryan theorem.

spk_0 I just said that.

spk_0 We can find the components from trigonometry knowing the length and the angle.

spk_0 Using basic trigonometry.

spk_0 So, let's do a problem.

spk_0 A ferrous wheel spins such that

spk_0 the carriages on the rim have a radial acceleration of point four meters per second squared

spk_0 and a tangential acceleration of point three meters per second squared.

spk_0 Find the magnitude of the total acceleration.

spk_0 Which direction does it act?

spk_0 So, we'll draw a little picture of what's going on.

spk_0 It doesn't tell me where on the ferrous wheel the car is.

spk_0 So, I'll just pick a point.

spk_0 The radial acceleration, radial means in the towards the center.

spk_0 And if this thing is going around in a circle, I know that acceleration needs to be inwards.

spk_0 And that has the magnitude of point four.

spk_0 Yep, centripetal and radial.

spk_0 That's the same.

spk_0 Pangential acceleration of point three meters per second squared.

spk_0 Pangential is perpendicular to the radial direction.

spk_0 Okay, so we can find the magnitude of the total acceleration.

spk_0 We have two components.

spk_0 Magnitude is what we get when we add up those components.

spk_0 So, this is the total acceleration.

spk_0 And it's magnitude you can find from the Pythagorean theorem.

spk_0 Alright, so let's consider this triangle right here.

spk_0 I'll rewrite this vector over here.

spk_0 And apply the Pythagorean theorem.

spk_0 And when I evaluate that, I will get point five meters per second squared.

spk_0 And so, that's how long this is vectoring.

spk_0 I'm also asking which direction does it act?

spk_0 Okay, so I can draw a diagram for these vectors.

spk_0 I've already done that.

spk_0 Let me draw it down here where it's maybe more clear.

spk_0 I can pick an angle in this diagram, solve for that angle.

spk_0 And then I need a way to express what that numeric value mean.

spk_0 In this case, it's going to mean an angle that's in the tangential direction relative

spk_0 to the radial direction.

spk_0 Okay.

spk_0 So, I can say that tangent of that angle is opposite over adjacent.

spk_0 And so, I can solve for a theta.

spk_0 I think that's 27 degrees.

spk_0 Well, they don't have it worked out.

spk_0 And it's not enough just to say 27 degrees.

spk_0 I have to somehow explain what that means.

spk_0 And it said usually angles are measured with respect to the x-axis,

spk_0 but I haven't drawn an x-axis yet.

spk_0 So, I have to say that 27 degrees.

spk_0 From the radial direction.

spk_0 Towards the direction of the tangential acceleration.

spk_0 So, that was chapter 1.

spk_0 Chapter 2, we talked about different parameters in motion.

spk_0 Velocity displacement and acceleration is the three quantities we dealt with a lot.

spk_0 Displacement, we represent by the vector x, the location of an object.

spk_0 Velocity is the time rate of change of displacement.

spk_0 And acceleration is the time rate of change of velocity or the second derivative of displacement.

spk_0 So, all these things are vectors.

spk_0 The displacement is a position.

spk_0 So, as a position in two or three-dimensional space,

spk_0 velocity is the time rate of change of that.

spk_0 So, it's also a vector.

spk_0 It has a magnitude and a direction.

spk_0 You're asked for a velocity.

spk_0 You need to give the magnitude and the direction associated with it.

spk_0 Likewise for acceleration.

spk_0 That's important.

spk_0 Acceleration is important because it relates.

spk_0 The forces that act on an object to how the object moves.

spk_0 Newton second law said f equals mA.

spk_0 We could use pre-body diagrams to help us determine what the net force on an object is.

spk_0 We could set them equal to mass times acceleration.

spk_0 Oftentimes, we ended up solving for the acceleration.

spk_0 Using these relationships to figure out how the object is moving.

spk_0 So, when you have constant acceleration, there were some equations that come from those relationships.

spk_0 The equations, we call these either the projectile motion equations

spk_0 or the equations for constant acceleration.

spk_0 They relate velocity, time, and position

spk_0 when the acceleration is constant.

spk_0 If you know two of these quantities, if you know how fast something is moving

spk_0 at say two different times,

spk_0 you can use this relationship to relate those initial and final velocities for the time that

spk_0 elapses between them.

spk_0 If you know two different velocities, the two different positions in space for an object,

spk_0 you can relate those two velocities to how far the object moves.

spk_0 And likewise, you can use the bottom equations to relate changes in time to changes in position.

