Understanding Circular Motion and Gravitation

Okay, good afternoon and welcome to Physics 301, Module 5. The topic is Uniform Circular Motion and Gravitation, and this is based on Chapter 5 of the textbook for this class. So the topics will include rotational kinematics, that is the description of motion for a particle that is moving in a circular path or a curved path. And then we also consider the situation of uniform circular motion and the forces that act on the particle. undergoing uniform circular motion. The first section begins with a description of rotational angle and angular velocity. So when describing the motion of an object that is moving in a circular path, it is more useful to consider the angular displacement as opposed to the linear coordinates x and y. So in this example, a particle starting from a and then ending up at b undergoes an angular displacement shown by the the variable theta. Theta is used to denote angles. It's a Greek symbol. At that same interval, the object covers a linear displacement along the curve equivalent to delta s. So delta s here is known as the arc length. The arc length is simply length of any curved path. So the radius of the circle is denoted by r. And the relationship between the angular displacement and the accolades and radius is shown here. If you recall in MAT 107, you surely studied under geometry, the relationship between accolades and the the angular displacement. So the arc length divided by the radius gives you the angular displacement. So from here you can deduce that the arc length is equal to the radius times the angular displacement. So in module one and module two and three, this delta S will correspond to what we call delta X. So essentially what we're saying is that if this arc was very short, it will correspond to like a straight line, a short straight line segment delta X. and delta x is related to the angular displacement by this simple equation. So multiply the radius by the angular displacement and you get the linear displacement. Okay, so you could also say x equal to r times theta. Okay, but in In mathematics textbooks, it's more common to represent the acland by the symbol s instead of x. So don't get, you know, thrown off if you see the acland denoted by s instead of x. It means the same thing. Now the... the relationship between the angular displacement and the sorry the relationship between the the the angular measure is shown in the is shown right here let me highlight it for you so we know that one complete round on the cycle is called a revolution. And one revolution is equivalent to 360 degrees. Now in terms of radians, this is equivalent to 2 pi radians. So 2 pi radians is equal to one revolution, and that is equal to... 360 degrees. Now if you divide both sides by 2 pi, you would obtain the relationship between radians and degrees. So one one radian will equal to 360 degrees divided by 2 pi okay so but both but fundamentally one rev so one rev is equal to 2 pi radians so these are all conversion factors that you will need to use when solving problems involving rotations. So if the angle is given in degrees, you can use this the conversion factor between radiance and degrees to find to do the conversion. If the angle is given in terms of revolutions, then you would use the conversion between revolutions and radiance. And if you do the division 360 divided by 2 pi, you get 57.3. Therefore, 1 radian equals 57.3 degrees. Now, take note of these abbreviations. So the first abbreviation is REV. That stands for revolutions. So whenever I say The abbreviation REV, that means revolution. And revolution, like we saw in the previous slide, is equal to 360 degrees. So another common abbreviation is rev per minute. So rev per minute means revolutions per minute. So for instance, if they say... The angular speed of an object is 20 revs per minute. That means 20 revolutions per minute. And we know that one revolution equals two pi radians. So we can convert from revolutions per minute to radians per minute and we can convert to radians per second because we know one minute is 60 seconds. Another common abbreviation is RPM. If you look at your car's speedometer, you would see the speed is either graduated or calibrated in RPMs. I mean, the rotations of your tire is calibrated in RPMs. That means the revolutions per minute. So if you look at that number, it tells you how many times your car's tires are rotating. And this could be indicative of the health of your car's rotation system. Now the first concept we're going to define is that of angular velocity after having established this background. So angular velocity is simply a measure of how fast an object is rotating. And similar to the linear velocity, the angular velocity is the change in angle divided by change in time. So final angle minus initial angle divided by time. If you recall, the linear velocity was defined as what? The initial position, final position minus initial position divided by time. Okay and So this is called omega. It's a Greek symbol. Looks like a W. Okay. So that's a symbol for angular velocity. And what I was saying earlier was that linear velocity was defined as change in x. divide by change in t which was x minus x initial divided by delta t. So if you look carefully you will notice that what we just did here is just re-level x as theta and re-level v as omega. It's important to remember the names because sometimes students mix up theta and omega. Okay, so one useful approach to study rotational kinematics is to view the variables as a re-leveling of the kinematic equations we already studied. So if you already knew that kinematic equations. What you need to do in this chapter is just rename the x, v, and a variables to the angular names theta, omega, and alpha. Alpha is the angular acceleration. And as we saw previously, The arc length delta s is the same as r times change in theta. So if you divide the arc length by time, so that gives you the linear speed. This equivalent leads to this relationship between the linear speed and the angular velocity. So the linear speed V is equal to r times omega because right here, changing theta divided by changing t is omega. So v equals r times omega, or omega is equal to v divided by r. So these are the two equations that link the... angular velocity to the linear velocity. So if you come across any problem that requires you to find the angular velocity given the linear velocity, then you would need to use one of those two equations. And the units of angular velocity is radians per second, that's the SI units, just like for linear velocity, the units were meters per second. So whenever the speed was given in different units like kilometers per