 Hi, I'm Zor. Welcome to Unizor Education. Let's solve a few problems on kinematics. This is the second series of problems. And this is all part of the Physics 14 course of advanced physics for high school students and high school level students, I would say. It's all presented on Unizor.com. Thank you to watch the lecture from this website because it contains notes for every lecture. And for lecture as this one, for instance, where the problems are solved, I usually put the problems themselves, I state the problems and have an answer. So you can try to solve the problems yourself and check against the answer, which I definitely recommend you to do before you listen into this lecture. It's very important to think about the problems before you listen to a solution, right? All right, so let's go. Problem number one is you have two wheels with teeth and they are locked into each other, which means that this one rotates this way, this rotates this way. Now, I know the radiuses, r and r, and I know the angular velocity of one of them. What's my angular velocity of another one? Well, that's an easy thing because obviously you have learned, there is a concept of linear velocity of the point which is rotating, right? So linear velocity is basically the magnitude of the velocity vector, and it's the same for these, but since they are locked into each other, it's exactly the same as this. So linear velocity of points on the rim of this wheel and on the rim of this wheel exactly the same. So what is my linear velocity? Well, it's radius times angular velocity, as we know, right? Because whenever my rotation, whenever my vector, let's say, moves by a certain angle phi, then this is, the length of this arc is r times phi. And the speed, linear speed along this movement is basically a derivative of this, and the derivative of this is obviously omega because phi is equal to omega times time, right? Obviously we are assuming that rotation is uniform, which means omega is a constant. Alright, so basically we can find out our, let's say this is omega 1. So this is r times omega 1, right? Because this is linear speed of the point on this wheel and this is linear speed on the rim of this wheel, and that must be the same from which w1 is equal to r over r omega. So as we see, the angular velocities are inversely proportional to radiuses. The bigger the radius, the smaller angular velocity, it's slower. And the smaller radius, obviously, the rotation is faster. And that's how all the different machinery is working whenever we want to change the number of rotations, let's say, per second or whatever. Okay, next. Next I would like to know how fast I'm rotating around the Earth's axis. So basically we are talking about Earth. Now this is its axis. Now this is equator. Now let's assume that I am at certain latitude. Now latitude is an angle from the equator to the point where I am locating. Now I am basically circulating around this trajectory, right? So this is the radius of my circulation. Now what is this radius? If I know the latitude, phi, then obviously my radius is equal to, sorry, my radius is equal to radius of the Earth, which is this one, times which is this one as well. And this is r times cosine of my latitude. So if my latitude is given, then I can find out the radius on which I am circulating around the Earth. Now what's my angular velocity? Well angular velocity is one rotation in 24 hours, right? T, which is actually 2 phi over 24 in radians per hour. Now the radius of the Earth, I also know kilometers. So I know the radius, I know my latitude. Well let's say my latitude is 60 degree of north, let's say. 60 degree of north. Alright, so that's pretty much sufficient to find out my linear speed as I am circulating, because I know the radius, I know the angular speed omega. So it's radius times omega is equal to r cosine phi times 2 phi over 24. And that is my kilometers per hour. Now I have actually calculated this based on the numbers and the result is 837 kilometers per hour, which is pretty fast. So I am on the 60s parallel, rotating around the Earth. My speed of rotation, well relative to stars, let's say, is something comparable to an airplane. That's my speed. As I'm stanging, well right now it's even more because right now I'm in New York. New York is about 40 something degrees, so it's a little bit less than that, which means my radius is greater, so I mean even faster than this. Something like 1000 kilometers per hour. I'm moving right now around the Earth's axis. Now obviously I'm also moving with the Earth somewhere else around sun, but that's not my problem right now. Alright, so that's my movement around the Earth's axis. Now next problem is similar. So let's consider the same Earth. And at some parallel, I don't know which one, on the height of 10 kilometers, I'm in the plane. So this airplane is moving east to west. Now I'm noticing that sun stands still in my window. So I'm basically moving exactly the same as the sun from east