spk_0 So, use these for the projectile motion type of problems.

spk_0 A simple example of that is what's the speed of a golf ball hit at 30 degrees if it travels 200 meters

spk_0 and we neglect any drag or air resistance.

spk_0 So, we have a situation that we can diagram. It usually helps to draw a little picture.

spk_0 Here's what's going on. The balls initially shot at 30 degrees with some unknown speed

spk_0 and we know that it travels 200 meters, meaning it's height.

spk_0 As it goes up, it returns to its starting height after some amount of time.

spk_0 And at that instant time, 200 meters away.

spk_0 So, projectile motion problems. We're going to use those projectile motion equations.

spk_0 And we're going to relate what's going on in the x direction and what's going on in the y direction.

spk_0 And the thing that relates those is time.

spk_0 So, usually set up an expression in one direction.

spk_0 Solve for the time. It takes that expression to be true.

spk_0 And use that in the other direction.

spk_0 Here, you don't know the initial velocity.

spk_0 But whatever it is, it determines how long it's in the air and it determines how far it goes.

spk_0 So, we can write an expression for how long it's in the air and how far it goes.

spk_0 We can require that it go 200 meters and then solve for what the initial velocity must be.

spk_0 So, in that y direction, you can say

spk_0 the height must return to zero, started zero and ended zero after some time C.

spk_0 Initial velocity in the y direction.

spk_0 This is theta for 30 degrees.

spk_0 The y component of velocity is going to be v sine of 30 degrees.

spk_0 The acceleration is going to be minus g.

spk_0 And we can use this, for example, to solve for how long the object is in the air.

spk_0 If we knew the velocity, we don't know the velocity.

spk_0 If we knew the time, we could solve this for the velocity.

spk_0 So, we have two things we don't know.

spk_0 We need a second equation.

spk_0 So, we'll consider the same general expression in the x direction.

spk_0 Okay, because we know something about how far it goes and we want to relate that to it time.

spk_0 Okay, so it goes 200 meters, starting from zero.

spk_0 Initial velocity in the x direction.

spk_0 You can find from this diagram, it's v cosine of 30 degrees.

spk_0 The acceleration in the x direction is zero.

spk_0 Gravity points down.

spk_0 That's our y direction, not along x.

spk_0 Okay, so I've got two equations and two unknowns.

spk_0 I don't know v and I don't know t.

spk_0 But given that I've got two equations, I can solve them.

spk_0 So, I can, for example, solve this one for t.

spk_0 Plug it into here and solve this for v.

spk_0 So, this tells me that t is equal to 200 meters over v cosine 30 degrees.

spk_0 Plug that in.

spk_0 Up here, solve the remaining expression for v.

spk_0 So, this v can't solve that v.

spk_0 So, the sine 30 over cosine 30, I can write as tangent of 30 degrees.

spk_0 If I bring this term, I'll leave it where it is.

spk_0 I'll bring this term to the other side and make it positive.

spk_0 I have a value for g.

spk_0 And,

spk_0 I'll bring the numerical values here.

spk_0 I have to take this v squared and put it over on this side if I want.

spk_0 And then I can solve this for v.

spk_0 Which I'll leave as an exercise for you.

spk_0 You can check.

spk_0 If that gives you 4.85 meters per second.

spk_0 And it's worked out in the notes, so you can reference that.

spk_0 And, questions then?

spk_0 Chapter 1, 2 and 3 that we just covered.

spk_0 Okay, centripetal acceleration is a different class of motion.

spk_0 We talked about projectile motion just now.

spk_0 That's been things that are constant acceleration.

spk_0 A lot of times though, things aren't moving in parabolic trajectories.

spk_0 They're instead moving in circles.

spk_0 Lots of examples of this.

spk_0 And when something moves in a circle,

spk_0 we know that its radial acceleration has to be v squared over r.

spk_0 You see on a test an object moves in a circular path,

spk_0 right down v squared over r as you're starting point.

spk_0 You know that's the acceleration that the object has.

spk_0 And it will likely be important in solving for any of the properties of motion

spk_0 or the forces that act on it that cause it to move in a circle.

spk_0 Okay, chapter 4 and 5, we talked about Newton's law.

spk_0 Newton's first law.

spk_0 An object that's moving will keep moving unless it's acted on by an outside force.

spk_0 It'll keep moving in the same direction with the same speed.

spk_0 Likewise, if it's not moving, it will remain at right.

spk_0 So Newton's second law quantifies that a little bit.

spk_0 It says that the net force on an object equals mass 10th acceleration.