hour, miles per hour, or anything different from meters per second, then we have to convert to meters per second. So here in module 5, you would need to convert to radians per second whenever the angular velocity is not given in the SI units of radians per second. The direction for angular velocity is counterclockwise or clockwise. Okay, so something moving in a circle has only two possible directions in the sense of the clock. which we call clockwise, or opposite the sense of the clock, which is known as counterclockwise. The positive direction is by default, or the convention is that counterclockwise is positive, clockwise is negative. So if you wanted to make an analogy to the linear case, counterclockwise will correspond to the positive x-axis and clockwise will... correspond to the negative x-axis. So clockwise this is negative and then counterclockwise. This is considered positive. So clockwise, this is clockwise. And this is counterclockwise. So if an object is rotating like the the hands of a clock, you would treat its angular velocity as negative and if it was rotating counter to the clock you would treat the angular velocity as positive. The direction for linear velocity is tangential to the path of motion. the direction of the for an object going around the circular or curved path its velocity is always directed along the tangent to that path at the point of interest okay and to summarize again the v is equal to r times omega or omega equals v divided by r and in this relationship We can deduce that V is proportional to the distance from the center of rotation, so V is proportional to R, and it does largest for a point on the rim of an object that is rotating. So for instance, if you consider different points on the tire that is rotating, the point that is at the rim, that is at the circumference of the tire, We'll have the lodgers. linear velocity compared to the one that is closer to the center of the tire. So in this example, it says we should calculate the angular velocity of a 0.3 meter radius car, a car tire, when the car travels at 15 meters per second. which is about 34 kilometers per hour. So to solve this we would need to, sometimes it's always useful to do a sketch. So they said this is the tire and the problem says that The car is going forward at 15 meters per second. So that means every point on the tire is also going forward at 15 meters per second. We know that the radius of the tire is 0.3 meters and there's a fine angular velocity so we're looking for omega. We know v. we know r and we don't know omega but we from our relationship we know that v equals r times omega or omega equals b divided by r so in this case it will just be 15.0 meters per second divided by 0.300 meters per second. Sorry, meters. The radius is measured in meters, not meters per second. And if you work this out, you will obtain... The meters will cancel out and you will have it 50.0 per second of... you could say 50.0 radians per second. Okay, so this is the solution. The next concept after Angular velocity is angular acceleration. So the angular acceleration is defined in a similar fashion. It's the rate of change of angular velocity. So alpha equals change in omega divided by change in t. So omega final minus omega initial divided by change in t. And if you recall, when we discuss linear motion, the linear acceleration, a, was equal to change in v divided by change in t. So v minus v initial divided by change in t. So what we did here is just rename v as omega. So in fact we're really using the same equation as from module two and again alpha so this is alpha pronounced alpha okay it's a greek greek symbol The units of angular acceleration is radians. per second divided by seconds because omega is measured in radians per second then you divide again by t which is seconds so you get radians per second squared similar to the linear counterpart which is meters per second squared so in the next example suppose the teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5 seconds. Calculate the angular acceleration in radians per second squared and if if she now slams on the brakes causing an angular acceleration of negative 87.3 meters per second squared, how long does it take the wheel to stop? So here we have a problem that requires us to find the angular acceleration and to find the time it takes the spinning wheel to come to a complete stop. So let's read the problem carefully and pay attention to some keywords. In the first part we know that the wheel starts spinning from rest to a final angular velocity of 250 rpms. So a closer look at this problem reveals that the keyword rest means the initial angular velocity is zero and then a final angular velocity is 250 rpm that's omega and remember rpm means revolutions per minute so we will need to convert this to radians per second and then the time is five seconds and the first part they say find the angular acceleration which is alpha now if she now slams on the brakes causing an angular acceleration, so this is your alpha. How long does it take the wheel to stop? So how long refers to time and to stop means the final angular velocity in path B will be zero if the wheel comes to a stop. Okay so having identified this information The first step is to convert the, you know, the RPMs to radians per second. So 250 revs per minute. One rev is 2 pi radians. One minute is 60 seconds. So if you line up the conversion factors correctly, you would end up with 26.2 radians per second for which is equivalent to the 250 RPMs. And then alpha is simply the change in omega divided by T. So 26.2 minus zero divided by five gives you 5.24 radians per second squared. And then part B, we know that the wheel comes to, starts at 250 RPM and stops. which means the final angular velocity is zero. And so the change in angular velocity is zero minus 250, which is negative 250 RPM. Alpha is minus 87.3 radians per second squared. And so looking at this equation for alpha, you can manipulate the equation to isolate T on one side because that's what we're looking for. And then after converting the RPMs to radians per second, again plug it into the equation, negative 26.2 divided by negative 87.3 and you get 0.3 seconds as the time required to bring the wheel from 250 RPM to zero with an acceleration of negative 87.3 radians per second. square. So let's look at angular acceleration and how it relates to tangential acceleration. So we already defined angular acceleration as the rate of change of angular velocity. In this diagram, pay attention to the direction of the tangential acceleration. So the tangential acceleration is directed in the same direction as the tangential velocity. So this is your tangential acceleration. Now