Comprehensive Physics Final Exam Review

In this video, we'll be reviewing for a comprehensive physics final. We'll solve problems that cover the big topics that are covered during the first semester of physics, and then we'll review some concepts that are not covered until the end of the semester, namely oscillations and fluids. So we'll start off with problems and then move to concepts. So for the first problem, we have a tennis ball that is hit at 23 meters per second at 58 degrees above the horizontal, and we want to find the horizontal and vertical components of velocity, max height. time to max height, velocity at its highest point, acceleration at its highest point, and how long the ball is in the air and how far away does it hit. So this is a projectile motion problem and it really covers just about everything someone could ask for when solving a problem like this. So the first thing you want to do is write what you know and we're given a vector. So we're given the initial velocity, 23 meters per second. and it was given at an angle of 58 degrees, so this is not the scale. And so as soon as I see this, the first thing that I want to do is get my components of this initial velocity vector, but that turns out to be the first thing that I'm asked for anyway. So horizontal and vertical components just refer to the x and y components of the vector. So vh, so v horizontal, or v not x, really the same thing. So if I draw my right triangle, then I have the magnitude of the vector on the hypotenuse. The vertical component is y, the horizontal component is x, and I like to just go ahead and label these. This is going to be my opposite side, opposite of this angle. This is my adjacent side, and as always, your vector is the hypotenuse. If I want v-x, that's going to be the adjacent side. So adjacent. is cosine, so use your SOHCAHTOA, so this is going to be V naught cosine of 58 degrees, or 23 meters per second, it's cosine, and that's going to give me 12.2 meters per second. And then once you know one component, you immediately know the other one is just the opposite trig function, so the vertical component, or V naught y, it's opposite, so it will be found using the sine function. So v-naught sine 58, so again 23 meters per second, sine of 58 degrees, and that gives 19.5 meters per second. So the first part was just vector math. Now we're actually getting into the physics, so we want to find the max height. So I want to start listing all of the things that I know. I know v-naught x, that means I also know vx because it's not going to be accelerating. in the x direction. So I can just write that ax is zero and ay is minus g since this is a projectile motion problem. And then I just found v not y. So the first thing that I notice is I'm looking for max height, so that's delta y, and I don't have time. Okay, so if I don't have time, the first equation that I'm going to go look at in terms of my kinematic equation Is this one. The one that does not have time in it. Okay, and Vy is going to be zero at the top at max height, so it simplifies a bit. And then if I solve for delta y, I get minus V0y squared divided by minus 2g. So my minus signs are just going to go away. And then I plug in what I got for V0y, 19.5 meters per second, square that, divide by 2 times 9.8 meters per second squared. And I get 19.4 meters. So now we want to find the time to max height. And we have even more information, so I could add delta y to my list of knowns. Let's put that here. But it turns out I don't need delta y. If I want to find the time to max height, I can just use my short kinematic equation. So vy equals v not y. Minus gt. I know v not y because I found it in my first step I know v y is going to be 0 at the top just like it was in the second part and then I just can solve For time so if I do that I get time and again if you set v y equal to 0 that makes That is what makes this v max t max so if you solve you get minus v not y Divided by minus g and you should notice those minus signs cancel They would need to because your time should be positive. So you get 19.5 meters per second divided by 9.8 meters per second squared and we get 2 seconds. Okay, now the next few questions are a little bit easier. Some things you should just know. So the velocity at the highest point. So this is one that you should just be able to look at and tell. So if this is going in a parabolic trajectory that it does in projectile motion then at the top it's only moving in X. So we know Vy is zero at the top but that means that all of our velocity is in the X direction and the velocity in our X direction does not change because the acceleration in the X direction is zero. So we know that the velocity at the top must just be the initial velocity in the X direction that we started with. So I'll put at the top, that's just equal to v-naught x, which we found to be 12.2 meters per second. The y component of velocity went away, but the x component stayed there, and it stayed at its same value. So the next one, what is the acceleration at the highest point? Well, we know this. There's no acceleration in the x direction. There's only acceleration in the y direction, so a is just minus g. So now we want to find how long the ball is in the air. So if we know how long it takes the ball to get to the top, It's going to spend the same amount of time going down. So our total time is just twice the time to max height, which is just going to be 2 times 2, or 4 seconds. And then finally, we want to see how far it goes. So we really only have one equation in x. So delta x equals v0x times time. Once we know how long it's in the air, so t total, We already know how fast it's going in x, and so we can use this equation to figure out how far it goes. And so this is going to be 12.2 meters per second. It's going to be going that fast for 4 seconds, and so that means it's going to go 48.8 meters. So now we're moving to a force problem, a Newton's second law problem. We have a 500-Newton crate that is pulled upward by a rope at 30 degrees above the horizontal. How strong does the upward pull need to be to keep the crate moving at constant velocity? The coefficient of kinetic friction is 0.4. So I always start problems, especially force problems, with a picture. So I have a crate. It's going to be pulled upward at an angle 30 degrees above the horizontal. So if it says something like above the horizontal, wherever you put your force, just draw a little dashed horizontal line or something and then draw that angle. And then it's on a rough surface because there's friction. So I like to indicate that so I don't forget friction. And now that I have that, I want to draw a free body diagram. So I want to know all the forces that are acting on this crate. So all objects that we deal with have mass. So they're going to have a weight. It's touching a surface. So that means there's going to be a normal force perpendicular to the surface. I have this force that's up and to the right at 30 degrees. And so as soon as I see this, I know I'm going to need to break. this force into its components. So maybe I call this a tension force. And so this tension force is going to try to move this crate to the right. And that must mean that friction is going to act to the left. And since it's asking me about when the crate's moving at constant velocity, since the crate is moving, it's going to be kinetic friction. And it's also the coefficient that it gave me. So now I want to start summing up the forces and I always start friction problems by summing up the forces in the y direction because I know they're going to depend on the normal force. And so if I do that, I have a positive normal force. And then I need to find the y or upward component of this tension force. So if I just draw my triangle somewhere else, put my angle, this is the magnitude. And so the y component will be on the opposite side, the x component will be on the adjacent side. And so that means ty is going to be t sine of 30. So plus, and it's upward, so it's positive. So t sine of 30 degrees. And then the weight is down, so it's going to be negative. So minus the weight. And I'm not going to substitute mg in for my weight because it told me the weight in newtons. And this is going to equal mass times acceleration in the y direction. But I know that's going to be zero because it's not going up off the ground or through the ground. So this is just equal to zero. And now I can solve for the normal force. I'm just going to move my weight over. It'll become positive. And then subtract this y component of tension, T sine 30. Okay, and so I know the weight. I could go ahead and plug that number in. I don't know what the normal force is because I don't know tension. So these are two unknowns. So I'm just going to leave that there for right now. Move on to my x direction some of the forces in x now I have the x component So just T cosine of 30 degrees of this pulling force and then I have friction working the other way to minus FK And again, it's going to be moving at constant velocity So that tells me ax must be zero. It's not accelerating. Okay, so I read that in the problem And now I want to substitute in my formula for FK since I know the normal force. So T cosine of 30 degrees minus mu K times N equals zero. And now I can take this N and I can substitute it in. OK, so I can put it in there. So I'll give myself a little bit more room. So I'm going to write T cosine 30 degrees. minus mu k and I'm going to put all of n inside parentheses so the weight minus t sine 30 and that equals zero. Okay so now I want to solve for the pulling force or the tension so I just need to do a little bit of algebra so I need to distribute my coefficient of friction to these two terms and then get it by itself. Okay so if I do that I have t t. cosine of 30 minus mu k times the weight minus minus is plus mu k t sine 30 and that equals zero now if i factor out my pulling force my tension i have t times cosine 30 so that's the first term and then plus mu k sine 30 and then i'm going to move this term to the other side let me go ahead and do that make it positive so mu k times the weight and now to isolate tension i just divide by everything in front of it so i get mu k times the weight over cosine 30 plus mu k sine 30 if i plug all that in i get 0.4 times 500 newtons you all over cosine 30 plus 0.4 times sine of 30. And if you plug that in, you get 188 newtons. And from there, if you wanted to go back and find the normal force, so the normal force by itself, you could just go plug in to this equation. Or if you had a different coefficient of friction or something like that, then you could go and plug that in. But really at... At this point, you have done all the physics and it's just algebra. Then you can plug your numbers in if you have a different case scenario or anything like that. The next problem, we have a book weighing 3 kilograms that is at rest on a horizontal table where the coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.2. Find the normal force, whether or not there is friction, the minimum force to cause the book to slide, the type of force that would cause it to stop if it was given a hard push, and the magnitude of that force. So again, we have another force problem. I want to draw a picture. So this is on a rough surface since there's friction. So at first... Just have the mass. It's not going to go anywhere. It's 3 kilograms. Our mu s is 0.4. Mu k is 0.2. And initially, if we just draw the free body diagram, there's only a normal force acting up and the weight acting down. So there's no horizontal forces. So if we want to find the normal force, we just sum up the forces in the y direction. We get normal minus the weight. It's not going anywhere, so put mass times acceleration, that's just zero. And then we don't know the weight, but we know the mass, so we can substitute mg in for the weight. And if we plug that in, 3 kilograms times 9.8 meters per second squared, we get 29.4 newtons. And then the next question is find whether or not there is friction and the answer is no. There's no forces in the x direction. It's not moving and there's no impending motion so it's not gonna move. Okay so you could have forces that are trying to get it to move and that's causing friction to kind of keep it from moving. But since there's no motion and no impending motion then we have no friction. However once we start to add some pushing force so some fx now there is friction in the other direction and if we're looking for the minimum force to get it started then it's still at rest and this is going to be static friction so we can add that to our free body diagram and it didn't tell me that it was just going to the right however i know if i want all of my force to overcome friction then i need to make it parallel to the surface or horizontal if i were to push up or down you That's going to, that upward force is not going to fight friction. So I know to be efficient, I need my force to be horizontal. And so I'll just write this. There's no friction at this part because that requires horizontal force. And now I want to find out what that horizontal force would need to be. So in the x direction, I just have this minimum pushing force. I'm just going to call it fx. So don't confuse it with the total forces in the x direction. So this is my pushing force. You may want to call it f sub push or something. And then minus my friction force. And if I'm pushing just hard enough to get it to move, then it's not moving yet. It's still at rest. If I push any harder than that, that's when it's going to start to move, so I can set my acceleration equal to zero. And so I can substitute, I don't know the friction force, so I can substitute in the formula for it. So you should always ask