spk_0 So the first law basically says that the net force is zero, the acceleration is zero.

spk_0 We use the second law all the time.

spk_0 Okay, that is the one thing.

spk_0 If you only took one thing away from this class,

spk_0 you'd probably be taking the class again next semester.

spk_0 But I hope it hopes that that would be it.

spk_0 The net force is mass 10th acceleration.

spk_0 Whenever we have problems to deal with forces,

spk_0 we draw for everybody diagram.

spk_0 We add up the forces in the x and y direction.

spk_0 That's what we call the net force.

spk_0 We set that equal to the mass 10th acceleration in the x and y direction.

spk_0 And that gives us some equations that we can start from to try to solve for the things we want to know.

spk_0 Okay, so we use Newton's second law all the time.

spk_0 Newton's third law tells us that if we push on something, it pushes back.

spk_0 And if we push hard on it, it pushes back just as hard.

spk_0 And the word that's what it means mathematically,

spk_0 it says the force of A and B is always equal enough to force of B on A.

spk_0 That is often useful in understanding how forces between different objects interact.

spk_0 So let's do an example of that.

spk_0 Physics student stands in a hot air balloon.

spk_0 What are the forces on the student?

spk_0 What are the reaction forces?

spk_0 Meaning which of the forces obey Newton's third law?

spk_0 So let's draw a picture of the hot air balloon.

spk_0 So what are some of the forces that act on the balloon on the person

spk_0 the weight?

spk_0 So let's say the weight of the person

spk_0 comes w and that points down.

spk_0 What is before we go any further?

spk_0 If the weight is pulling the person down, Newton's third law says

spk_0 the person must be pulling something up.

spk_0 Not the normal force.

spk_0 Not tension.

spk_0 There is a buoyant force, but that's not

spk_0 what's pulling the person down?

spk_0 Gravity.

spk_0 What's causing the gravity?

spk_0 What object?

spk_0 The earth.

spk_0 The earth is pulling on the person.

spk_0 So Newton's third law says the person must be doing what to the earth.

spk_0 Pulling on the earth.

spk_0 Newton's third law relates the weight of the person

spk_0 to the force that the earth feels as it's attracted towards the person.

spk_0 Right?

spk_0 The person is attracted towards the earth.

spk_0 Therefore the earth is attracted towards the person.

spk_0 Okay?

spk_0 There is a buoyant force on the hot air balloon

spk_0 and that might be in the opposite direction and have the equal magnitude as the weight of the person

spk_0 as this thing is an equilibrium.

spk_0 But that is not due to Newton's third law.

spk_0 The third law says the person is pulled down by the earth.

spk_0 So the earth must be pulled up by the person.

spk_0 What about the normal force?

spk_0 So there's a normal force pushing up on the person.

spk_0 The basket pushes up on the person.

spk_0 What does the person do to the basket?

spk_0 Pushes down.

spk_0 There's a normal force on the basket

spk_0 caused by the person pushing on it.

spk_0 So there are a couple examples of pairs of forces that obey Newton's third law.

spk_0 And when we're dealing with forces we start with free body diagrams.

spk_0 We should be expected, I expect you to be able to draw free body diagrams even for some wacky

spk_0 situations.

spk_0 So you should expect that you'll be tested on this.

spk_0 Let's consider a wacky situation.

spk_0 A balloon.

spk_0 That is, you can think of this as like a balloon that goes up to the top of a

spk_0 slanted ceiling.

spk_0 It hits the ceiling.

spk_0 And it's sitting there against the ceiling.

spk_0 What are the forces that act on it?

spk_0 Let me make this a little easier and just consider like a helium balloon.

spk_0 And I collect all the stuff at the basket.

spk_0 What forces act on this balloon?

spk_0 Go what?

spk_0 Okay, what do we call that force of the ceiling pushing on the balloon?

spk_0 Normal force.

spk_0 And which way does it act?

spk_0 Yeah.

spk_0 It's perpendicular to the surface.

spk_0 So it depends on the direction of the surface.

spk_0 In this case, the planted surface, so it's the planted normal force.

spk_0 What are their forces act on the balloon?

spk_0 There's three more.

spk_0 Okay, so the balloon has some weight.

spk_0 There's friction.

spk_0 Which way is friction going to act?

spk_0 Yeah, opposite.

spk_0 So any time there's two surfaces in contact, there can be friction.

spk_0 If you ask if there were no friction, what would happen?

spk_0 Well, the balloon would want to slide up towards the top of the ceiling.

