 Hi, I'm Zor. Welcome to Unisor Education. Today we will solve a couple of problems related to forces and obviously before going into the forces we have to know the acceleration, the velocity, etc. etc. Now this lecture is part of the course called Physics for Teens presented on Unisor.com website. On the same website you can find math for teens, which is actually a mathematical prerequisite to this course, and math for teens is a relatively complete course of mathematics for high school level students, and this is intended as a relatively complete course of physics for teenagers. Now the site is free, by the way, there are no advertisements, so free for all, no problems. There are exams and there are very detailed notes for each lecture, well except those lectures which are the problems. Problems are stated on the website and I'm actually encourage you to try to solve these problems just by yourself. There are answers and you can check it, but the solutions basically I will present usually during the lecture about this. So today we are solving the problems. Now these problems are very simple. Two problems related to movement with acceleration along the straight line. The third problem is about the circular movement. Now they are simple, but they are relatively, how should I say, universal because most of the simple problems related to this type of motions are basically the same thing as these ones, maybe slight deviations, but again these problems cover most of the simple problems when you are considering the motion and the forces related to this motion if there is a constant acceleration, either straight line or circular rotation. Okay, so the first problem is the following. You have the car which is going straight from 0 to maximum speed and it takes that time. So from 0 to maximum speed along straight line, so it's an acceleration, the acceleration is constant by definition of this problem and there is a time, so there is a maximum speed and there is a time it achieved and there is a mass of the car. And now I have lots of questions about this situation. Now question number one, what should be chosen as the most convenient reference frame, the systems of coordinates for this particular problem? Now obviously the best system which I believe in this case should be chosen is the one which has origin exactly at the point where the car starts. The x-axis should be directed towards the straight line where the car is moving the same direction. Now y and z are perpendicular to this but they're irrelevant because the coordinates will always be along the straight line, so x would be changing but y and z will be still on 0. And obviously this system of coordinates is grounded, it's basically fixed on the ground, assuming that the ground is an inertial frame of reference. To a certain degree of precision it is actually true because our planet is rotating very uniformly and it's so big that you can basically consider the movement within this relatively short distance as being a straight line. So my first question is answered. The reference frame is supposed to be in the beginning, should be origin and x along the direction. Now my second question is what is acceleration? Well acceleration is a vector, right? However in this particular case this vector will have only x-coordinate because y and z-coordinates will always be 0. We are moving along the x-axis and also I said that the acceleration is supposed to be constant. Now if acceleration is constant, acceleration is the derivative of the velocity, right? So if acceleration a is constant it means that velocity as a function of time is supposed to be a linear function and this is my linear function. That's the only function the derivative of which is equal to a, right? That we know from the calculus. Now since at time 0 my velocity is 0 this should not be there. So the formula for velocity is this from which I can find out the a because I know that at time t maximum velocity is equal to v maximum, right? So this is a times t maximum from which a is equal to v maximum divided by t maximum. Okay, that's my velocity and it's constant, fine, done. Now I know my acceleration and I know my mass. My next question is what's the force? Force as a function of t as a vector but again this vector will have only x coordinate so that's why I have only one number. y and z numbers are always 0. Now this is by the second law of Newton is supposed to be m times a of t but a of t is actually constant so it's just plain a and I know the value of a. This is m v max divided by t max. So that's my force as a vector. This is the magnitude of this vector and direction is along the x. So this is a constant force which pushes the car up, up, up, accelerate, accelerate, accelerate with a constant acceleration and at time t max it reaches the point, reaches the speed v max. Now my next question is what is my velocity as the function of time? Well actually we already did this and plus constant but the constant is equal to 0 because the time t 0 t is equal to 0 my v is supposed to be equal to 0 so this is the formula for my velocity and again this is a vector in theory right but again a is a vector which is in one direction only in the direction of the x. There are no y and z component that's why my magnitude basically characterize the movement completely. Whenever I'm talking about the straight line movement I need only one code and it's one number so no need actually to go to vector risk. But this number can be positive or negative by the way if it's within the one line it can be either towards the increasing of the x as in this particular case or opposite but we have we have chosen direction of x towards where the car is going so it's always a is positive and that's why v is positive. And the last one how is my distance from point 0 will be related to whatever I know about this. Well again what is the distance? Distance is a position right it's a position at the x axis so it's basically x of t right now we know that velocity is derivative of the position right now we are talking about one code in the formula so basically this function this function is derivative of this. Now if I know the derivative of x of t is equal to at what is my x of t at square plus bt plus c actually b is equal to 0 we know that and the point t is equal to 0 again I have chosen my origin of coordinate 0 so this is all 0 and this is my formula or if you wish you can put it v max times t square divided by 2 t max so that's the answer to all my problems for this very simple problem. Again this is this is basically the manipulation with all the parameters all the characteristics of the movement we have position we have velocity we have acceleration and then we have mass and we have the force so that that's all the components anybody is dealing with whenever any kind of a simple dynamic problem actually is offered now my second problem is almost exactly the same except one little detail my little detail is that at time t is equal to 0 my speed is already something so we started acceleration at this moment so the car before this moment car was doing whatever it was doing doesn't matter but at moment time is equal to 0 my speed was the minimum and then