remember that v is equal to r times omega. So change in v is equal to r times change in omega. We divide both sides by t. Change in v divided by change in t is the tangential acceleration. And then delta omega divided by delta t is alpha, so r times alpha. So tangential acceleration is equal to angular acceleration times radius. We're going to come back to centripetal acceleration later, so just ignore that for now. Okay, so this is the equation I just derived. You need to remember this relation, and the units for tangential acceleration is meters per second squared. The units for angular acceleration, alpha, is radians per second squared. Okay, so in the yellow box, you have the two equations that... that are useful in converting alpha to a sub t or a sub t to alpha. So if you know alpha you can find a sub t by multiplying alpha by the radius and if you know a sub t you need to divide it by r to get alpha. So kinematics of rotational motion. So as we saw for one-dimensional and two-dimensional kinematics kinematics is simply concerned with description of motion without regard to force or mass. So in the next slide we're going to see that all the translational or linear kinematic quantities such as displacement, velocity, and acceleration have direct analogs in rotational motion. So in this table The first column shows the rotational variables, the second column shows the translational or linear variables, and the third column shows the relationship between the linear and translational variables. So theta is measured in radians, omega is measured in radians per second. alpha is measured in radians per second squared, x is measured in meters, v is measured in meters per second, a is measured in meters per second squared. And the relationship between the linear and translational is shown in the third column. You could also rearrange the equations to have Now x equal to r times theta, b equal to r times omega, a sub t equal to r times alpha. So these are all equivalent relations between translational or linear and rotational. Okay take note I'm using x here, some books will use s for the to highlight the fact that we're dealing with curves and not just straight lines, okay? So remember this relationship. If you can make a mental association between the linear and angular variables, then to an extent, module five, most of module five is just like a review of module. two and three of module two because rotational you know the rotational variables are kind of like one one-dimensional except in the case of acceleration where you have two types the translational and the centripetal acceleration. So if you remember the three main equations we use in module two, just re-label your x as theta, your v as omega and your a as alpha and you get the equations for rotational motion. And these are the equations. So the first equation says the linear displacement is equal to the average velocity times time. And so the angular variable or angular displacement is equal to the average angular velocity times time. So there should be like a delta here to emphasize the fact that you're referring to displacement, okay? Here as well as here. So v equals v sub o plus a t omega equals omega sub o plus alpha t. delta x equals v o plus a t, delta theta equals omega naught t plus one half alpha t squared, and v squared equals v sub o squared plus two a x, omega squared equals omega sub o squared plus two alpha times change in theta. So exactly the same equation except we level the three variables theta. omega and alpha. And the subscripts you know a subscript of zero means initial value and the variables without subscripts means the final value. So the problem solving strategy for rotational Kinematics is as follows. The first step is to examine the situation to determine that rotational kinematics is involved. If rotational is not involved, then you cannot use the rotational equations. But note that you don't need to consider forces or masses because when you're doing kinematics, you're purely looking at the displacement, the velocity. the acceleration and the time. Next, identify exactly what needs to be determined in the problem. Sometimes it's useful to sketch the situation. Third, make a list of what is given or what you can infer from the problem. Like in the previous two examples, we could infer that if something starts from rest, that means the initial velocity is zero. In the same problem, we could deduce that if something started at a certain speed and came to a stop, that means the final speed is zero. So you need to always pay attention to those keywords. So make sure you list everything that's given and everything that you can infer from the keywords in the problem. And then the fourth step is to identify the appropriate. the equation or equations for the quantity to be determined. Sometimes you may need to use two equations. You may need to use one equation first as a transition and then get the value from that equation and then feed it into a second equation to find the unknown. And so you're gonna apply the same strategies we use for translational. in a motion. Then step number five, once you identify the equation, always isolate the unknown to one side and then substitute the values with the correct units and then simplify to get the final answer. Always pay attention to the units. The SI unit for angular measure is radians. and check your answer to see if it's reasonable in terms of units. in terms of magnitude. Ask yourself, does this make sense? If you're calculating the speed of plane or the acceleration, ask yourself if the number you got makes sense, you know, for given the, you know, the numbers in the problem. Don't just, you know, write down whatever your calculator shows you without, you know, pausing to ask yourself if it makes sense. Oftentimes you can make a simple algebraic error and by checking your answer in step number six it's easy to correct that you know that error. So now we're going to talk about centripetal acceleration which is the acceleration that is present whenever circular motion is is taking place. So whenever an object is moving along a curve, the most be an acceleration that is directed towards the center of the curve. As you can see in this diagram, if we go make a flashback to concept of vector addition. So this is v1 the initial velocity this is the final velocity. Take note v1 and v2 could have the same magnitude but because they're directed at different directions so they're not the same. So v2 minus v1 cannot be zero because even though they have the same magnitude they don't have the same direction. the vector minus v1 will be a vector directed in the opposite to v1. Okay so negative v1 will be directed