spk_0 It's like, the ceiling balloon or some sort of balloon that's in the back.

spk_0

spk_0 Floating up.

spk_0 If it wants to slide up, but it's not, or even if it is, friction has to oppose that.

spk_0 It has to lung the surface.

spk_0 And that leads one more force.

spk_0 Boinsy.

spk_0 Yeah, we need buoyant force.

spk_0 Right now, all the forces are pointing down.

spk_0 So there has to be something causing to go up.

spk_0 That's the buoyant force.

spk_0 Likewise, we encounter lots of things that are not bumping up

spk_0 up against this plant, but sitting on top of this plant,

spk_0 so you could have a similar diagram that looks like this.

spk_0 Work.

spk_0 So that was chapter 4 and 5.

spk_0 Jack for six and seven involved work in energy.

spk_0 So work is what you have when a force acts over a distance.

spk_0 So if the force and the displacement are in the same direction,

spk_0 like here, you see there's positive work done.

spk_0 That means energy is being added to the system.

spk_0 So the weight lifter is adding energy to the barbell by lifting it up.

spk_0 Holding the barbell, although it may be difficult, doesn't involve any work.

spk_0 No work in the physics sense.

spk_0 No energy is being added to the barbell.

spk_0 There may be a force applied to hold it up, but it's not moving.

spk_0 The displacement is zero.

spk_0 So from work,

spk_0 it's force dotted with displacement.

spk_0 You need a force and a displacement in order to have work.

spk_0 In here, rather than just drop the barbell and have it smash through the floor,

spk_0 the weight lifter is setting it down gently.

spk_0 So we're still applying it upward force, as you set it down.

spk_0 But it moves down.

spk_0 So a force and a displacement in opposite directions

spk_0 give us a negative amount of work.

spk_0 That means energy is being removed from the system.

spk_0 It's lowering the object down slowly, removing the potential energy that it had

spk_0 without converting that to kinetic energy, which is what would happen if he just dropped it.

spk_0 You can understand work in energy conservation a little bit by considering this diagram.

spk_0 There's sort of three different forms of energy that we deal with in physics 50.

spk_0 There's potential energy over here.

spk_0 And there's kinetic energy over here.

spk_0 There's also heat.

spk_0 Heat is what you get when there's friction or any non-conservative forces that act.

spk_0 That's this red area over here.

spk_0 And so if there's no, those are the different types of energy.

spk_0 Work just moves energy around.

spk_0 So work can transfer potential energy to kinetic energy and back,

spk_0 or it can transfer kinetic energy to heat.

spk_0 If work transfers kinetic energy to heat, we call that a non-conservative force,

spk_0 non-conservative because there's no way to get the heat back into kinetic energy.

spk_0 It's a one-way energy drain.

spk_0 If you consider just the kinetic and potential energy, it's dotted line.

spk_0 Then any energy that flashes back and forth between kinetic and potential,

spk_0 we say is conserved.

spk_0 It's still part of the total energy.

spk_0 The system can be transferred back and forth at will by the work done by conservative forces.

spk_0 So work is a changing of energy in one form to another.

spk_0 Conservative forces do work in converting kinetic and potential.

spk_0 Non-conservative forces convert kinetic energy to potential energy.

spk_0 So with this diagram, we can write the conservation of energy equation.

spk_0 It says the initial potential in kinetic energy,

spk_0 total energy in this dotted line, these are the things that we can actually measure.

spk_0 Very difficult to measure the thermal energy of the system.

spk_0 So the energy that we can measure plus any work that drains energy from the system,

spk_0 that's how much energy must be left at some final time.

spk_0 So the final potential plus final kinetic.

spk_0 So that's the conservation of energy equation.

spk_0 We've used this quite a lot both in class and in lab.

spk_0 It's important that you not forget about this term,

spk_0 work due to non-conservative forces.

spk_0 Do you have friction or drag of any sort?

spk_0 That's going to be a term here.

spk_0 It's always negative.

spk_0 Non-conservative force is always negative work.

spk_0 They always take energy away.

spk_0 So we have this on the left side.

spk_0 It says your initial energy minus whatever is taken away by friction is what you're left with.

spk_0 Okay.

spk_0 Okay, we have three problems.

spk_0 I think we have time to do one of them.

spk_0 Let me show them to you.

spk_0 And you can decide which one will do.

spk_0 There are 16 multiple choice questions.

spk_0 Three short answer questions.

spk_0 16 one from each chapter.