yourself, do I need to substitute in the formula for this force, or do I already have it? So it's going to be mu s times the normal force, and I just found the normal force in the previous part. So that's mu s times the weight, or mg. If I plug in my numbers, I get coefficient of static friction 0.4. And I could just put in my number for the weight, or I could put in mg again. It doesn't matter. Since I've written mg, I'll do 3 kilograms times 9.8 meters per second squared. And this is going to be 11.8 newtons. Okay, so this is the minimum force needed to overcome friction. And then if you were to give it a hard push and it started sliding, then once you have already got it moving, you're no longer pushing on it anymore. The only thing that's slowing it down or acting horizontally is going to be kinetic friction. So at this point, we've kind of already pushed it, and so our pushing force is gone, and now the only force we have is kinetic friction. So our free body diagram goes to this. And so since there's only a force of kinetic friction, if you wanted to find out... what that is well it's just going to be mu k times the weight and we know mu k is 0.2 Times again the weight. If you wanted to plug in for N, you could do that, or you could do MG. So if you plug all this in, you get 5.9 newtons. And then from there, if you wanted to calculate the acceleration and how fast it would slow down, you could do that using Newton's law. Okay, so we've gotten through projectile motion and a couple of force problems. Now we're going to move on to a collision problem. So we're going to use conservation of momentum. So we have a 1,000 kilogram car that travels north at 15 meters per second and collides with a 2,000 kilogram truck traveling east at 10 meters per second. If the cars stick together after the collision, find their velocity and direction after the wreck. Okay, so this is a two-dimensional problem. One's moving up, one's moving to the right. So we want to keep track of everything. So our truck is going to be moving to the right, and it is 2,000 kilograms. And then our car... is moving up car and it is 1,000 kilograms and then so this is kind of my before the collision state and then afterwards I'm going to draw them of mixed together so they've hit each other and now they're stuck and they're moving up at some up and to the right at some angle and we want to find the final velocity of both cars as they move together after the collision. So this is any type of collision. We want to use conservation of momentum. So this says that the total initial momentum is equal to the total final momentum of both the cars. And so if we can find one, we can find the other. We know the final momentum has to be exactly the same as the initial momentum. And we know a lot about the initial state. We know the mass, and we know their velocity. So we can find the initial momentum. So the magnitude of the initial momentum is just going to be the square root, so we're going to use Pythagorean theorem, of the initial x component plus the initial y component. And since we know the initial momentum has to equal the final momentum, as soon as we find that, then we know the magnitude of the final momentum. They have to be the same. So we just need to find these components. the initial momentum. Okay, so if we do that we'll do the X component PIX. Well, that's just the truck the initial momentum of the truck because the car isn't moving in the X direction initially. So this is going to be the momentum initially of the truck. So that's going to equal the mass of the truck times the initial velocity of the truck. And again notice my subscripts. I'm doing it in a way that makes sense to me. If you need to put them in a different order or call things other things like object A, object B, your subscripts are really there for you to organize the problem. Just label everything appropriately and keep it organized. And so this is going to be 2,000 kilograms times the initial velocity of the truck is 10 meters per second. And so that's just going to be 20,000 kilogram meters per second. And if we do P-I-Y, This is pretty easy because we only have one object moving in each direction, so this is the initial momentum of the car. This will be the mass of the car times the initial velocity of the car, which is 1,000 kilograms, moving at 15 meters per second, or 15,000 kilogram meters per second. So now we're ready to plug into our Pythagorean theorem. So this is... We're really wanting the magnitude of our final momentum, and we know that's equal to the magnitude of our initial momentum, which is given by the Pythagorean theorem, so we need to put 20,000 kg m per second squared plus 15,000 kg m per second squared. And that's going to give... 25,000 kilogram meters per second. Okay, so we have the magnitude of the final momentum because we know it has to equal the magnitude of the initial momentum. So now we can just use our formula for momentum. So we know the magnitude of the final momentum is just going to equal the total mass of whatever is moving. So in this case, the car plus the truck times the final velocity. And so if we want to find the final velocity, we just need to rearrange. So this is P, F. Divided by the total mass since it's a completely inelastic collision, so we do 25,000 kilogram meters per second Divided by the total mass 1,000 plus 2,000 kilograms is 3,000 kilograms and that gives 8.3 meters per second So now we want to find the direction of the final momentum So we know it's going to go up a little bit because the car is initially moving upward We know it's going to the right a little bit because the truck was initially going to the right. And the momentum in both of those directions must be conserved. So initially, it had some momentum. And that was a combination of the X component from the truck and the Y component from the car. And again, the momentum is conserved. So it's the same initially as it is finally. So if we find it at the beginning, we know that's the same for at the end. So we want to find this angle. And so we can just use these components that we found. So if we do that, we need to plug into inverse tangent. I'll separate this a little bit. And it's going to be, if we want the angle above our x-axis, our opposite over adjacent, that's going to be P-I-Y. over PIX and we've already found these. So this is tan inverse of 15,000 kilogram meters per second over 20,000 kilogram meters per second. Try to close that and this is going to give me an angle of 36.9 degrees. Okay, so we have another momentum problem, another collision problem. Here there are three box cars connected together moving at a steady 24 meters per second. They hit a fourth box car that is at rest and stick together. Find the speed of the four cars and how much kinetic energy is dissipated in the collision. Okay, so it sounds like a completely inelastic collision. Initially, the box cars are moving at 24 meters per second. And we want to find VF, so the speed at the end. And notice, anytime we're finding momentum, momentum is mass times velocity, but we're not given the mass. But we're still, anytime you do that, you don't want to let it fluster you. You want to keep solving the problem as if you did know it and see what happens. So this is a collision problem. It's a conservation of momentum problem. And really, anytime you use a conservation law, you want to draw a picture of what happens in the beginning versus what happens at the end. So initially, we have three boxcars moving at some initial speed, and they're going to hit a fourth boxcar that's not moving. And then our final state after the collision is they've all hit each other, and now they're all moving together. some final velocity. Okay, so now if we use conservation of momentum, we know the initial momentum of the system is equal to the final momentum of the system. Initially, we have three boxcars moving to the right and that's it. So we have, I'm gonna call this, so I'm gonna say each boxcar has a mass m. And so initially I have three of these boxcars, 3m, moving at initial speed vi. Okay, and then at the end, I have four boxcars, so 4m moving at speed vf. And then you can immediately see what happens. The mass is just going to go away. It doesn't matter because it's the same. If it was different masses, you couldn't do this. So then if I want to solve for vf, I just need to divide by 4. So this is going to be 3 fourths of vi, or 3 fourths of 24 meters per second. which is just 18 meters per second. And now it asks how much kinetic energy is dissipated. Again, kinetic energy is something where you typically need the mass, so we want to see what happens. So if we want to know how much is dissipated, we want to know the change relative to what was initially there. So we want to know how much we lost. So that's going to be the change in kinetic energy, so how much was lost, over what we started with. So this will give us a fraction of the energy that was lost. So if we do that, change in kinetic energy, that's just kf minus ki, your final minus your initial. So at the end, there was one half. And then the mass, well, there were four masses times VF squared minus initially, it was one-half times three masses VI squared. So this is our one-half MV squared at the end minus one-half MV squared in the beginning. And then this is just, again, one-half 3M times VI squared. Okay, so again, you notice I can get... rid of all these halves because they're in every term and there's a mass in every term so I can get rid of it and that simplifies a good bit so now I have 4 vf squared minus 3 vi squared all over 3 vi squared okay so be careful do not try to cancel these two out okay because you're subtracting on top and there's not a 3 vi squared in every term you can't get rid of that just yet Okay, so we need to plug it in. But one thing we could do is we could separate this. So I could separate this into two fractions and have 4VF squared over 3VI squared. And then if I separate this, then 3VI squared over 3VI squared, that's just one. Okay, so if I plug this in, I get 4 times 18 meters per second. I have to square that, divided by... 3 times 24 meters per second. Square that and then subtract 1. And I get minus 0.25. Okay, and so this is a decimal. What it means is minus 25%. So there was 25% of the initial energy was lost in this collision. So our next problem, we have a conservation of energy problem. So we have a bicycle that is moving at 10 meters per second downhill when one of its 2.16 kilogram wheels comes off when it is 50 meters above the bottom of the hill. The diameter of the wheel is 85 centimeters and we want to find the speed of the wheel when it gets to the bottom. Okay so I'm going to write everything that I know. Okay so vi is 10 meters per second. The mass of the wheel is 2.16. Kilograms and since it's going to be rotating I know I'm going to treat it like a rigid body So I'm going to use capital M. The height above the ground is 50 meters the diameter of the wheel is 0.85 meters so that means the radius which is usually what we use is half of that or 0.425 meters and just to draw a quick sketch of what we've got going on initially We have this wheel that's on a hill. So it is rolling. So when you know from being on a bicycle that your wheel is not only spinning, it's also translating. So it's rolling without slipping down the hill. And it's going to come off. And then at the end, it's going to be at the bottom of the hill. And it's still going to be rolling. So we need to use conservation of energy. And the way you know to use conservation of energy, if it doesn't explicitly tell you to, is notice what's changing. So one, we know it's probably going to speed up. If you've ever rolled down a hill, you know you change your speed. So that's kinetic energy changing or increasing. You're also changing your height. So you're changing your gravitational potential energy. So when you notice that height is changing and speed is changing, you should think, oh, potential energy is changing. Kinetic energy is changing. Let me use conservation of energy. which tells me the energy I start with has to be equal to the energy I end with. Okay, so initially, the wheel is not only rolling without slipping, but it's off the ground. Okay, so it has potential energy as well. So I'm going to write the kinetic energy. It has 1 half mv i squared, so initial translational kinetic energy, plus 1 half i omega squared since it's spinning or rolling. And then it has gravitational potential energy because it's some amount off the ground, 50 meters. And then at the end, it's still going to be rolling. Now it's going to be rolling probably a little faster if we had to guess. So this should be 1 half mvf squared and still rolling 1 half i omega f squared. But now it doesn't have any potential energy because it's on the ground, which we will call y equals 0. So now we need to substitute. in for i and omega and try to pull out our linear speed since that's what we're looking for okay it didn't ask for angular speed so if we do that we have one half mbi squared plus one half the moment of inertia of a wheel well a wheel is kind of a thin walled cylinder and we're going to neglect the little spokes because they don't have much mass so really this is just a thin walled cylinder and that has moment of inertia of nr mr squared okay and then so that would either be given to you or you would look it up And we can also use the fact that omega is v over r from our relationship v equals r omega. And so this is going to be v initial squared over r squared plus mgh. And I'm going to do the same thing on the other side. So plus one half. m r squared and then now it's going to be vf squared over r squared so make sure you put your initial and final on both of