in this direction. So if you wanted to add, so you wanted to find delta v which is v2 minus v1 of v final minus v initial. According to the graphical tip-tail method, you will write, let's say, your v. So technically what you're doing is like you're adding minus v1 to v2. So minus v1, this is minus v1, and then v2. tip to tail is this way and then your delta v which is your v one minus v one plus v two or v two minus v one is this vector so this is delta v this is negative v one and this is v two So delta v is the same as v2 minus v1 or minus v1 plus v2. Okay, and as you can see from this vector diagram, if I translated this triangle to the circle, you notice that delta v is directed towards the center of the circle. So your your centripetal acceleration which comes from the change in direction of the velocity vectors v1 and v2 is always directed you know towards the center of the circle and So, okay, so let's review this. So speed and velocity around a circle. So in uniform circular motion, the velocity changes direction but the magnitude of V is constant. So when you hear the word uniform circular motion, that means the magnitude of V is the same, although the direction is changing as the particle goes around the circle. So the total distance around the circle is 2 pi r, which is the circumference of the circle. The time required to go around a circle or to complete one cycle of motion is called the period, is denoted by uppercase T. Again, do not confuse this with tension, because tension is also represented by T. So always pay attention to the context. So in this diagram, the circumference is 2 pi r. which is the distance. The time taken to cover the circumference is the period. Therefore, the speed for uniform circular motion is 2 pi radius divided by t, which is the period. Therefore, for an object like the Earth, we know the period of the Earth's rotation is 24 hours. It takes the Earth 24 hours, that's one day to completely rotate around its axis. So if we know the radius of the earth, we could compute the translational speed of the earth by simply plugging in the radius for r and then plugging in 24 hours. Of course you need to convert to seconds to get the rotational speed or the tangential speed of the earth. in meters per second. Now if you wanted to find the angular speed of the earth it would simply be changing theta divided by change in time and we know of course for one complete revolution the angle is simply two pi radians so two pi radians divided by time. So again you can see that this two pi divided by time is like your omega. So again the relation we derived earlier r omega still holds okay. So this is the same as r times omega where omega is defined as two pi divided by the period. So if they ask you to find the angular velocity of the earth, all you need to do is do 2 pi divided by 24 hours. Convert the 24 hours to seconds and then plug it into this equation. So is it possible for an object traveling at constant speeds to have a non-zero acceleration? You can press yes or no. There's a yes or no button. So let me see your feedback by pressing yes or no. Is it possible for an object that is traveling in a circular path with a constant speed, is it possible for it to have a non-zero acceleration? If you think the answer is yes, you know, press yes. on the feedback buttons. So if we click on participants you see two feedback buttons. Yes it's a green button and no it's a red circle with an x in it. So if the answer is yes click on the green circle. If the answer is no click on the red circle. Okay so Some, yes, some class members have responded and I think most of the class says yes. Okay, so this is, the current answer is yes, indeed, because even though the speed is constant, As long as the direction is changing, which is the case if something is moving in a circle, the direction changes every time because velocity is defined as the vector quantity, which is characterized by two parts, magnitude and direction. So for any object that is traveling at constant speed around a circular path, the direction is changing all the time even though the magnitude of the speed remains constant. Okay, so this is something students find hard to digest. Because common sense tells us that if the speed is constant, then the acceleration should be zero. But that is faulty thinking because acceleration is not defined in terms of speed. It's defined as the change in velocity divided by change in time. So speed is just one part of the velocity. It's just a magnitude. And so the magnitude could be constant if the direction changes. then there will be an acceleration. Okay, so again, these are just reminders. The velocity vector is always tangent to the path and is perpendicular to the radius. Similarly, the centripetal acceleration vector is perpendicular to the velocity vector and points towards the center of the circle. So the The centripetal acceleration is directed along the radius and is therefore perpendicular to the velocity vector because the velocity vector is perpendicular to the radius. So this is the expression for the centripetal acceleration. We're not going to go into the detailed derivation. that is not necessary for this course, but you need to remember that the centripetal acceleration a sub c is equal to v squared divided by r and v is the the tangential velocity, r is the radius of the circular path and we replace v by 2 pi r divided by t which we derived in the previous slide. Then the centripetal acceleration could be written as 4 pi squared r divided by period squared. We can simplify this expression. Now following the concept of acceleration, we can introduce the concept of centripetal force, which is the force that results from the acceleration. Remember Newton's law. implies that whenever there's an acceleration, there must be a net force. That's the second law. F net equals m a. So if a is not zero, which is the case for circular motion, we just established that for circular motion, there must be an acceleration directed towards the center, which we call centripetal acceleration. So that means for every circular motion there must be a force directed towards the center, which we call centripetal force. So F net in the direction towards the center equals m a c and since a c is v squared over r, so that means f c equals m v squared over r. Now take notes, FC is not like a single force, like it's not like a normal force, tension force, friction force and so on, rather it's the net force towards the center. So FC could be the result of adding multiple forces, it could be the result of just a single force if it's... only a single force is acting towards the center. So in terms of radius and period, the equation