spk_0 Okay.

spk_0 Here we have two cars colliding.

spk_0 And we want to figure out

spk_0 after the collision which direction they move.

spk_0 Here we have an elevator.

spk_0 And we want to figure out how fast it moves.

spk_0 When we release it from rest.

spk_0 And here we have a collision.

spk_0 Two objects collide.

spk_0 They bounce off of each other.

spk_0 And we want to figure out

spk_0 what direction they move after the collision.

spk_0 So all three of these problems in some form or another will be in the final.

spk_0 We have time to go over one of them now.

spk_0 Who wants to go over the first one?

spk_0 Second one.

spk_0 Third one.

spk_0 Second one, one out.

spk_0 We'll do that.

spk_0 I told you very early on in the course

spk_0 that you would benefit greatly from understanding

spk_0 Atwood's machine.

spk_0 There's basically this problem.

spk_0 And there's many different incarnations

spk_0 of this problem that I can ask.

spk_0 And I've already asked one or two versions of this

spk_0 on various exams.

spk_0 Basically it's an elevator or a mass.

spk_0 It's not three to fall.

spk_0 Rather it's got a rope tied to it.

spk_0 And depending on the problem there might be a block that it's dragging.

spk_0 Or another block that it's lifting up.

spk_0 And this problem.

spk_0 There's another block that it's lifting up.

spk_0 So the buckets are massless.

spk_0 They're loaded with these various loads.

spk_0 There are these pulleys that have some mass and some radius.

spk_0 And they're solid dip.

spk_0 The system has zero energy when the loaded bucket is at rest

spk_0 and it's 10 meters off the ground.

spk_0 What is the, I'm sorry, when this bucket is at rest and on the ground

spk_0 how much energy does it have when it's loaded and 10 meters up.

spk_0 Okay, so that's some sort of initial energy.

spk_0 Okay, so compared to when the bucket is at the ground,

spk_0 this bucket is higher up.

spk_0 This bucket over here is 10 meters lower.

spk_0 Okay, so.

spk_0 I call this point one.

spk_0 And this point zero.

spk_0 The energy at point one is entirely potential energy.

spk_0 This bucket has some additional potential energy from being lifted up.

spk_0 And this one has lost some potential energy to being lowered.

spk_0 So the potential energy looks like MGH.

spk_0 There's the, we call it the bucket and the counterweight.

spk_0 I'll call it the bucket.

spk_0 Plus the counterweight both has some potential energy.

spk_0 The bucket has a mass of 400 kilograms.

spk_0 It has been lifted up 10 meters.

spk_0 The counterweight has a mass of 300 kilograms.

spk_0 And it has been lowered 10 meters.

spk_0 Therefore, it's potential energy decreased.

spk_0 Well, it's at the starting point.

spk_0 So I have 10 and 9.8 meters per second squared.

spk_0 That's 98.

spk_0 I'm 400 minus 300.

spk_0 Kilogram meters per second, meter squared per second squared is a dual.

spk_0 The system has 9,800 duals

spk_0 when it's lifted up 10 meters.

spk_0 What is the total kinetic energy of the entire system?

spk_0 And the bucket is moving with its speed V.

spk_0 So if the bucket,

spk_0 I'll release this and this is going to fall down at its speed V.

spk_0 This bucket is falling.

spk_0 What else is moving?

spk_0 If this bucket falls.

spk_0 The other side goes up.

spk_0 And what else is moving?

spk_0 The pulley is moving.

spk_0 And because they have mass, we can't neglect that.

spk_0 They rotate with the speed of V over R.

spk_0 They both have the same radius.

spk_0 Both are going to rotate at the same rate.

spk_0 So there's four things in this problem that move together.

spk_0 So when I ask what the total kinetic energy the system is, I need to include all four.

spk_0 All four effects.

spk_0 So something moving linearly as the kinetic energy of one half and V squared.

spk_0 So let's take one half the mass of the bucket and this velocity squared.

spk_0 We have one half the mass of the counterweight.

spk_0 It's moving at the same speed as the bucket.

spk_0 This is the patch by a rope.

spk_0 If this is moving down, this is moving up.

spk_0 They're moving at the same rate.

spk_0 Which is V.

spk_0 And then the pulleys.

spk_0 There's two of them.

spk_0 They're identical.

spk_0 So I'll just write twice the kinetic energy of one pulley.

spk_0 And the kinetic energy of a disc or any rotating object is one half i omega squared.

spk_0 Here I have to recognize that pulleys are shaped like a disc.