those and now if you notice we have a mass in every single term so that goes away and then our r squareds here there's one on top and bottom so it just goes to one so that disappears and we're left with something much simpler so we have vi squared over two plus vi squared over two plus gh equals vf squared over 2 plus vf squared over 2. Okay, so this is kind of like half vi squared plus half vi squared, so that's going to equal just vi squared. So half and a half is a whole plus gh equals vf squared. And so now we just want to solve for vf. So we're going to take the square root of both sides. I'll flip which side it's on. Okay, and then we can just plug in, so square root of the initial speed, 10 meters per second, square that, plus 9.8 meters per second squared. Let me move this down a little bit, and our height's going to be 50 meters. So if we plug that in, we get 24.3 meters per second. And so notice that we didn't even need the radius or the diameter because it canceled out. We didn't need the mass. And so for a lot of these problems, it ends up happening that if you don't have a spring, your mass is going to cancel out because there's a mass in both of your kinetic energy terms and your potential energy term. And so if you're not given it, just like in previous problems, keep going, keep solving the problem as if you did know the mass, and then you'll see that maybe it cancels out. The next problem, we have a diver with her arms and legs straight. She has a moment of inertia of 18 kilogram meters squared. She tucks her arms and legs in to decrease her moment of inertia to 3.6 kilogram meters squared. If she spins fast enough to make two revolutions in one second, find the number of revolutions she would have made in 1.8 seconds if she had kept her arms and legs straight. So the first thing you should notice here is that your moment of inertia is changing. So initially, it's 18 kilogram meters squared. And at the end, when she tucks in, it goes down to 3.6 kilogram, if I can write, kilogram meter squared. And so any time your moment of inertia changes, you should think, use conservation of angular momentum. So the only other thing that I know is omega F, she makes two revolutions in one second. So this is two revolutions in one second. And now I can use conservation of... angular momentum, which says the angular momentum we start with has to equal the angular momentum we're left with. And the angular momentum for a rigid body, which is what she is, is I initial times omega initial equals I final times omega final. So angular momentum is L equals I omega. So this looks a lot like linear momentum. P equals MV. You just replace mass with moment of inertia. linear velocity with angular velocity. Okay. And now we know I initial, we know I final, and we know omega final. And we're looking for how many revolutions she would have made in 1.8 seconds if she had kept her arms and legs straight. Okay. So it's asking how many revolutions would she make if her initial velocity, angular velocity didn't change. So we need to be able to solve for it. Okay. So this is, we're just going to divide by... the initial moment of inertia, so move it to the other side, times omega f. And now this is going, if we leave our omega f in revolutions per second, this is going to give us our initial angular velocity in revolutions per second. But that's okay because it's asking for the number of revolutions. Okay, so you have to be careful, know when you can and can't leave your angular quantities in revolutions versus radians. So if we do this, this is going to be the final moment of inertia, 3.6. kilogram meter squared over 18 kilogram meter squared times this is going to be two revolutions per second so what pops out should be 0.4 revolutions per second okay and now we know how long we know our time we want to know how many revolutions she does in 1.8 seconds and we know how fast she's spinning so our What we're looking for is delta theta. So delta theta is a measurement of angle that you go through. And so revolution is a type of delta theta. So this is just going to be our omega times time. Okay, so you could, this comes from omega equals delta theta over time. Okay, speed equals distance over time. You could think of it like that. And so if we plug in, we go 0.4 revolutions per second for 1.8 seconds, and that gives us 0.72 revolutions. The next problem, we have some rotational equilibrium examples. So for each object, find the magnitude of the unknown force needed for rotational equilibrium about the pivot. So our pivot is shown by this little green dot, and in all of these we have an unknown force F. And we want to find its magnitude if these rods, or whatever they are, if they're to be in equilibrium. So in order for them to be in rotational equilibrium, that must mean that the sum of the torques is zero. So we want to sum up the torques for each of these. So in the first one, in A, we have two forces that are going to cause a rotation. So the first one, I'd like to look... See which ones are going to cause positive rotations and which ones are going to cause negative So I like to put all my positives first So if we just look at our unknown force F and imagine that's all that's there if we push up at the right end That's going to cause it to rotate about this green point. It's going to rotate counterclockwise. So that's positive so I'm going to go ahead and list it and the way we find each individual tort is RF sine phi. Okay, so we need to put r, and r is the distance from the axis of rotation. So the distance from the axis of rotation, that's just the distance from f to the pivot, which is given as 0.1 meters. Okay, and then our force, well we don't know that, so we'll just write f, and then times sine phi. Well, phi is the angle between r and f, so if you were to draw r, it's going from the force to the left towards the pivot and then if you go back to the same point and draw the force it's going to be up it's going to be straight up well this angle is just 90 degrees and if you look at all of these all of your r values for every force in all of these examples it will be going left or right all of your forces will be going either up or down so our phi for all of these will be 90 degrees. And so sine of phi just goes to one. Okay, so that simplifies things a great deal. Okay, so now we're ready to look at our next force. This 50 newton force, if you imagine it's the only thing there, it's going to cause a clockwise rotation. So it's going to be negative. So you have to remember to put that in as your first step. And so it is 0.3 meters from the pivot and it is a 50 newton force. Okay, and that needs to sum to zero. Okay, so we ran out of room. And now if you solve for F, you need to move this term to the other side and make it positive. Okay, so that'll be 0.3 meters times 50 newtons. And then we want to divide by 0.1. Okay, so if you do that, you get a force of 150 newtons. So one thing to notice. This kind of makes sense. This force is closer, so it's a third of the distance, and so it has to be three times as strong to cause the same rotation as this 50 newton force that's three times further away. So if we move on to our next one, we're going to do it the same way. So we sum up the torques. We know the torques on the right are going to cause counterclockwise rotation, so they're going to be positive. So I'm going to go right to left. So again, This unknown force is going to be, now it has moved, since the pivot moved, our r changed to 0.3. times our unknown force F and now our 50 Newton force is the one that's 0.1 meters away okay and this sums to zero and so now again we just solve for F doing it the same way as we did before so 0.1 meters times 50 Newtons divided by 0.3 and this is going to equal 16 0.7 newtons okay so now same sort of thing it is three times further away so the force can be a third as big to have the same torque as this 50 newton force that it's much closer. So move on to part C, sum up the torques. So our unknown force is 10 centimeters from the pivot. So it's back to 0.1 times F. And then there's two forces to the left of the pivot. So that's going to cause clockwise rotations. So they're negative. So the 40 Newton force is 20 centimeters away. So 0.2 times 40 Newtons. And then the 50... Newton force is 20 plus 10 so that's 0.3 meters times 50 Newtons again that sums to zero so now we have F equals we got to move both of these to the other side So 0.2 meters times 40 newtons plus, it's going to become positive, 0.3 meters times 50 newtons, all divided by 0.1 that's in front of F, and that equals 230 newtons. Okay, so now we have more force on the left, so we need a greater force, greater than our 150, to cause the same rotation. So we'll do part D. So part D looks just like part B except we have a 40 Newton force at the pivot. So just kind of copy that. We have 0.3 meters times F and then minus 0.1 meters times our 50 Newton force. And then the question is, what is this 40 newton force going to do? Well, this force is applied at the pivot or at the axis of rotation. So the R for the 40 newton force is just zero. So it goes away. It doesn't cause a torque. Okay, so this sums to zero. And then if you look at the math, it's going to be exactly the same as part B. And so this is 16.7 newtons. And so that force, that 40 newton force being at the pivot, doesn't do anything in terms of rotation. So the last one, we have our unknown force 0.1 meters away, and then we have a 100-Newton force that's going to cause a clockwise rotation, it's a minus, and it is 15 centimeters away, so 0.15 meters times 100 Newtons. And now our 50-Newton force has jumped. to the bottom and now it's facing down so that's just going to cause a rotation in the opposite direction which means it's going to be positive now because it would cause counterclockwise rotation and it is 15 plus 15 centimeters so 0.3 meters times 50 newtons and again that sums to 0 so we get f equals 0.15 meters times 100 newtons minus 0.3 meters. When we move this last term to the other side, it goes negative, divided by 0.1. And so for this last one, if you plug it in, you actually get 0 newtons. And so this 50 newton force is twice as far away as this 100 newton force from the pivot. And 100 newtons is twice as big as 50 newtons. So these are actually already canceling each other out. If you were to add any force to the right side, you would cause it to not be in rotational equilibrium, and you would cause counterclockwise rotations. So you actually don't want to apply a force here. So in this last problem, we have a 0.3 kilogram mass that oscillates on the end of a horizontal spring with an amplitude of 0.15 meters and period of 0.45 seconds. And we want to find the velocity at equilibrium. the total energy, the spring constant, and the max acceleration. So this is an example problem from periodic motion, and it really asks you for really all of the different things that you could ask for in terms of an object that's oscillating back and forth. And so if we write what we know, which is really important for problems like this where there's so many different types of quantities, and they're all changing at different locations. We have the mass is equal to 0.30 kilograms. The amplitude, which is a type of distance, or the maximum distance away, 0.15 meters. And the period, so capital T. 0.45 seconds. So the velocity of an object oscillating on a spring at some point is given by the equation square root of k over m times square root of a squared minus x squared. And you could combine these radicals, okay, so you could have them all up under one, but it's written like this for a reason. And you may have seen it derived in a textbook or in class, but it should be given to you. This square root of k over m out front is actually just equal to omega. And that's good because we don't yet know k. And so if we don't know v and we don't know k, we can't solve for this just yet. And then we want to find the velocity at equilibrium. Well, that's when x equals zero. So this is going to simplify even more. So if we put in omega for square root of k over m, we get omega times, well, if x is 0, this becomes square root of a squared, which is just a. So this simplified a lot. And then omega is just 2 pi times the frequency, or 2 pi over the period. So frequency is 1 over the period. And so we can write this in terms of things we know, 2 pi. times the amplitude over the period. Okay, so I'll get rid of this. And so now since we know amplitude and we know period, we can solve this. And not only that, but the velocity at equilibrium we know is going to be the maximum velocity. So I'll put this as Vmax, and this is 2 pi times the amplitude, 0.15 meters, over the period, so 0.45 seconds. And And this comes out to be 2.1 meters per second. So now we want to find the total energy. And we know the energy looks different in terms of an equation depending on where you're at. If the object is at the amplitude, so the spring is fully stretched or fully compressed, then all of the energy is potential. But if it's at equilibrium, it's all kinetic. And so since we know the velocity at equilibrium, we know that when it's at equilibrium, the energy is all kinetic or 1 half m V max square so we're going to use what we just found to find the total energy so this is 1 half times 0.3 kilograms times 2.1 meters per second square that and we get total energy of 0.66 joules so now we want to find the spring constant and so we can go through look at our equations and we have one for the the period of oscillation is just 2 pi times square root of m over k. If you weren't given this one, but you were given the equation frequency equals 1 over 2 pi square root of k over m, then you could use the relationship that period