to west. So I'm moving along this parallel. Let's say my speed is 900 kilometers per hour, which is the normal speed for the plane. Now my question is, where exactly am I as far as the parallel is concerned? What's my latitude? If sun stands still, if I'm basically moving around the Earth's axis, it's exactly the same angular speed as the sun. Because that's exactly what it means. If sun stands still in my window, it means I'm moving east to west with exactly the same angular velocity as the sun. So I'm making one full circle in 24 hours. Alright, so what's my angular velocity? Well, we know this is 2 pi over 24 hours, right? That's my angular velocity. Let's put it omega. Omega is equal to pi over 24. Now what I don't know, I don't know the radius because I don't know my latitude. That's what's necessary to do, to find out. Now, what exactly is this radius? Well, this radius is equal to r times cosine phi, where phi is my latitude, right? Now, I'm actually flying a little bit higher. I'm not on the surface of the Earth. I'm 10 kilometers higher than that. So let's say this is my radius around which I circulate. Basically, it doesn't really mean much because this is 6400. And relative to this, this is minuscule. So our height over the Earth is not really that important. But anyway, let's add it in. So this is my radius. This is my omega. This is my angular velocity, right? And that's my speed. And the speed is, again, no. So all I have to do is to find omega from this. Oh, no, sorry, omega is given. All I have to find is phi. So omega I know, v I know, h I know, r I know. So all I have to do is to find cosine, which is a very easy equation, basically here, right? So finding cosine first, and then you do the arc cosine to find out the angle itself, which is the latitude. And I did make some calculations here, and the result is approximately 57 degrees latitude. Okay? Next. Okay, now we have a clock. And at 12 noon, both hands are vertical, and they point to 12. They coincide. The question is, when is the next time they will coincide? Now, obviously, it will be somewhere after one o'clock, right? Now, at one o'clock it will be this, and this is the hand for the hours, this is the hand for a minute, but a little bit after one o'clock, my minute will catch up with the hour, right? Okay, so how can we find out this time interval during which they will coincide again? Well, we do know their angular velocity. Our hand will make one circle in 12 hours, so it's 2 pi over 12 hours. My minute hand will have angular velocity 2 pi in one hour, right? Which is 2 pi. So I have angular velocities. Now, if I have an angular velocity, I obviously have their linear velocity, r times omega h, and the minute is equal to r times omega minute, where r is the radius of the clock. Unknown, but it's completely irrelevant, because the length of the hands is irrelevant. All I'm interested in in coinciding that, right? So it doesn't really matter what's the length. They can be of the same lengths, doesn't really matter, equal to the radius of the circle. All right, so I have linear speeds. Now, if this is this little thing, this little arc, let's call its length L. Now, if it lengths L, it means that my minute hand made a full circle plus this L, while my hour hand made only the distance of this L, right? So I have the linear speed of one, I have the linear speed of another. So it means that Vh, that's the hour times unknown time t, covered L, right? Now, at the same time, my minute covered the full circle plus L, right? So Vm speed of minute, linear speed, times t is equal to 2 pi r plus L. So I can substitute this into this, and I have an equation for time t. So Vm is r omega m times t equals 2 pi r plus Vh times t. Vh is 2 pi over 12 L Vh, r times, one second, L is equal to Vh times t. Vh is this, which is r times omega h times t. And obviously r goes out. Now, omega m is equal to 2 pi. This is equal to 2 pi over 12, from which 2 pi also down. So we have t equals 1 plus t divided by 12, from which 12t is equal to 12 plus t, from which t is equal to 12 divided by 11. This is the time in hours, basically. Everything is in hours here. This is one hour and a little bit more. So one hour will be when the hour hand will be at one, and a little bit more, that's exactly the time needed to catch up. So that's the answer. This is the time when they will need a game. Not by the way the whole number of minutes. Okay. Now we have a river, which goes straight with width s. It has certain flow of the water, which is unknown. Now, I have a boat, which is supposed to cross. So I'm always going perpendicularly to the banks of the river. I know that my speed of the boat is twice as big as the speed of the current. So the current goes this way, my boat goes always this way. But obviously, during the time I'm reaching, my water will move me from this point somewhere here. So my real trajectory would be this, which is the combination of my movement always perpendicularly plus