could be written like this. Okay and just to add we already know that v is equal to r times omega. So v squared is equal to r squared omega squared. So you could write this as m times r squared omega squared divided by r and one of the r's will cancel and so you would have m r omega squared. So this is another form of the centripetal force. So you could use this equation. You could use this equation. You could also use this equation to find the centripetal force on an object undergoing circular motion. The difference is what is given. So if you know V and R, then you use the first box. If you know, so here we know V and R. Now if on the other hand you know, so I'm assuming the mass is given in all cases. Now if you knew the radius and the period, then you would use this equation. And if on the other hand you knew the radius and the angular velocity, then you would use the third equation. Okay, so you have like three different tools and you can pull out each tool depending on what is given in the problem. So let's apply this to the case of an object moving in a horizontal circle. So for an object moving in a horizontal circle as shown in this diagram, imagine something tied to a string and then caused to move in a horizontal circle. The force of the string pulling the solid inward is known as the tension force. And this is the force responsible for the object moving in a circular path. And since this is the only force acting in the central direction, the net force in the central direction is just the tension. And so if you wrote Newton's law, if you wrote Newton's second law for this problem, you would have... sum of forces plus the center equals m a c and the sum of forces is just the tension force equals m and the ac is v squared over r so that means for this case the tension force is equal to m v squared on r if the arcs you define v then you could simply say v squared equals R times T, multiply both sides by R, then divide by M, and then your V would equal square root of R times the tension, divide by M. On the other hand, it could ask you to find the radius. So the radius will just be R equals, multiply both sides by R, so R times T equals MV squared, so R equals MV squared. divide by T. Okay, so the tension on the string provides the centripetal force directed towards the center and therefore you can say T equals mv squared over r. Take note T here is the tension, it's not the period. So don't confuse this T with other T that stands for period. Okay so Here's an example. If a car with a mass m travels at constant speed around a circular curve with radius r, in what direction does the centripetal force point? Okay, so obviously in this case the The centripetal acceleration will point towards the center of the circular curve and that corresponds to the second option. Which statement below best describes the centripetal force on the satellite that is orbiting the Earth in a uniform circular path with constant speed? The satellite is 300 miles above the Earth. So we have four options A, B, C and D. So you can type in A if your answer is the first option and B for second option, C for the third option, D for the fourth option. okay so if you select that D, then that's the correct answer because in this case, so let's say the satellites, let's say this is the orbit of the satellites. The satellite is located somewhere here. And this is the center of this. This orbit is the Earth. The force pulling the satellite inwards towards the Earth is the weight of the satellite or the gravitational force. So since this is the only force pulling the satellite inwards, if you apply the sum of forces towards the center equals mv squared. over r. So this the net force towards the center is just f of g equals mv squared over r. Okay, so and this force is mass times gravity equals mv squared over r. You can cancel out the m. and then you just have v squared equals r times g of v equals square root of r times g. So if you knew the radius of the orbit then you could calculate the satellite speed from this equation, the radius times gravity square root. Okay so you'll be surprised that these are some of the the simple equations. Simple calculations that have to be done to put a satellite into a certain orbit. The next problem, so a roller coaster has a mass of 5 times 10 to the 2 kilograms. When fully loaded with passengers, if the vehicle is... if the vehicle is so low that it travels at 20 meters per second, what is the force of the track on the vehicle at A? Note G equals 9.8 meters per second squared. So at A, the first thing we need to do is to draw a free body diagram, okay? Because we're looking for a force. And the free body diagram will look like this. The weight of the car directed downwards. the force of the track which is also known as the normal force directed upwards and the acceleration since the car is going the circular path the centripetal acceleration is v squared divided by r and we're looking for the force n so you're going to apply newton's second law f net equals m a So sum of forces in the central direction. Remember the center of curvature is located above A. So that's why AC is pointing upwards. Okay so sum of FC equals MAC and the sum of forces we have N upwards minus mg which is downwards equals mv squared over r and so n is equal to mg plus mv squared over r or you could factor the m out g plus b squared over r and then just plug in the numbers 500 kilograms for the mass five times ten to the two and G is 9.8, V is 20 ms, R is 10 ms. Okay? Okay. And this is the summary of the solution. And then just for completeness sake, this is Newton's universal law of gravitation. Any two masses m and uppercase M will experience a force of attraction equal to g times m times m divided by r squared. Some books will use m1 m2 to differentiate the masses. And if you notice what we call m times g is actually this. this equation g small m uppercase m over r squared so the g that we use as 9.8 is actually this expression g is a constant the gravitational constant m sub e is the mass of the earth and then r would be the radius of the earth so picture this so this is the earth and then this is a small mass m on the earth. This is m e. So the force of attraction from the center of the earth, which is r e, to the object m, that force is equal to g small m then m e divided by r e squared. and we just write this as m times g, okay? So the g that is 9.8 is actually the g constant times m e divided by r e squared. So if this is a good take-home problem, go to the book and look up the value of... g, the value of the mass of the Earth is given, or you can Google it from your internet search. Also, look up the radius of the Earth, plug in those numbers, and