is 1 over frequency, and you could just flip everything. So this is just a reciprocal relationship. So make sure, if you're not given every single equation, make sure you can use the ones that you do have. to get what you want. And so now if we solve for K, we need to square both sides, move K up and T squared down. Okay, so that's going to give us with some little algebra, 2 pi over T all squared times M. Okay, so you may have to go through that algebra, just go slow. And you get 2 pi over the period, 0.45 seconds squared, times the mass, 0.3 kilograms. and that gives us a spring constant of 58 Newtons per meter. Okay, so now we want to find the max acceleration. And to find the acceleration, we just use Newton's laws, and we get minus K over M times X. Okay, so this is derived very easily just using Newton's laws. force in the x direction is the spring force, so minus kx, and set that equal to mass times acceleration. And we know the acceleration must be max when the force is max, and the force of a spring is maximum when it's stretched or compressed the most, or at the amplitude. So A max is equal to minus K over M times the amplitude. And I'm going to put in minus the amplitude because that just means the acceleration will be going to the right, so it's positive. But the magnitude of the maximum acceleration occurs at either plus or minus the amplitude. So if I plug that in, I get minus 58 newtons per meter for K over 0.3 kilograms. times minus 0.15 meters and I get 29 meters per second squared Okay, so that's it for the problems we've gone through. Projectile motion and kinematics, forces, conservation of energy, conservation of momentum, conservation of angular momentum, some rotational equilibrium and rotational dynamics, and now oscillations and periodic motion. So now we're going to move to a few concepts that are typically talked about at the end of the semester. You may not solve that many problems on them, but they're good to know. So the first concept I want to review is... periodic motion since we just finished that problem. So periodic motion. And for this, we mostly focus on the example, much like from the problem where we have an object that's attached to a spring. So some mass attached to a horizontal spring and it's oscillating left and right. And it's oscillating about some equilibrium position, x equals zero. So it's going negative and positive. And so we want to know its motion at every different point in its oscillation. So we're going to look at each point. So when it's at the amplitude, so when x is plus or minus the amplitude, that's going to mean that the spring is stretched or compressed its maximum amount. So the force is going to be maximum. And then if the force is maximum, that's going to mean that the acceleration is also maximum. So these are also some ideas that can be useful in solving problems. So it's really Good to understand. So we also know at the amplitude that your velocity is zero. Okay, so if your spring stops stretching or stops compressing, that means that you stopped. And so if you stopped, that means your velocity is zero. If you kept going, then the spring would continue to stretch or continue to compress and the force and acceleration would increase. And so you would still be moving. So you know whenever these two are maximum or you're at the amplitude that V must also be zero. Okay, and If V is zero, that means you have no kinetic energy, but because the spring is stretched, you have some elastic potential energy, so your total energy is all potential. So it's one-half kx squared, but our x is just going to be the amplitude. So here, the energy is all elastic potential. So that's pretty much all the quantities we're interested in, so now let's go back. and look at what they are when we're at x equals zero, which is another way of saying at equilibrium. Okay, so we're at our equilibrium position, and that means that our force, our spring is neither stretched nor compressed, so the force is going to be zero. By Newton's second law, if the force is zero, acceleration is also zero. Well, up until this point, when we stretched it or compressed it and then released it, this object, it's been accelerating the entire time it's been on its way to equilibrium. And so now it's as fast as it's going to go. So V equals V max. Okay. Once it passes equilibrium, it's going to overshoot. Now the spring is going to stretch or compress in the opposite direction and start to slow it down. So we know velocity is max at equilibrium. Well, If velocity is max, that means it definitely has kinetic energy. And if the spring is neither stretched nor compressed, all this elastic potential energy is gone. It's turned into kinetic. So our energy is now 1 half mv squared, where the velocity is maximum. Okay, so now it's all kinetic. And now we can look at the case. of anywhere else. Okay, so those are two special locations, plus or minus the amplitude and equilibrium. But if we wanna know these quantities anywhere else, one, we may want to use what we know about the motion at the amplitude or at equilibrium to solve for some other quantities. But once we have those and we wanna find these quantities anywhere else, well then the force is just given by the spring force. Okay, so at any point the spring is Stretched or compressed, it's just equal to minus kx. So here, I guess I can say at the amplitude, that would make it f equals minus ka. Well, that means the acceleration, if we set this equal to ma and solve for the acceleration, it would be minus k over m times x. So for the acceleration, Maximum would be minus k over m times a or minus a. So I'll put minus a so it's positive. And then our velocity. This would be square root of k over m times square root of a squared minus x squared. So this is another one that's derived, and it's derived from an energy equation. So what is that equation? Well, we said at equilibrium, it's all kinetic. At amplitude, all of the energy is potential. Anywhere else, it has a little bit of both. So the energy is a little bit of 1 half mv squared plus 1 half. kx squared. Okay? And then if you were to set this equal to the total energy at the amplitude, one half ka squared, since energy should be conserved, these numbers should be the same, then you can solve for this expression for the velocity at some point. Okay? But we won't do that here. know that the velocity at some point is given by this expression and the energy at any other point other than equilibrium in the amplitude is some combination of kinetic energy and elastic potential energy. The next idea I want to talk about is pressure and Pascal's principle in fluid mechanics or fluid statics. So typically when we look at Pascal's principle we have this equation pressure equals the initial pressure plus Rho gh. Okay, so just to define what these are, this is the pressure at some point of interest, pressure that you're looking at. So I'll put pressure at point of interest. And then everything to the right is what that pressure depends on. Well, it depends on some reference point or some initial pressure. Okay, so this may be the pressure at the top, and you're looking for the pressure at the bottom. And then this other term has the density of the fluid that you're in. And then G is just gravitational acceleration like we're used to. And H, this is the height or the distance between... this reference pressure and the pressure you're looking at. Okay, so let me draw a picture. If I can. We have some fluid. And typically we take this reference pressure to be the pressure at the top. Well, the pressure at the top of the fluid, well, that's where the air is meeting the surface of the water. So what's causing the pressure? Well, the pressure is... caused by the perpendicular force from the air pushing down on the water. So the pressure here is just air pressure, or what we would call atmospheric pressure, the pressure from the air in the atmosphere. Well, that means if we want the pressure anywhere down here, Some other P, so this would be P-naught. To get the pressure at this point, we need to add the pressure from the air that's pushing on the top of the fluid plus all of the pressure from the fluid that's above this point. Okay, so you can think about this in terms of forces. The air is pushing down, that's applying some pressure. And now as you go lower, you have more and more water or more and more fluid above you that's adding pressure down on top of you. of you okay so the flesh the pressure depends on how deep your pressure depends on depth The easiest way to think about that is the deeper you go, the more fluid above you that is pushing down on you, and that's more you're having to support. So pressure increases based on the depth. The important thing is if you look at the equation, it only depends on depth. Okay, so once you know the density of the fluid, the only thing that matters is the pressure at the top and how deep you are. Okay, the other thing that we get from Pascal's principle, and this is really the heart of Pascal's principle, is that increasing the pressure at the top, so the initial pressure, this is going to increase the pressure at all other points. So at all other depths, H, by the same amount. Okay, so the easiest way to think about this is, say you have a bucket. So you're holding this bucket. And then you get on, maybe you get on your friend's shoulders. Okay, well you feel the pressure. pressure from what's ever in the bucket. Maybe we put some fluid in here. If we go a little bit lower, your friend feels the weight of the bucket plus the weight of you on their shoulders. So that's kind of like the point in this fluid. It's having to support not only the air, okay, so you are the surface at the top just supporting the air. Your friend is a little bit lower. It's having to support whatever's in the bucket plus it's having to support your weight. Well now imagine that we add more water. Okay, so if we add even more water to the bucket, well you feel however much water you added. So if you added 10 pounds of water, now your pressure increased by that much. So that force times the area. Your friend felt the same increase. So he was supporting your weight plus the bucket. When you added 10 pounds, that added 10 pounds to what he felt. So if you increase the pressure at the top, all the points below it will experience that same increase. So if I add 10 pounds of pascals of pressure at the top, all the points below the top will feel an increase of 10 pascals of pressure. The last concept we're going to talk about is Archimedes'principle. And Archimedes told us that there are buoyant forces when we're in fluid. So if we're partially or completely immersed in a fluid, There's going to be a buoyant force pushing us up, and the magnitude of that buoyant force is going to equal the weight of the displaced fluid. So this doesn't look like a weight, but if you think about it, weight is equal to mass times gravitational acceleration, so that's our g. And mass is also, or actually let's use our definition of density. Density is mass per unit volume. So if we solve this for mass, we get mass is equal to row times volume. And then we plug this in here. And we see this is just another way of writing the weight. So this is the weight of the fluid that you displaced. So if you want to talk about how to get an object to float, in order to float, that object needs to... displace its weight in water or in the fluid so let's use water as an example it needs to displace its weight in that fluid and so if you float when you're in a swimming pool it's because you have displaced enough water equal to your weight okay and so if you wanted to talk about how a ship floats so we know that ships are far more dense than water and if we just throw a chunk of metal in the water it's most likely going to sink so how do ships float Well, ships and their hulls, they are designed where they're hollow. They only have air in them. And so combined, the amount of volume that it displaces in water is equal to the ship's weight. So it displaces water equal to, maybe I'll put its weight. plus everything on board. Okay, so it needs to be able to displace enough water not only to support itself, but all of the people that you put on it. So everything on board. Maybe that's one word. I don't know. Okay, and so the other question would be, how do you know how much of the object is going to be above the surface? We don't want... our ship to float where the top of it is at the top of the water. We want a good portion of the ship to be above the surface so that we can walk around and not be underwater. Okay, and so the amount above the surface, it's the amount of an object floating above, well this depends on the amount of that object needed to equal its weight in water. So it depends on the volume needed to displace its weight in water. Okay, so as soon as an object displaces its weight in that fluid, well, if you look at a free body diagram, Its weight is going to be pulling down and the buoyant force is going to be pushing up. As soon as these are equal it stops moving. So the acceleration is zero and whatever is left That's what is going to be above the surface. So you can solve for how much volume is needed in that object depending on its density in order for your buoyant force to equal the weight. And so that's just using Newton's laws and solving for the volume. As soon as your volume gives you a buoyant force equal to your weight, you could subtract that much volume from your total volume and whatever's left over that's going to be on top of the surface. So that's it for our concepts and our review problems. I hope this helped. Good luck on your finals. And if you have any questions, feel free to let me know and I'll see you in the next video.