spk_0 And that a disc.

spk_0 It's a moment of inertia of one half and r squared.

spk_0 So I can plug all this and plug this in to the inertia of the disc.

spk_0 I can plug this in for its angular velocity.

spk_0 One half times one half.

spk_0 The r squared cancel.

spk_0 One of these twos cancel one of the one halves.

spk_0 The term here looks like one half the mass of the pulley times V squared.

spk_0 I add with the kinetic energy the bucket and the kinetic energy the counterweight.

spk_0 That's the total kinetic energy of the system when the bucket's moving at speed V.

spk_0 And I could plug in the specific masses that I'm given to the pulley, the bucket and the counterweight.

spk_0 And then finally in part C I say at the bucket's released from the height of 10 meters how fast is it going when it crashes into the ground.

spk_0 Well so we've already done all the work here.

spk_0 We've looked at how much energy it has when it's at 10 meters.

spk_0 If that all gets converted into the kinetic energy and we can set this energy equal to this expression and solve for the velocity.

spk_0 So I had a number 9800 joules for the total energy saying E1 equal kinetic energy at point zero.

spk_0 And that kinetic energy at point zero.

spk_0 I'm going to factor out the one halves and the V's.

spk_0 One half and some quantity and V squared.

spk_0 That quantity looks like the mass of the bucket, 400 kilograms plus the mass of the

spk_0 counterweight 300 kilogram plus the mass of the pulley 50, that's total 750 kilograms.

spk_0 And I don't have this worked out.

spk_0 I don't know the number.

spk_0 I can write this.

spk_0 Like that.

spk_0 There are.

spk_0 So there's a factor of two over here but I've already cancelled that.

spk_0 Okay so a couple of comments as you go through your final exam really as you go through the rest of your career either in physics or in engineering or whatever you're studying.

spk_0 When you're working out the solutions to problems.

spk_0 It's very tempting to take the information you're given, plug numbers into your calculator and write down values.

spk_0 You get the wrong value.

spk_0 You're not going to get any partial credit if you don't show your work.

spk_0 It's important to show your work as well if you have the right answer because it's useful for others to see what you've done.

spk_0 It's useful for yourself to see what you've done when you go back and look at stuff.

spk_0 And while that may not, while that may not seem that important now when you're working.

spk_0 It's a professional.

spk_0 It's very important because people don't trust anybody.

spk_0 Everything you do is going to get reviewed and checked and confirmed and you're going to confirm other people's work.

spk_0 So if you haven't documented what you did then it's useless someone has to redo it.

spk_0 On the exam documenting what you do allows you to get some partial credit.

spk_0 So if I can see what you were trying to do I can give you partial credit for the steps you did along the way.

spk_0 It will also help you avoid mistakes.

spk_0 You write things out.

spk_0 Any numbers that you write down need to have units.

spk_0 Otherwise this doesn't represent anything physically.

spk_0 Okay this is an energy it has to have units of energy.

spk_0 If you wrote this I could say well they're trying to use conservation of energy.

spk_0 You don't include that jewel this is no longer a conservation of energy.

spk_0 Well that's also important because it allows you to see sometimes that there's discrepancies.

spk_0 If you're trying to add two quantities and their units don't match

spk_0 that's a sign you've done something wrong and you can go back and try to fix it.

spk_0 If you haven't written the units down you'll never get that.

spk_0 Okay so that's the one thing that people could do that I think would most improve their scores.

spk_0 I mean one little thing you could do.

spk_0 So I would do that.

spk_0 Also writing down sort of your starting expression.

spk_0 Writing them out allows you to make sure you've got all the different terms considered.

spk_0 If you later want to go back and check you might realize all I didn't consider

spk_0 the fact that this problem has pullies in it that have math.

spk_0 You might have neglected a term like that.

spk_0 It's easy to see if you've included that.

spk_0 If you write out all the individual terms at some starting point and work from there.

spk_0 Okay so I'd encourage you to do that.

spk_0 Hopefully if you do that if you learn from the mistakes you made on the previous midterms.

spk_0 If you do the practice midterms which I posted online and if you either do the extra credit

spk_0 by putting together solutions for the multiple choice in the midterms or review the solutions

spk_0 that other people put together hopefully that will help you on the final and hopefully that

spk_0 will improve your grade and I won't be seeing you next year if you'll be in 51 or 52.

spk_0 That's all for me.

spk_0 Good luck on the final.

spk_0 Yeah.

spk_0 You

Review of Chapters 1-8

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