the water pushes me to the right in this case, and I will end up down the river, down the flow of the river. The question is, what is this distance? Well, let's just think about it. Now, what is the time I can theoretically cross the river? Well, that's the s divided by 2v. That's the time. So I'm actually covering this distance across the flow of the river in this time. Now, with this time, using the flow during this time, it is always moving me to the right with speed v. And what's the distance it will move me further? This one, which is s over 2. So if my speed is twice as big as the speed of the flow of water, then I will be pushed by the water by half of the widths of the river, right? Okay, next. I have the system of coordinates, x, y. So the whole thing happens on a plane, x, y plane. I have one car here at A0. A is this length along the x-coordinate, and 0 is because I am on the x-axis. Another car is here. It's x-coordinate is 0, but y-coordinate is b. Now these cars are moving towards each other, towards the center, towards the origin. Speed u and speed v. Uniform speed, obviously. My question is, well, obviously they are in certain distance right now. As we start moving to the origin of coordinates, obviously they are making their distance between them smaller and smaller. Now these speeds are different and distances are different, as well they don't really have to meet. They will go through the center and further down to the negative side, and this one goes through the origin of coordinates and goes down to the negative side on the y. And then the distance between them becomes greater and greater. So first the distance is diminishing, and then after a certain time where it was actually the minimum, it will start growing again. The second question is, how long will it take for them to reach the minimum distance between them given their positions and speeds? Well, all we have to do basically is to calculate as a function of time the distance between them, right? So if there is a time t, which is past since t is equal to 0, this car is here and this car is there. Now after that time, if we have time t past, we have uniform speeds, right? So my new position of the first car, let's call it A of t, that's the position, is equal to, well, obviously x coordinate is 0, but this one would be A minus u t A. And then as the t is increasing, every second I'm moving to the left by u and that's why this is a position of my first car. Y coordinate is 0. Now, position of the other car is, now it goes this way, so x coordinate is always 0. Y coordinate starts with B and then it diminishes by V every second. So that's the position. So I know positions at time t of both cars. Well, let's find out the distance between them. Now, if I know two points on the coordinate plane, on the system of coordinates, I can very easily find out the distance between them. So if you remember it's x1 minus x2 square plus y1 minus y2 square and square root of them all, right? All right, so square of the distance, let's say this way. I will minimize the square of the distance instead of the square root of that distance because it doesn't really matter since I'm minimizing. If I'm minimizing square root, I have to minimize whatever is under the square root, right? It's monotonic. So what is the distance between them? Square of the distance. It's this minus this square, which is A minus u t square minus 0, right? Plus this minus this square, which is B minus v t square. And, well, lo and behold, this is just a quadratic polynomial, right? How to minimize quadratic polynomial? Well, if you have a quadratic polynomial which looks like this and A is positive, then it has a minimum at point minus B over 2A. Well, these A and B have nothing to do with these A and B, just the coincidence. I'm talking about coefficients of the polynomial. Well, we have the polynomial, so let's just change it slightly so we will have a clear understanding of what is our coefficients. Okay, what's at t square? u square plus v square, right? Now, what's at t? Minus 2A u t, right? So plus t minus 2A u and minus 2B v. That's the t. And my free member, and nothing depends actually on this free member, would be A square plus B square, right? So this is my polynomial. Again, minus B over 2A, this is my B. So minus B would be 2 times A u plus B v. Divided by 2A, which is 2A square plus B square. So this is the time t, and obviously I can cancel this too, and I don't need these brackets. So this is my answer. This is the time it takes to reach the minimum distance between these two cars. Well, that was my last problem. I do suggest you to go through these problems again. Go to Unizor.com. That's, you have to go through physics, for teens, mechanics, kinematics, frames of reference, and then the problem number two. And just try to solve these problems again, just by yourself with a note pad and a pen and see if you get the same answers. Other than that, that's it. Thank you very much and good luck.