I started accelerating I uniformly accelerate so again this is maximum so my a is still a constant and I'm accelerating during this time from moment t is equal to 0 to t is equal to max I'm accelerating uniformly with constant acceleration from this speed to this speed so this speed is greater and now I have exactly the same questions now starting from the beginning first question is my reference frame the same thing obviously I choose the origin at the point when I start acceleration and I direct my x-axis towards the movement of the car now what's my acceleration now in this case let me again start from the moment when a is equal to constant that means that v is supposed to be a t plus b right same same same same same considerations since my acceleration is the derivative of my velocity and my derivative is equal to constant my velocity should be a linear function but now I have a slightly different condition here because at time t is equal to 0 my velocity is equal to v minimum right which means what well it means that b is supposed to be equal to v minimum now at time t is equal to t maximum I have v maximum so v max equals a t max plus v min from which a is equal to v max minus v min divided by t max so my difference between the speeds divided by the time it took to gain the speed from minimum to maximum which is actually an average of anything in this case it's average acceleration so if you have a function at one point and the function to another point we subtract the difference and divide by the argument that would be the average change during the some unit of time in this particular case but since my acceleration is constant the average is equal to this constant obviously but I just do it slightly differently but we can obviously apply the considerations of the average if I know that the something is constant the constant the acceleration is constant then I obviously can find the average acceleration on this period of time which is v max minus v min divided by by by total time and the average would be equal to instantaneous acceleration because it's always the same at all the places so we got the acceleration now since we got the acceleration we got the function the the force okay the force is equal to m times a and this is the same formula for a so it's m v max minus v min divided by t max so that's the force which is acting now acceleration and velocity and the force are all vectors I did not put the the bar on the top because they're all directed along the x-axis and there is no real different direction when you have to really take into consideration the vector character of the of these characteristics they're all directed in the same direction which means we can just operate with them as was with scholars what's next now the speed well the speed I have already defined that's the speed and that's based on basically the fact that if you have a derivative equal to constant then my function is supposed to be this way the linear function with this a is a is a coefficient at t and we have defined this free member based on the condition that at time t is equal to zero if this is zero this is supposed to be we mean right so that's why the function is we mean and fun and finally the distance covered again the distance is the the x coordinate and I know that that this is first derivative of this function now if this is the first derivative of the function then the function itself is a t squared divided by 2 right to get this one derivative of this is equal to t and divided by 2 it would be just t a times t now this would be women times t and then I have some kind of a free member which I don't know yet what it is but again at time t is equal to zero now this is my distance of time at time t is equal to zero I'm supposed to be at the beginning of the coordinates right which means I have to get zero which means c should be equal to zero otherwise my if I put t is equal to zero if this is not zero I will not get zero here right so this is my formula for distance from from the beginning of acceleration and this is the end my of my second problem that my third problem is about rotation so the first two problems were about straight line with constant acceleration my the third problem would be about rotation with constant acceleration now this is slightly more difficult but very slightly basically we have from kinematics we have all the apparatus which we need for this so what's given you have a planet of mass m which is rotating on the distance r from its star well you can consider the earth around Sun now we are considering the trajectory to be an ideal circle and I have a period t zero which means one circumference of that orbit of that trajectory is done in time t zero and then I have a huge number of problems from a to I okay now problem number one is what is the best reference frame now let me think now this is my Sun and this is my planet which goes around it okay so this is some this is Earth now what is the best frame of reference well I think that the most convenient frame of reference is heliocentric where origin is in the middle in the center of the Sun now the x-axis should be directed towards the beginning of my watching observing this planet so that would be x y should also be within the same plane so the orbit of the of the planet is within some plane so XY XY axis should be within that plane x goes to the planet planets beginning point and why is perpendicular within that plan within that plane and Z should be perpendicular to this all to the plane to XY to XY plane and obviously whenever the planet is moving within this plane it's Z co-ordinated is always zero so we can completely disregard the Z coordinate so now we have a plane and let's look at this plane from above so this is my Sun and this is my planet so this is our now my first my second question after I have chosen the frame my second question is what is my angular speed so how how much regions I cover per unit of time well if t0 time covers my entire circle and entire circle is 2 pi region so my omega which is angular speed is equal to 2 pi divided by t0 okay that's my angular speed okay next question what's my position now this is the x this is y right so what's my position at times t I'm here now what is this angle now if I have an angular speed then the angle I turn from the x counter clockwise would be phi of t is equal to omega t since omega is an angle by angle covered in unit of time then omega times t would be the angle which I'm turning by in the time t now if I have this angle okay let's wipe out my ugly picture and put something more decent this is x this is y this is my phi this is my planet this is my Sun okay so what's my coordinates this is our now obviously my x of t is equal to our cosine of phi which is omega t and y of t is equal to our sine of omega t okay so we've got the coordinates this is a position of this point position vector if you wish this is position vector this is x y okay got it so what's my velocity now velocity is the first derivative from the position right by definition so velocity x velocity is okay derivative of cosine is minus sine and there is a mega which goes as an inner function so it's minus r sine omega t derivative