see if it gives you a number close to 9.8 meters per second squared. Okay, so f is the... Magnitude of the gravitational force, g is the gravitational constant, the value is given, 6.674 times 10 to the minus 11 Newton meters squared per kilogram squared. And let's go back. So this equation implies that any two objects, that means any two objects in the universe, regardless of size, experience a force, a mutual force of attraction. But because of the gravitational constant, which is of the other 10 to the minus 11, that means the force between everyday objects will be so small that you cannot even measure it using, you know, a force sensor. So even though you're experiencing a pull or push from different masses around you, the attraction is so small that you don't notice it. Okay. But when you start thinking in terms of planets, because of the large mass of the planet, you know, the force of attraction or the gravitational force becomes important. Okay. And then Kepler's laws. So this is again, this is for completeness sake, you need to know what is Kepler's law. Kepler's law simply relates the periods and the radius of orbit for satellites or any objects orbiting, you know, a massive planet. And we're going to run out the class today with the concept of, the concept of banking in terms of road design. So banking is just the process of raising one side of the road so that it's inclined or it's not flat, so the road is not flat. And this is useful because without friction, the only way a car can negotiate a bend or a curve is if the car can use part of its weight. to provide a centripetal force, you know, to enable it to travel in a curved path. So consider this situation, this is a curved road. Okay, you want to negotiate this curve. If the track is flat, then the only way you can negotiate this curve is if there is friction between the tires and the road. So you would need a force of friction directed towards the center of the curve to keep the car going in the curve path. Otherwise, the car will skip. of the road. Okay. So this is the aerial view. If you look at the car from the air, that's what you see. Now, if you look from the side, then you simply see this, the normal force, that's the force of the road on the car. And then the weight of the car, which is the force of gravity. So this is viewed from the side. Now, if we look at this. two cases, we know that from here the net force is equal to zero because the car is not moving up and down. So that means F sub n minus F sub g equal to zero or F n equals F g which is n times g. Now here we know that the force of friction is equal to mu s times Fn, so which is mu s times mg. We already saw this in chapter four. Now if you apply Newton's law for this case, sum of forces in the central direction equals m a c. The only force acting in the central direction is the force of friction. So that means we know the force of friction which is mu s times mg equal to mv squared over r. You can cancel the m's and then this means that your v squared is just equal to r times mu times g. Or your v is equal to square root of... R times mu times g. So for a flat road, friction must be present for you to be able to negotiate the bend. Otherwise, the car will skate off the track. That is why it's so hazardous to drive under icy conditions if the road doesn't have some kind of banking. That's why. You will notice most exits on most curved segments of a road, you will notice that the road is not flat. So the engineers designing the roads must incorporate some banking into the road to provide the safety force required to safely negotiate the bend. And that ideal angle will be... will also be known as ideal banking. Now let's look at the second case. So if friction is not present, then there must be some banking or some inclination to help the car safely negotiate a curve. So here I'm going to call the I'm going to call the angle of inclination theta. So this is the angle of inclination. Of course, we know that the weight of the car will be directed downwards. The normal force will be directed perpendicular to the... inclination as shown to the right. So this is the free body diagram. So this is your m times g, this is your f sub n, and if this angle is theta, the angle between the normal force and the vertical will also be theta. If you complete this triangle, you notice that this is the Y component and this is the X component. So this is Fn, opposite is sine, so Fn sine of theta, and then Fn cosine of theta. Now to keep this curve safely on this curve, the force that... that provides the centripetal force in this case is the the Fn sine of theta. Fn cosine of theta just helps keep the car you know balanced on the road. So if we apply Newton's second law in the direction of the curve the center force which is the force directed towards the love is Fn sine of theta equals mv squared over r. And of course Fn cosine of theta equals m times g because the car doesn't move, you know, up and down. So Fn cosine of theta minus mg equals zero or you could directly say fn cosine of theta equals m times g. So you can solve from the second equation, you could say fn equals to mg divided by cosine. And then plug this into the first equation and you have, where you have fn, you say mg divided by cosine times sine. equal to m b squared over r. You can cancel the m's and remember that sine divided by cosine is equal to tangent from trigonometry. Okay so this means that the tangent of the angle of banking is equal to v squared over r. So tan of theta equals v squared over r. And then from here, you can, sorry, g times the tangent. Only m cancel out, g did not cancel. So g times the tangent equals v squared over r. And So basically what we establish for this is g tangent of theta equals v squared over r. So that means v is equal to, or you could say the tangent of the angle of banking is equal to v squared over r times g, and then the angle of banking is equal to the tan inverse of b squared over r times g. So if the axis is to calculate the angle of banking, all you need to know is the speed of the car, the radius of curvature, and then g. Alternatively, if the axis is to find v, then v will just be the square root of r times g times the tangent of the angle of banking. Our time is up, so we'll end here today. And I'll make this recording available to the class in Blackboard by the end of the day tomorrow. So you should go back and review the concepts and then apply this to problem solving. And next week, we're going to devote the classes to solving. you know, practice problems from chapter five, and then we'll have a review session before the test. Remember to, you know, set up your proctor appointment for the test.