time to max height, velocity at its highest point, acceleration at its highest point, and how long the ball is in the air and how far away does it hit. So this is a projectile motion problem and it really covers just about everything someone could ask for when solving a problem like this. So the first thing you want to do is write what you know and we're given a vector. So we're given the initial velocity, 23 meters per second.

and it was given at an angle of 58 degrees, so this is not the scale. And so as soon as I see this, the first thing that I want to do is get my components of this initial velocity vector, but that turns out to be the first thing that I'm asked for anyway. So horizontal and vertical components just refer to the x and y components of the vector. So vh, so v horizontal, or v not x, really the same thing. So if I draw my right triangle, then I have the magnitude of the vector on the hypotenuse.

The vertical component is y, the horizontal component is x, and I like to just go ahead and label these. This is going to be my opposite side, opposite of this angle. This is my adjacent side, and as always, your vector is the hypotenuse.

If I want v-x, that's going to be the adjacent side. So adjacent. is cosine, so use your SOHCAHTOA, so this is going to be V naught cosine of 58 degrees, or 23 meters per second, it's cosine, and that's going to give me 12.2 meters per second. And then once you know one component, you immediately know the other one is just the opposite trig function, so the vertical component, or V naught y, it's opposite, so it will be found using the sine function. So v-naught sine 58, so again 23 meters per second, sine of 58 degrees, and that gives 19.5 meters per second.

So the first part was just vector math. Now we're actually getting into the physics, so we want to find the max height. So I want to start listing all of the things that I know. I know v-naught x, that means I also know vx because it's not going to be accelerating. in the x direction.

So I can just write that ax is zero and ay is minus g since this is a projectile motion problem. And then I just found v not y. So the first thing that I notice is I'm looking for max height, so that's delta y, and I don't have time.

Okay, so if I don't have time, the first equation that I'm going to go look at in terms of my kinematic equation Is this one. The one that does not have time in it. Okay, and Vy is going to be zero at the top at max height, so it simplifies a bit.

And then if I solve for delta y, I get minus V0y squared divided by minus 2g. So my minus signs are just going to go away. And then I plug in what I got for V0y, 19.5 meters per second, square that, divide by 2 times 9.8 meters per second squared. And I get 19.4 meters.

So now we want to find the time to max height. And we have even more information, so I could add delta y to my list of knowns. Let's put that here. But it turns out I don't need delta y.

If I want to find the time to max height, I can just use my short kinematic equation. So vy equals v not y. Minus gt.

I know v not y because I found it in my first step I know v y is going to be 0 at the top just like it was in the second part and then I just can solve For time so if I do that I get time and again if you set v y equal to 0 that makes That is what makes this v max t max so if you solve you get minus v not y Divided by minus g and you should notice those minus signs cancel They would need to because your time should be positive. So you get 19.5 meters per second divided by 9.8 meters per second squared and we get 2 seconds. Okay, now the next few questions are a little bit easier. Some things you should just know.

So the velocity at the highest point. So this is one that you should just be able to look at and tell. So if this is going in a parabolic trajectory that it does in projectile motion then at the top it's only moving in X.

So we know Vy is zero at the top but that means that all of our velocity is in the X direction and the velocity in our X direction does not change because the acceleration in the X direction is zero. So we know that the velocity at the top must just be the initial velocity in the X direction that we started with. So I'll put at the top, that's just equal to v-naught x, which we found to be 12.2 meters per second.

The y component of velocity went away, but the x component stayed there, and it stayed at its same value. So the next one, what is the acceleration at the highest point? Well, we know this.

There's no acceleration in the x direction. There's only acceleration in the y direction, so a is just minus g. So now we want to find how long the ball is in the air.

So if we know how long it takes the ball to get to the top, It's going to spend the same amount of time going down. So our total time is just twice the time to max height, which is just going to be 2 times 2, or 4 seconds. And then finally, we want to see how far it goes.

So we really only have one equation in x. So delta x equals v0x times time. Once we know how long it's in the air, so t total, We already know how fast it's going in x, and so we can use this equation to figure out how far it goes. And so this is going to be 12.2 meters per second. It's going to be going that fast for 4 seconds, and so that means it's going to go 48.8 meters.

So now we're moving to a force problem, a Newton's second law problem. We have a 500-Newton crate that is pulled upward by a rope at 30 degrees above the horizontal. How strong does the upward pull need to be to keep the crate moving at constant velocity?

The coefficient of kinetic friction is 0.4. So I always start problems, especially force problems, with a picture. So I have a crate.

It's going to be pulled upward at an angle 30 degrees above the horizontal. So if it says something like above the horizontal, wherever you put your force, just draw a little dashed horizontal line or something and then draw that angle. And then it's on a rough surface because there's friction.

So I like to indicate that so I don't forget friction. And now that I have that, I want to draw a free body diagram. So I want to know all the forces that are acting on this crate.

So all objects that we deal with have mass. So they're going to have a weight. It's touching a surface.

So that means there's going to be a normal force perpendicular to the surface. I have this force that's up and to the right at 30 degrees. And so as soon as I see this, I know I'm going to need to break. this force into its components.

So maybe I call this a tension force. And so this tension force is going to try to move this crate to the right. And that must mean that friction is going to act to the left.

And since it's asking me about when the crate's moving at constant velocity, since the crate is moving, it's going to be kinetic friction. And it's also the coefficient that it gave me. So now I want to start summing up the forces and I always start friction problems by summing up the forces in the y direction because I know they're going to depend on the normal force. And so if I do that, I have a positive normal force. And then I need to find the y or upward component of this tension force.

So if I just draw my triangle somewhere else, put my angle, this is the magnitude. And so the y component will be on the opposite side, the x component will be on the adjacent side. And so that means ty is going to be t sine of 30. So plus, and it's upward, so it's positive.

So t sine of 30 degrees. And then the weight is down, so it's going to be negative. So minus the weight.

And I'm not going to substitute mg in for my weight because it told me the weight in newtons. And this is going to equal mass times acceleration in the y direction. But I know that's going to be zero because it's not going up off the ground or through the ground. So this is just equal to zero. And now I can solve for the normal force.

I'm just going to move my weight over. It'll become positive. And then subtract this y component of tension, T sine 30. Okay, and so I know the weight. I could go ahead and plug that number in. I don't know what the normal force is because I don't know tension.

So these are two unknowns. So I'm just going to leave that there for right now. Move on to my x direction some of the forces in x now I have the x component So just T cosine of 30 degrees of this pulling force and then I have friction working the other way to minus FK And again, it's going to be moving at constant velocity So that tells me ax must be zero. It's not accelerating. Okay, so I read that in the problem And now I want to substitute in my formula for FK since I know the normal force.

So T cosine of 30 degrees minus mu K times N equals zero. And now I can take this N and I can substitute it in. OK, so I can put it in there.

So I'll give myself a little bit more room. So I'm going to write T cosine 30 degrees. minus mu k and I'm going to put all of n inside parentheses so the weight minus t sine 30 and that equals zero. Okay so now I want to solve for the pulling force or the tension so I just need to do a little bit of algebra so I need to distribute my coefficient of friction to these two terms and then get it by itself.

Okay so if I do that I have t t. cosine of 30 minus mu k times the weight minus minus is plus mu k t sine 30 and that equals zero now if i factor out my pulling force my tension i have t times cosine 30 so that's the first term and then plus mu k sine 30 and then i'm going to move this term to the other side let me go ahead and do that make it positive so mu k times the weight and now to isolate tension i just divide by everything in front of it so i get mu k times the weight over cosine 30 plus mu k sine 30 if i plug all that in i get 0.4 times 500 newtons you all over cosine 30 plus 0.4 times sine of 30. And if you plug that in, you get 188 newtons. And from there, if you wanted to go back and find the normal force, so the normal force by itself, you could just go plug in to this equation. Or if you had a different coefficient of friction or something like that, then you could go and plug that in. But really at...

At this point, you have done all the physics and it's just algebra. Then you can plug your numbers in if you have a different case scenario or anything like that. The next problem, we have a book weighing 3 kilograms that is at rest on a horizontal table where the coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.2.

Find the normal force, whether or not there is friction, the minimum force to cause the book to slide, the type of force that would cause it to stop if it was given a hard push, and the magnitude of that force. So again, we have another force problem. I want to draw a picture. So this is on a rough surface since there's friction.

So at first... Just have the mass. It's not going to go anywhere. It's 3 kilograms.

Our mu s is 0.4. Mu k is 0.2. And initially, if we just draw the free body diagram, there's only a normal force acting up and the weight acting down.

So there's no horizontal forces. So if we want to find the normal force, we just sum up the forces in the y direction. We get normal minus the weight. It's not going anywhere, so put mass times acceleration, that's just zero. And then we don't know the weight, but we know the mass, so we can substitute mg in for the weight.

And if we plug that in, 3 kilograms times 9.8 meters per second squared, we get 29.4 newtons. And then the next question is find whether or not there is friction and the answer is no. There's no forces in the x direction. It's not moving and there's no impending motion so it's not gonna move.

Okay so you could have forces that are trying to get it to move and that's causing friction to kind of keep it from moving. But since there's no motion and no impending motion then we have no friction. However once we start to add some pushing force so some fx now there is friction in the other direction and if we're looking for the minimum force to get it started then it's still at rest and this is going to be static friction so we can add that to our free body diagram and it didn't tell me that it was just going to the right however i know if i want all of my force to overcome friction then i need to make it parallel to the surface or horizontal if i were to push up or down you That's going to, that upward force is not going to fight friction. So I know to be efficient, I need my force to be horizontal.

And so I'll just write this. There's no friction at this part because that requires horizontal force. And now I want to find out what that horizontal force would need to be. So in the x direction, I just have this minimum pushing force.

I'm just going to call it fx. So don't confuse it with the total forces in the x direction. So this is my pushing force.

You may want to call it f sub push or something. And then minus my friction force. And if I'm pushing just hard enough to get it to move, then it's not moving yet.

It's still at rest. If I push any harder than that, that's when it's going to start to move, so I can set my acceleration equal to zero. And so I can substitute, I don't know the friction force, so I can substitute in the formula for it. So you should always ask yourself, do I need to substitute in the formula for this force, or do I already have it?

So it's going to be mu s times the normal force, and I just found the normal force in the previous part. So that's mu s times the weight, or mg. If I plug in my numbers, I get coefficient of static friction 0.4.

And I could just put in my number for the weight, or I could put in mg again. It doesn't matter. Since I've written mg, I'll do 3 kilograms times 9.8 meters per second squared. And this is going to be 11.8 newtons.

Okay, so this is the minimum force needed to overcome friction. And then if you were to give it a hard push and it started sliding, then once you have already got it moving, you're no longer pushing on it anymore. The only thing that's slowing it down or acting horizontally is going to be kinetic friction.