of sine is a cosine and the mega is inner function so it's r cosine omega t so that's the answer that's my velocity now geometrically speaking velocity is a vector right now velocity at that point should be pictured as a vector which has certain direction and magnitude so what's the direction of this vector well let's just do very simple thing and we did it before in one of the lectures let's just multiply so this is my velocity vector now this is my velocity vector this is position this is position vector of t and this is my velocity vector of t now let's do scalar product p of t times v of t now scalar product is this x times x and y times y now this times this would be minus r square sine cosine and this would be plus r square sine cosine so it will be zero now we know from vectors that if the scalar product of two vectors is zero they are perpendicular to each other so first what I can say is that this vector is the vector of velocity which is basically showing the direction of movement it's perpendicular to the position now position rate is this radius this is position vector and my velocity would be here perpendicular and we know that perpendicular to the radius is a tangential line right so it's basically when it circulates the direction of the of the velocity vector is always tangential to the circle now the magnitude of this the magnitude of v of t equals to square root of x square coordinate plus y square coordinate now x square is basically a square of this which is wait a moment I made a mistake here it should be minus r I said something about inner function but I didn't write down this omega I'm sorry about this yeah the derivative is r is just a multiplier from cosine is minus sine and then derivative of the inner function omega I was talking about inner function but I didn't write down the mega so mega is here as a multiplier so what is the square of this plus square of this well it's r square r square omega square sine square and r square omega sir a square cosine square and they're added together sine square plus cosine square is wine is one so this is r omega so r omega is my magnitude of this velocity vector got it next acceleration okay acceleration is let me put this here is a second derivative of x and y or first derivative of the velocity so second derivative of this is equal to minus r omega as multiplier from sine derivative is a cosine and again inner function omega so it will be minus r omega square cosine of omega t now why t would be okay r omega or multiplier derivative of cosine is minus sine so I will have minus again r omega now will be omega plus in inner function will always be another omega so it's a mega square and and sine of omega t now this is also very interesting so this is my acceleration vector now let's compare it to the vector of position position vector is from here to here acceleration as you see is directed exactly opposite the only thing is there is a extra multiplier so the vector would be shorter or longer depending on omega but it will always be since this is minus this is cosine this is cosine this is sine this is so so sine r and r r and r so everything is the same except direction which is minus and the magnitude which is so it's basically here but I want to say is that acceleration always directed towards the center that's the gravity that's the direction of the force so the force of gravity is keeping my planet on the orbit and this force is always directed towards the sun and that's why acceleration always directed back from the position to the center of the of the circle so this is basically my acceleration vector now what's its direction I have already done that and the magnitude so magnitude is again square root of so it would be square root of r square omega to the force cosine square and sine square they will be one so it's r omega square so this is my magnitude of my acceleration now next question is express magnitude of acceleration in terms of magnitude of velocity where obviously a of t is equal to v square divided by r right of t square of this is r square omega square divided by r would be r omega square so this is the relationship between linear speed because the magnitude is a linear speed by definition and the acceleration which by the way is a constant it does not depend on t actually it's a constant there is no t here velocity depends on the t but acceleration does not I mean the magnitude of acceleration the direction obviously depends so direction would be always changing but the magnitude is the same and in case of velocity again the direction would be changed but the magnitude is constant so this is how I so this is big basically also can be a scratched alright so and the last is how to express the force of gravity using whatever information we have right now well the the force is equal to again second law of Newton mass times acceleration acceleration is basically the magnitude is a constant but direction is changing which means my f my the force with which sun attracts the planet is also constant by magnitude but direction would be always along the acceleration right force and acceleration are vectors of the same direction but just different magnitude obviously so what's my function what's my f in this particular case it's mass times acceleration which is this one so it's mass times r times omega square or we can always replace instead of this we can use we can use this and that would be exactly the same so it would be m v square divided by r so it's either this for my force of gravity or this doesn't really matter this is in terms of angular speed this is in terms of linear speed okay so my point here is that probably most of the simple problems related to uniform rotation or in the previous problems uniform acceleration along the straight line all these problems are related to manipulation with the same characteristics again it's coordinates or position it's velocity it's acceleration it's the second Newton's law and they're all solved one from another maybe you maybe you have given the force and you have to define determine the mass or you have given the force in the mass you have to determine acceleration or if you given the force in the mass and and omega let's say then you have to define the linear speed I mean they're all interrelated and the formulas are exactly the same one is expressed in terms of another so all these little problems relatively simple problems would be out of all these characteristics something is given and something you have to determine using these formulas and the second law of Newton and the difference is only what's given and what have to be defined let's say the linear speed is given you have to define the period or you have an angular speed and you have to define the linear speed or something I mean they're all interrelated okay well I do suggest you to go to your to go to this website and this particular set of problems you can find it if you will go to physics mechanics dynamics forces and this is the problems for the forces just do it yourself see if you have exactly the same answers because the answers are on the website and that will be a very good exercise that's it for today thank you very much and good luck