undergoing uniform circular motion. The first section begins with a description of rotational angle and angular velocity. So when describing the motion of an object that is moving in a circular path, it is more useful to consider the angular displacement as opposed to the linear coordinates x and y. So in this example, a particle starting from a and then ending up at b undergoes an angular displacement shown by the the variable theta.

Theta is used to denote angles. It's a Greek symbol. At that same interval, the object covers a linear displacement along the curve equivalent to delta s.

So delta s here is known as the arc length. The arc length is simply length of any curved path. So the radius of the circle is denoted by r. And the relationship between the angular displacement and the accolades and radius is shown here.

If you recall in MAT 107, you surely studied under geometry, the relationship between accolades and the the angular displacement. So the arc length divided by the radius gives you the angular displacement. So from here you can deduce that the arc length is equal to the radius times the angular displacement. So in module one and module two and three, this delta S will correspond to what we call delta X. So essentially what we're saying is that if this arc was very short, it will correspond to like a straight line, a short straight line segment delta X.

and delta x is related to the angular displacement by this simple equation. So multiply the radius by the angular displacement and you get the linear displacement. Okay, so you could also say x equal to r times theta. Okay, but in In mathematics textbooks, it's more common to represent the acland by the symbol s instead of x. So don't get, you know, thrown off if you see the acland denoted by s instead of x.

It means the same thing. Now the... the relationship between the angular displacement and the sorry the relationship between the the the angular measure is shown in the is shown right here let me highlight it for you so we know that one complete round on the cycle is called a revolution. And one revolution is equivalent to 360 degrees. Now in terms of radians, this is equivalent to 2 pi radians.

So 2 pi radians is equal to one revolution, and that is equal to... 360 degrees. Now if you divide both sides by 2 pi, you would obtain the relationship between radians and degrees.

So one one radian will equal to 360 degrees divided by 2 pi okay so but both but fundamentally one rev so one rev is equal to 2 pi radians so these are all conversion factors that you will need to use when solving problems involving rotations. So if the angle is given in degrees, you can use this the conversion factor between radiance and degrees to find to do the conversion. If the angle is given in terms of revolutions, then you would use the conversion between revolutions and radiance. And if you do the division 360 divided by 2 pi, you get 57.3. Therefore, 1 radian equals 57.3 degrees.

Now, take note of these abbreviations. So the first abbreviation is REV. That stands for revolutions. So whenever I say The abbreviation REV, that means revolution. And revolution, like we saw in the previous slide, is equal to 360 degrees. So another common abbreviation is rev per minute.

So rev per minute means revolutions per minute. So for instance, if they say... The angular speed of an object is 20 revs per minute.

That means 20 revolutions per minute. And we know that one revolution equals two pi radians. So we can convert from revolutions per minute to radians per minute and we can convert to radians per second because we know one minute is 60 seconds. Another common abbreviation is RPM.

If you look at your car's speedometer, you would see the speed is either graduated or calibrated in RPMs. I mean, the rotations of your tire is calibrated in RPMs. That means the revolutions per minute. So if you look at that number, it tells you how many times your car's tires are rotating. And this could be indicative of the health of your car's rotation system.

Now the first concept we're going to define is that of angular velocity after having established this background. So angular velocity is simply a measure of how fast an object is rotating. And similar to the linear velocity, the angular velocity is the change in angle divided by change in time.

So final angle minus initial angle divided by time. If you recall, the linear velocity was defined as what? The initial position, final position minus initial position divided by time.

Okay and So this is called omega. It's a Greek symbol. Looks like a W. Okay. So that's a symbol for angular velocity.

And what I was saying earlier was that linear velocity was defined as change in x. divide by change in t which was x minus x initial divided by delta t. So if you look carefully you will notice that what we just did here is just re-level x as theta and re-level v as omega.

It's important to remember the names because sometimes students mix up theta and omega. Okay, so one useful approach to study rotational kinematics is to view the variables as a re-leveling of the kinematic equations we already studied. So if you already knew that kinematic equations.

What you need to do in this chapter is just rename the x, v, and a variables to the angular names theta, omega, and alpha. Alpha is the angular acceleration. And as we saw previously, The arc length delta s is the same as r times change in theta. So if you divide the arc length by time, so that gives you the linear speed.

This equivalent leads to this relationship between the linear speed and the angular velocity. So the linear speed V is equal to r times omega because right here, changing theta divided by changing t is omega. So v equals r times omega, or omega is equal to v divided by r. So these are the two equations that link the... angular velocity to the linear velocity.

So if you come across any problem that requires you to find the angular velocity given the linear velocity, then you would need to use one of those two equations. And the units of angular velocity is radians per second, that's the SI units, just like for linear velocity, the units were meters per second. So whenever the speed was given in different units like kilometers per hour, miles per hour, or anything different from meters per second, then we have to convert to meters per second.

So here in module 5, you would need to convert to radians per second whenever the angular velocity is not given in the SI units of radians per second. The direction for angular velocity is counterclockwise or clockwise. Okay, so something moving in a circle has only two possible directions in the sense of the clock. which we call clockwise, or opposite the sense of the clock, which is known as counterclockwise. The positive direction is by default, or the convention is that counterclockwise is positive, clockwise is negative.

So if you wanted to make an analogy to the linear case, counterclockwise will correspond to the positive x-axis and clockwise will... correspond to the negative x-axis. So clockwise this is negative and then counterclockwise.

This is considered positive. So clockwise, this is clockwise. And this is counterclockwise.