So at this point, we've kind of already pushed it, and so our pushing force is gone, and now the only force we have is kinetic friction. So our free body diagram goes to this. And so since there's only a force of kinetic friction, if you wanted to find out... what that is well it's just going to be mu k times the weight and we know mu k is 0.2 Times again the weight. If you wanted to plug in for N, you could do that, or you could do MG.

So if you plug all this in, you get 5.9 newtons. And then from there, if you wanted to calculate the acceleration and how fast it would slow down, you could do that using Newton's law. Okay, so we've gotten through projectile motion and a couple of force problems.

Now we're going to move on to a collision problem. So we're going to use conservation of momentum. So we have a 1,000 kilogram car that travels north at 15 meters per second and collides with a 2,000 kilogram truck traveling east at 10 meters per second.

If the cars stick together after the collision, find their velocity and direction after the wreck. Okay, so this is a two-dimensional problem. One's moving up, one's moving to the right. So we want to keep track of everything.

So our truck is going to be moving to the right, and it is 2,000 kilograms. And then our car... is moving up car and it is 1,000 kilograms and then so this is kind of my before the collision state and then afterwards I'm going to draw them of mixed together so they've hit each other and now they're stuck and they're moving up at some up and to the right at some angle and we want to find the final velocity of both cars as they move together after the collision.

So this is any type of collision. We want to use conservation of momentum. So this says that the total initial momentum is equal to the total final momentum of both the cars.

And so if we can find one, we can find the other. We know the final momentum has to be exactly the same as the initial momentum. And we know a lot about the initial state. We know the mass, and we know their velocity. So we can find the initial momentum.

So the magnitude of the initial momentum is just going to be the square root, so we're going to use Pythagorean theorem, of the initial x component plus the initial y component. And since we know the initial momentum has to equal the final momentum, as soon as we find that, then we know the magnitude of the final momentum. They have to be the same.

So we just need to find these components. the initial momentum. Okay, so if we do that we'll do the X component PIX. Well, that's just the truck the initial momentum of the truck because the car isn't moving in the X direction initially. So this is going to be the momentum initially of the truck.

So that's going to equal the mass of the truck times the initial velocity of the truck. And again notice my subscripts. I'm doing it in a way that makes sense to me. If you need to put them in a different order or call things other things like object A, object B, your subscripts are really there for you to organize the problem.

Just label everything appropriately and keep it organized. And so this is going to be 2,000 kilograms times the initial velocity of the truck is 10 meters per second. And so that's just going to be 20,000 kilogram meters per second.

And if we do P-I-Y, This is pretty easy because we only have one object moving in each direction, so this is the initial momentum of the car. This will be the mass of the car times the initial velocity of the car, which is 1,000 kilograms, moving at 15 meters per second, or 15,000 kilogram meters per second. So now we're ready to plug into our Pythagorean theorem.

So this is... We're really wanting the magnitude of our final momentum, and we know that's equal to the magnitude of our initial momentum, which is given by the Pythagorean theorem, so we need to put 20,000 kg m per second squared plus 15,000 kg m per second squared. And that's going to give... 25,000 kilogram meters per second.

Okay, so we have the magnitude of the final momentum because we know it has to equal the magnitude of the initial momentum. So now we can just use our formula for momentum. So we know the magnitude of the final momentum is just going to equal the total mass of whatever is moving. So in this case, the car plus the truck times the final velocity.

And so if we want to find the final velocity, we just need to rearrange. So this is P, F. Divided by the total mass since it's a completely inelastic collision, so we do 25,000 kilogram meters per second Divided by the total mass 1,000 plus 2,000 kilograms is 3,000 kilograms and that gives 8.3 meters per second So now we want to find the direction of the final momentum So we know it's going to go up a little bit because the car is initially moving upward We know it's going to the right a little bit because the truck was initially going to the right. And the momentum in both of those directions must be conserved. So initially, it had some momentum.

And that was a combination of the X component from the truck and the Y component from the car. And again, the momentum is conserved. So it's the same initially as it is finally. So if we find it at the beginning, we know that's the same for at the end.

So we want to find this angle. And so we can just use these components that we found. So if we do that, we need to plug into inverse tangent.

I'll separate this a little bit. And it's going to be, if we want the angle above our x-axis, our opposite over adjacent, that's going to be P-I-Y. over PIX and we've already found these. So this is tan inverse of 15,000 kilogram meters per second over 20,000 kilogram meters per second.

Try to close that and this is going to give me an angle of 36.9 degrees. Okay, so we have another momentum problem, another collision problem. Here there are three box cars connected together moving at a steady 24 meters per second.

They hit a fourth box car that is at rest and stick together. Find the speed of the four cars and how much kinetic energy is dissipated in the collision. Okay, so it sounds like a completely inelastic collision.

Initially, the box cars are moving at 24 meters per second. And we want to find VF, so the speed at the end. And notice, anytime we're finding momentum, momentum is mass times velocity, but we're not given the mass. But we're still, anytime you do that, you don't want to let it fluster you. You want to keep solving the problem as if you did know it and see what happens.

So this is a collision problem. It's a conservation of momentum problem. And really, anytime you use a conservation law, you want to draw a picture of what happens in the beginning versus what happens at the end.

So initially, we have three boxcars moving at some initial speed, and they're going to hit a fourth boxcar that's not moving. And then our final state after the collision is they've all hit each other, and now they're all moving together. some final velocity.

Okay, so now if we use conservation of momentum, we know the initial momentum of the system is equal to the final momentum of the system. Initially, we have three boxcars moving to the right and that's it. So we have, I'm gonna call this, so I'm gonna say each boxcar has a mass m.

And so initially I have three of these boxcars, 3m, moving at initial speed vi. Okay, and then at the end, I have four boxcars, so 4m moving at speed vf. And then you can immediately see what happens.

The mass is just going to go away. It doesn't matter because it's the same. If it was different masses, you couldn't do this. So then if I want to solve for vf, I just need to divide by 4. So this is going to be 3 fourths of vi, or 3 fourths of 24 meters per second. which is just 18 meters per second.

And now it asks how much kinetic energy is dissipated. Again, kinetic energy is something where you typically need the mass, so we want to see what happens. So if we want to know how much is dissipated, we want to know the change relative to what was initially there.

So we want to know how much we lost. So that's going to be the change in kinetic energy, so how much was lost, over what we started with. So this will give us a fraction of the energy that was lost. So if we do that, change in kinetic energy, that's just kf minus ki, your final minus your initial.

So at the end, there was one half. And then the mass, well, there were four masses times VF squared minus initially, it was one-half times three masses VI squared. So this is our one-half MV squared at the end minus one-half MV squared in the beginning. And then this is just, again, one-half 3M times VI squared.

Okay, so again, you notice I can get... rid of all these halves because they're in every term and there's a mass in every term so I can get rid of it and that simplifies a good bit so now I have 4 vf squared minus 3 vi squared all over 3 vi squared okay so be careful do not try to cancel these two out okay because you're subtracting on top and there's not a 3 vi squared in every term you can't get rid of that just yet Okay, so we need to plug it in. But one thing we could do is we could separate this.

So I could separate this into two fractions and have 4VF squared over 3VI squared. And then if I separate this, then 3VI squared over 3VI squared, that's just one. Okay, so if I plug this in, I get 4 times 18 meters per second. I have to square that, divided by... 3 times 24 meters per second.

Square that and then subtract 1. And I get minus 0.25. Okay, and so this is a decimal. What it means is minus 25%.

So there was 25% of the initial energy was lost in this collision. So our next problem, we have a conservation of energy problem. So we have a bicycle that is moving at 10 meters per second downhill when one of its 2.16 kilogram wheels comes off when it is 50 meters above the bottom of the hill. The diameter of the wheel is 85 centimeters and we want to find the speed of the wheel when it gets to the bottom. Okay so I'm going to write everything that I know.

Okay so vi is 10 meters per second. The mass of the wheel is 2.16. Kilograms and since it's going to be rotating I know I'm going to treat it like a rigid body So I'm going to use capital M. The height above the ground is 50 meters the diameter of the wheel is 0.85 meters so that means the radius which is usually what we use is half of that or 0.425 meters and just to draw a quick sketch of what we've got going on initially We have this wheel that's on a hill. So it is rolling.

So when you know from being on a bicycle that your wheel is not only spinning, it's also translating. So it's rolling without slipping down the hill. And it's going to come off.

And then at the end, it's going to be at the bottom of the hill. And it's still going to be rolling. So we need to use conservation of energy.

And the way you know to use conservation of energy, if it doesn't explicitly tell you to, is notice what's changing. So one, we know it's probably going to speed up. If you've ever rolled down a hill, you know you change your speed. So that's kinetic energy changing or increasing.

You're also changing your height. So you're changing your gravitational potential energy. So when you notice that height is changing and speed is changing, you should think, oh, potential energy is changing. Kinetic energy is changing. Let me use conservation of energy.

which tells me the energy I start with has to be equal to the energy I end with. Okay, so initially, the wheel is not only rolling without slipping, but it's off the ground. Okay, so it has potential energy as well.

So I'm going to write the kinetic energy. It has 1 half mv i squared, so initial translational kinetic energy, plus 1 half i omega squared since it's spinning or rolling. And then it has gravitational potential energy because it's some amount off the ground, 50 meters.

And then at the end, it's still going to be rolling. Now it's going to be rolling probably a little faster if we had to guess. So this should be 1 half mvf squared and still rolling 1 half i omega f squared.

But now it doesn't have any potential energy because it's on the ground, which we will call y equals 0. So now we need to substitute. in for i and omega and try to pull out our linear speed since that's what we're looking for okay it didn't ask for angular speed so if we do that we have one half mbi squared plus one half the moment of inertia of a wheel well a wheel is kind of a thin walled cylinder and we're going to neglect the little spokes because they don't have much mass so really this is just a thin walled cylinder and that has moment of inertia of nr mr squared okay and then so that would either be given to you or you would look it up And we can also use the fact that omega is v over r from our relationship v equals r omega. And so this is going to be v initial squared over r squared plus mgh. And I'm going to do the same thing on the other side. So plus one half.

m r squared and then now it's going to be vf squared over r squared so make sure you put your initial and final on both of those and now if you notice we have a mass in every single term so that goes away and then our r squareds here there's one on top and bottom so it just goes to one so that disappears and we're left with something much simpler so we have vi squared over two plus vi squared over two plus gh equals vf squared over 2 plus vf squared over 2. Okay, so this is kind of like half vi squared plus half vi squared, so that's going to equal just vi squared. So half and a half is a whole plus gh equals vf squared. And so now we just want to solve for vf.

So we're going to take the square root of both sides. I'll flip which side it's on. Okay, and then we can just plug in, so square root of the initial speed, 10 meters per second, square that, plus 9.8 meters per second squared. Let me move this down a little bit, and our height's going to be 50 meters.

So if we plug that in, we get 24.3 meters per second. And so notice that we didn't even need the radius or the diameter because it canceled out. We didn't need the mass.

And so for a lot of these problems, it ends up happening that if you don't have a spring, your mass is going to cancel out because there's a mass in both of your kinetic energy terms and your potential energy term. And so if you're not given it, just like in previous problems, keep going, keep solving the problem as if you did know the mass, and then you'll see that maybe it cancels out. The next problem, we have a diver with her arms and legs straight.

She has a moment of inertia of 18 kilogram meters squared. She tucks her arms and legs in to decrease her moment of inertia to 3.6 kilogram meters squared. If she spins fast enough to make two revolutions in one second, find the number of revolutions she would have made in 1.8 seconds if she had kept her arms and legs straight. So the first thing you should notice here is that your moment of inertia is changing. So initially, it's 18 kilogram meters squared.

And at the end, when she tucks in, it goes down to 3.6 kilogram, if I can write, kilogram meter squared. And so any time your moment of inertia changes, you should think, use conservation of angular momentum. So the only other thing that I know is omega F, she makes two revolutions in one second.

So this is two revolutions in one second. And now I can use conservation of... angular momentum, which says the angular momentum we start with has to equal the angular momentum we're left with.

And the angular momentum for a rigid body, which is what she is, is I initial times omega initial equals I final times omega final. So angular momentum is L equals I omega. So this looks a lot like linear momentum.

P equals MV. You just replace mass with moment of inertia. linear velocity with angular velocity. Okay.

And now we know I initial, we know I final, and we know omega final. And we're looking for how many revolutions she would have made in 1.8 seconds if she had kept her arms and legs straight. Okay.

So it's asking how many revolutions would she make if her initial velocity, angular velocity didn't change. So we need to be able to solve for it. Okay. So this is, we're just going to divide by...

the initial moment of inertia, so move it to the other side, times omega f. And now this is going, if we leave our omega f in revolutions per second, this is going to give us our initial angular velocity in revolutions per second. But that's okay because it's asking for the number of revolutions.

Okay, so you have to be careful, know when you can and can't leave your angular quantities in revolutions versus radians. So if we do this, this is going to be the final moment of inertia, 3.6. kilogram meter squared over 18 kilogram meter squared times this is going to be two revolutions per second so what pops out should be 0.4 revolutions per second okay and now we know how long we know our time we want to know how many revolutions she does in 1.8 seconds and we know how fast she's spinning so our What we're looking for is delta theta. So delta theta is a measurement of angle that you go through. And so revolution is a type of delta theta.

So this is just going to be our omega times time. Okay, so you could, this comes from omega equals delta theta over time. Okay, speed equals distance over time.

You could think of it like that. And so if we plug in, we go 0.4 revolutions per second for 1.8 seconds, and that gives us 0.72 revolutions. The next problem, we have some rotational equilibrium examples.

So for each object, find the magnitude of the unknown force needed for rotational equilibrium about the pivot. So our pivot is shown by this little green dot, and in all of these we have an unknown force F. And we want to find its magnitude if these rods, or whatever they are, if they're to be in equilibrium.

So in order for them to be in rotational equilibrium, that must mean that the sum of the torques is zero. So we want to sum up the torques for each of these. So in the first one, in A, we have two forces that are going to cause a rotation.

So the first one, I'd like to look... See which ones are going to cause positive rotations and which ones are going to cause negative So I like to put all my positives first So if we just look at our unknown force F and imagine that's all that's there if we push up at the right end That's going to cause it to rotate about this green point. It's going to rotate counterclockwise. So that's positive so I'm going to go ahead and list it and the way we find each individual tort is RF sine phi. Okay, so we need to put r, and r is the distance from the axis of rotation.

So the distance from the axis of rotation, that's just the distance from f to the pivot, which is given as 0.1 meters. Okay, and then our force, well we don't know that, so we'll just write f, and then times sine phi. Well, phi is the angle between r and f, so if you were to draw r, it's going from the force to the left towards the pivot and then if you go back to the same point and draw the force it's going to be up it's going to be straight up well this angle is just 90 degrees and if you look at all of these all of your r values for every force in all of these examples it will be going left or right all of your forces will be going either up or down so our phi for all of these will be 90 degrees. And so sine of phi just goes to one.

Okay, so that simplifies things a great deal. Okay, so now we're ready to look at our next force. This 50 newton force, if you imagine it's the only thing there, it's going to cause a clockwise rotation. So it's going to be negative.

So you have to remember to put that in as your first step. And so it is 0.3 meters from the pivot and it is a 50 newton force. Okay, and that needs to sum to zero. Okay, so we ran out of room. And now if you solve for F, you need to move this term to the other side and make it positive.

Okay, so that'll be 0.3 meters times 50 newtons. And then we want to divide by 0.1. Okay, so if you do that, you get a force of 150 newtons. So one thing to notice. This kind of makes sense.

This force is closer, so it's a third of the distance, and so it has to be three times as strong to cause the same rotation as this 50 newton force that's three times further away. So if we move on to our next one, we're going to do it the same way. So we sum up the torques.

We know the torques on the right are going to cause counterclockwise rotation, so they're going to be positive. So I'm going to go right to left. So again, This unknown force is going to be, now it has moved, since the pivot moved, our r changed to 0.3.

times our unknown force F and now our 50 Newton force is the one that's 0.1 meters away okay and this sums to zero and so now again we just solve for F doing it the same way as we did before so 0.1 meters times 50 Newtons divided by 0.3 and this is going to equal 16 0.7 newtons okay so now same sort of thing it is three times further away so the force can be a third as big to have the same torque as this 50 newton force that it's much closer. So move on to part C, sum up the torques. So our unknown force is 10 centimeters from the pivot. So it's back to 0.1 times F. And then there's two forces to the left of the pivot.

So that's going to cause clockwise rotations. So they're negative. So the 40 Newton force is 20 centimeters away. So 0.2 times 40 Newtons. And then the 50...

Newton force is 20 plus 10 so that's 0.3 meters times 50 Newtons again that sums to zero so now we have F equals we got to move both of these to the other side So 0.2 meters times 40 newtons plus, it's going to become positive, 0.3 meters times 50 newtons, all divided by 0.1 that's in front of F, and that equals 230 newtons. Okay, so now we have more force on the left, so we need a greater force, greater than our 150, to cause the same rotation. So we'll do part D.

So part D looks just like part B except we have a 40 Newton force at the pivot. So just kind of copy that. We have 0.3 meters times F and then minus 0.1 meters times our 50 Newton force. And then the question is, what is this 40 newton force going to do?

Well, this force is applied at the pivot or at the axis of rotation. So the R for the 40 newton force is just zero. So it goes away.

It doesn't cause a torque. Okay, so this sums to zero. And then if you look at the math, it's going to be exactly the same as part B. And so this is 16.7 newtons.

And so that force, that 40 newton force being at the pivot, doesn't do anything in terms of rotation. So the last one, we have our unknown force 0.1 meters away, and then we have a 100-Newton force that's going to cause a clockwise rotation, it's a minus, and it is 15 centimeters away, so 0.15 meters times 100 Newtons. And now our 50-Newton force has jumped.

to the bottom and now it's facing down so that's just going to cause a rotation in the opposite direction which means it's going to be positive now because it would cause counterclockwise rotation and it is 15 plus 15 centimeters so 0.3 meters times 50 newtons and again that sums to 0 so we get f equals 0.15 meters times 100 newtons minus 0.3 meters. When we move this last term to the other side, it goes negative, divided by 0.1. And so for this last one, if you plug it in, you actually get 0 newtons. And so this 50 newton force is twice as far away as this 100 newton force from the pivot. And 100 newtons is twice as big as 50 newtons.

So these are actually already canceling each other out. If you were to add any force to the right side, you would cause it to not be in rotational equilibrium, and you would cause counterclockwise rotations. So you actually don't want to apply a force here.

So in this last problem, we have a 0.3 kilogram mass that oscillates on the end of a horizontal spring with an amplitude of 0.15 meters and period of 0.45 seconds. And we want to find the velocity at equilibrium. the total energy, the spring constant, and the max acceleration. So this is an example problem from periodic motion, and it really asks you for really all of the different things that you could ask for in terms of an object that's oscillating back and forth. And so if we write what we know, which is really important for problems like this where there's so many different types of quantities, and they're all changing at different locations.

We have the mass is equal to 0.30 kilograms. The amplitude, which is a type of distance, or the maximum distance away, 0.15 meters. And the period, so capital T.

0.45 seconds. So the velocity of an object oscillating on a spring at some point is given by the equation square root of k over m times square root of a squared minus x squared. And you could combine these radicals, okay, so you could have them all up under one, but it's written like this for a reason.

And you may have seen it derived in a textbook or in class, but it should be given to you. This square root of k over m out front is actually just equal to omega. And that's good because we don't yet know k. And so if we don't know v and we don't know k, we can't solve for this just yet.

And then we want to find the velocity at equilibrium. Well, that's when x equals zero. So this is going to simplify even more.

So if we put in omega for square root of k over m, we get omega times, well, if x is 0, this becomes square root of a squared, which is just a. So this simplified a lot. And then omega is just 2 pi times the frequency, or 2 pi over the period.

So frequency is 1 over the period. And so we can write this in terms of things we know, 2 pi. times the amplitude over the period.

Okay, so I'll get rid of this. And so now since we know amplitude and we know period, we can solve this. And not only that, but the velocity at equilibrium we know is going to be the maximum velocity.

So I'll put this as Vmax, and this is 2 pi times the amplitude, 0.15 meters, over the period, so 0.45 seconds. And And this comes out to be 2.1 meters per second. So now we want to find the total energy.

And we know the energy looks different in terms of an equation depending on where you're at. If the object is at the amplitude, so the spring is fully stretched or fully compressed, then all of the energy is potential. But if it's at equilibrium, it's all kinetic. And so since we know the velocity at equilibrium, we know that when it's at equilibrium, the energy is all kinetic or 1 half m V max square so we're going to use what we just found to find the total energy so this is 1 half times 0.3 kilograms times 2.1 meters per second square that and we get total energy of 0.66 joules so now we want to find the spring constant and so we can go through look at our equations and we have one for the the period of oscillation is just 2 pi times square root of m over k.

If you weren't given this one, but you were given the equation frequency equals 1 over 2 pi square root of k over m, then you could use the relationship that period is 1 over frequency, and you could just flip everything. So this is just a reciprocal relationship. So make sure, if you're not given every single equation, make sure you can use the ones that you do have.

to get what you want. And so now if we solve for K, we need to square both sides, move K up and T squared down. Okay, so that's going to give us with some little algebra, 2 pi over T all squared times M. Okay, so you may have to go through that algebra, just go slow. And you get 2 pi over the period, 0.45 seconds squared, times the mass, 0.3 kilograms.

and that gives us a spring constant of 58 Newtons per meter. Okay, so now we want to find the max acceleration. And to find the acceleration, we just use Newton's laws, and we get minus K over M times X. Okay, so this is derived very easily just using Newton's laws.

force in the x direction is the spring force, so minus kx, and set that equal to mass times acceleration. And we know the acceleration must be max when the force is max, and the force of a spring is maximum when it's stretched or compressed the most, or at the amplitude. So A max is equal to minus K over M times the amplitude.

And I'm going to put in minus the amplitude because that just means the acceleration will be going to the right, so it's positive. But the magnitude of the maximum acceleration occurs at either plus or minus the amplitude. So if I plug that in, I get minus 58 newtons per meter for K over 0.3 kilograms.

times minus 0.15 meters and I get 29 meters per second squared Okay, so that's it for the problems we've gone through. Projectile motion and kinematics, forces, conservation of energy, conservation of momentum, conservation of angular momentum, some rotational equilibrium and rotational dynamics, and now oscillations and periodic motion. So now we're going to move to a few concepts that are typically talked about at the end of the semester. You may not solve that many problems on them, but they're good to know. So the first concept I want to review is...

periodic motion since we just finished that problem. So periodic motion. And for this, we mostly focus on the example, much like from the problem where we have an object that's attached to a spring.

So some mass attached to a horizontal spring and it's oscillating left and right. And it's oscillating about some equilibrium position, x equals zero. So it's going negative and positive.

And so we want to know its motion at every different point in its oscillation. So we're going to look at each point. So when it's at the amplitude, so when x is plus or minus the amplitude, that's going to mean that the spring is stretched or compressed its maximum amount.

So the force is going to be maximum. And then if the force is maximum, that's going to mean that the acceleration is also maximum. So these are also some ideas that can be useful in solving problems.

So it's really Good to understand. So we also know at the amplitude that your velocity is zero. Okay, so if your spring stops stretching or stops compressing, that means that you stopped.

And so if you stopped, that means your velocity is zero. If you kept going, then the spring would continue to stretch or continue to compress and the force and acceleration would increase. And so you would still be moving.

So you know whenever these two are maximum or you're at the amplitude that V must also be zero. Okay, and If V is zero, that means you have no kinetic energy, but because the spring is stretched, you have some elastic potential energy, so your total energy is all potential. So it's one-half kx squared, but our x is just going to be the amplitude.

So here, the energy is all elastic potential. So that's pretty much all the quantities we're interested in, so now let's go back. and look at what they are when we're at x equals zero, which is another way of saying at equilibrium.

Okay, so we're at our equilibrium position, and that means that our force, our spring is neither stretched nor compressed, so the force is going to be zero. By Newton's second law, if the force is zero, acceleration is also zero. Well, up until this point, when we stretched it or compressed it and then released it, this object, it's been accelerating the entire time it's been on its way to equilibrium. And so now it's as fast as it's going to go. So V equals V max.

Okay. Once it passes equilibrium, it's going to overshoot. Now the spring is going to stretch or compress in the opposite direction and start to slow it down.

So we know velocity is max at equilibrium. Well, If velocity is max, that means it definitely has kinetic energy. And if the spring is neither stretched nor compressed, all this elastic potential energy is gone. It's turned into kinetic.

So our energy is now 1 half mv squared, where the velocity is maximum. Okay, so now it's all kinetic. And now we can look at the case.

of anywhere else. Okay, so those are two special locations, plus or minus the amplitude and equilibrium. But if we wanna know these quantities anywhere else, one, we may want to use what we know about the motion at the amplitude or at equilibrium to solve for some other quantities.

But once we have those and we wanna find these quantities anywhere else, well then the force is just given by the spring force. Okay, so at any point the spring is Stretched or compressed, it's just equal to minus kx. So here, I guess I can say at the amplitude, that would make it f equals minus ka. Well, that means the acceleration, if we set this equal to ma and solve for the acceleration, it would be minus k over m times x.

So for the acceleration, Maximum would be minus k over m times a or minus a. So I'll put minus a so it's positive. And then our velocity.

This would be square root of k over m times square root of a squared minus x squared. So this is another one that's derived, and it's derived from an energy equation. So what is that equation? Well, we said at equilibrium, it's all kinetic.

At amplitude, all of the energy is potential. Anywhere else, it has a little bit of both. So the energy is a little bit of 1 half mv squared plus 1 half. kx squared.

Okay? And then if you were to set this equal to the total energy at the amplitude, one half ka squared, since energy should be conserved, these numbers should be the same, then you can solve for this expression for the velocity at some point. Okay?

But we won't do that here. know that the velocity at some point is given by this expression and the energy at any other point other than equilibrium in the amplitude is some combination of kinetic energy and elastic potential energy. The next idea I want to talk about is pressure and Pascal's principle in fluid mechanics or fluid statics.

So typically when we look at Pascal's principle we have this equation pressure equals the initial pressure plus Rho gh. Okay, so just to define what these are, this is the pressure at some point of interest, pressure that you're looking at. So I'll put pressure at point of interest.

And then everything to the right is what that pressure depends on. Well, it depends on some reference point or some initial pressure. Okay, so this may be the pressure at the top, and you're looking for the pressure at the bottom.

And then this other term has the density of the fluid that you're in. And then G is just gravitational acceleration like we're used to. And H, this is the height or the distance between...

this reference pressure and the pressure you're looking at. Okay, so let me draw a picture. If I can.

We have some fluid. And typically we take this reference pressure to be the pressure at the top. Well, the pressure at the top of the fluid, well, that's where the air is meeting the surface of the water. So what's causing the pressure?

Well, the pressure is... caused by the perpendicular force from the air pushing down on the water. So the pressure here is just air pressure, or what we would call atmospheric pressure, the pressure from the air in the atmosphere.

Well, that means if we want the pressure anywhere down here, Some other P, so this would be P-naught. To get the pressure at this point, we need to add the pressure from the air that's pushing on the top of the fluid plus all of the pressure from the fluid that's above this point. Okay, so you can think about this in terms of forces.

The air is pushing down, that's applying some pressure. And now as you go lower, you have more and more water or more and more fluid above you that's adding pressure down on top of you. of you okay so the flesh the pressure depends on how deep your pressure depends on depth The easiest way to think about that is the deeper you go, the more fluid above you that is pushing down on you, and that's more you're having to support. So pressure increases based on the depth. The important thing is if you look at the equation, it only depends on depth.

Okay, so once you know the density of the fluid, the only thing that matters is the pressure at the top and how deep you are. Okay, the other thing that we get from Pascal's principle, and this is really the heart of Pascal's principle, is that increasing the pressure at the top, so the initial pressure, this is going to increase the pressure at all other points. So at all other depths, H, by the same amount. Okay, so the easiest way to think about this is, say you have a bucket. So you're holding this bucket.

And then you get on, maybe you get on your friend's shoulders. Okay, well you feel the pressure. pressure from what's ever in the bucket. Maybe we put some fluid in here. If we go a little bit lower, your friend feels the weight of the bucket plus the weight of you on their shoulders.

So that's kind of like the point in this fluid. It's having to support not only the air, okay, so you are the surface at the top just supporting the air. Your friend is a little bit lower.

It's having to support whatever's in the bucket plus it's having to support your weight. Well now imagine that we add more water. Okay, so if we add even more water to the bucket, well you feel however much water you added.

So if you added 10 pounds of water, now your pressure increased by that much. So that force times the area. Your friend felt the same increase. So he was supporting your weight plus the bucket.

When you added 10 pounds, that added 10 pounds to what he felt. So if you increase the pressure at the top, all the points below it will experience that same increase. So if I add 10 pounds of pascals of pressure at the top, all the points below the top will feel an increase of 10 pascals of pressure.

The last concept we're going to talk about is Archimedes'principle. And Archimedes told us that there are buoyant forces when we're in fluid. So if we're partially or completely immersed in a fluid, There's going to be a buoyant force pushing us up, and the magnitude of that buoyant force is going to equal the weight of the displaced fluid. So this doesn't look like a weight, but if you think about it, weight is equal to mass times gravitational acceleration, so that's our g.

And mass is also, or actually let's use our definition of density. Density is mass per unit volume. So if we solve this for mass, we get mass is equal to row times volume.

And then we plug this in here. And we see this is just another way of writing the weight. So this is the weight of the fluid that you displaced. So if you want to talk about how to get an object to float, in order to float, that object needs to... displace its weight in water or in the fluid so let's use water as an example it needs to displace its weight in that fluid and so if you float when you're in a swimming pool it's because you have displaced enough water equal to your weight okay and so if you wanted to talk about how a ship floats so we know that ships are far more dense than water and if we just throw a chunk of metal in the water it's most likely going to sink so how do ships float Well, ships and their hulls, they are designed where they're hollow.

They only have air in them. And so combined, the amount of volume that it displaces in water is equal to the ship's weight. So it displaces water equal to, maybe I'll put its weight. plus everything on board. Okay, so it needs to be able to displace enough water not only to support itself, but all of the people that you put on it.

So everything on board. Maybe that's one word. I don't know.

Okay, and so the other question would be, how do you know how much of the object is going to be above the surface? We don't want... our ship to float where the top of it is at the top of the water. We want a good portion of the ship to be above the surface so that we can walk around and not be underwater. Okay, and so the amount above the surface, it's the amount of an object floating above, well this depends on the amount of that object needed to equal its weight in water.

So it depends on the volume needed to displace its weight in water. Okay, so as soon as an object displaces its weight in that fluid, well, if you look at a free body diagram, Its weight is going to be pulling down and the buoyant force is going to be pushing up. As soon as these are equal it stops moving. So the acceleration is zero and whatever is left That's what is going to be above the surface.

So you can solve for how much volume is needed in that object depending on its density in order for your buoyant force to equal the weight. And so that's just using Newton's laws and solving for the volume. As soon as your volume gives you a buoyant force equal to your weight, you could subtract that much volume from your total volume and whatever's left over that's going to be on top of the surface.

So that's it for our concepts and our review problems. I hope this helped. Good luck on your finals. And if you have any questions, feel free to let me know and I'll see you in the next video.

Transcript for:Comprehensive Physics Final Exam Review

Transcript for:
Comprehensive Physics Final Exam Review