So if an object is rotating like the the hands of a clock, you would treat its angular velocity as negative and if it was rotating counter to the clock you would treat the angular velocity as positive. The direction for linear velocity is tangential to the path of motion. the direction of the for an object going around the circular or curved path its velocity is always directed along the tangent to that path at the point of interest okay and to summarize again the v is equal to r times omega or omega equals v divided by r and in this relationship We can deduce that V is proportional to the distance from the center of rotation, so V is proportional to R, and it does largest for a point on the rim of an object that is rotating. So for instance, if you consider different points on the tire that is rotating, the point that is at the rim, that is at the circumference of the tire, We'll have the lodgers. linear velocity compared to the one that is closer to the center of the tire.

So in this example, it says we should calculate the angular velocity of a 0.3 meter radius car, a car tire, when the car travels at 15 meters per second. which is about 34 kilometers per hour. So to solve this we would need to, sometimes it's always useful to do a sketch. So they said this is the tire and the problem says that The car is going forward at 15 meters per second.

So that means every point on the tire is also going forward at 15 meters per second. We know that the radius of the tire is 0.3 meters and there's a fine angular velocity so we're looking for omega. We know v. we know r and we don't know omega but we from our relationship we know that v equals r times omega or omega equals b divided by r so in this case it will just be 15.0 meters per second divided by 0.300 meters per second. Sorry, meters.

The radius is measured in meters, not meters per second. And if you work this out, you will obtain... The meters will cancel out and you will have it 50.0 per second of... you could say 50.0 radians per second. Okay, so this is the solution.

The next concept after Angular velocity is angular acceleration. So the angular acceleration is defined in a similar fashion. It's the rate of change of angular velocity. So alpha equals change in omega divided by change in t.

So omega final minus omega initial divided by change in t. And if you recall, when we discuss linear motion, the linear acceleration, a, was equal to change in v divided by change in t. So v minus v initial divided by change in t.

So what we did here is just rename v as omega. So in fact we're really using the same equation as from module two and again alpha so this is alpha pronounced alpha okay it's a greek greek symbol The units of angular acceleration is radians. per second divided by seconds because omega is measured in radians per second then you divide again by t which is seconds so you get radians per second squared similar to the linear counterpart which is meters per second squared so in the next example suppose the teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5 seconds.

Calculate the angular acceleration in radians per second squared and if if she now slams on the brakes causing an angular acceleration of negative 87.3 meters per second squared, how long does it take the wheel to stop? So here we have a problem that requires us to find the angular acceleration and to find the time it takes the spinning wheel to come to a complete stop. So let's read the problem carefully and pay attention to some keywords.

In the first part we know that the wheel starts spinning from rest to a final angular velocity of 250 rpms. So a closer look at this problem reveals that the keyword rest means the initial angular velocity is zero and then a final angular velocity is 250 rpm that's omega and remember rpm means revolutions per minute so we will need to convert this to radians per second and then the time is five seconds and the first part they say find the angular acceleration which is alpha now if she now slams on the brakes causing an angular acceleration, so this is your alpha. How long does it take the wheel to stop? So how long refers to time and to stop means the final angular velocity in path B will be zero if the wheel comes to a stop.

Okay so having identified this information The first step is to convert the, you know, the RPMs to radians per second. So 250 revs per minute. One rev is 2 pi radians.

One minute is 60 seconds. So if you line up the conversion factors correctly, you would end up with 26.2 radians per second for which is equivalent to the 250 RPMs. And then alpha is simply the change in omega divided by T. So 26.2 minus zero divided by five gives you 5.24 radians per second squared. And then part B, we know that the wheel comes to, starts at 250 RPM and stops.

which means the final angular velocity is zero. And so the change in angular velocity is zero minus 250, which is negative 250 RPM. Alpha is minus 87.3 radians per second squared.

And so looking at this equation for alpha, you can manipulate the equation to isolate T on one side because that's what we're looking for. And then after converting the RPMs to radians per second, again plug it into the equation, negative 26.2 divided by negative 87.3 and you get 0.3 seconds as the time required to bring the wheel from 250 RPM to zero with an acceleration of negative 87.3 radians per second. square. So let's look at angular acceleration and how it relates to tangential acceleration.

So we already defined angular acceleration as the rate of change of angular velocity. In this diagram, pay attention to the direction of the tangential acceleration. So the tangential acceleration is directed in the same direction as the tangential velocity.

So this is your tangential acceleration. Now remember that v is equal to r times omega. So change in v is equal to r times change in omega. We divide both sides by t.

Change in v divided by change in t is the tangential acceleration. And then delta omega divided by delta t is alpha, so r times alpha. So tangential acceleration is equal to angular acceleration times radius. We're going to come back to centripetal acceleration later, so just ignore that for now.

Okay, so this is the equation I just derived. You need to remember this relation, and the units for tangential acceleration is meters per second squared. The units for angular acceleration, alpha, is radians per second squared. Okay, so in the yellow box, you have the two equations that... that are useful in converting alpha to a sub t or a sub t to alpha.

So if you know alpha you can find a sub t by multiplying alpha by the radius and if you know a sub t you need to divide it by r to get alpha. So kinematics of rotational motion. So as we saw for one-dimensional and two-dimensional kinematics kinematics is simply concerned with description of motion without regard to force or mass. So in the next slide we're going to see that all the translational or linear kinematic quantities such as displacement, velocity, and acceleration have direct analogs in rotational motion. So in this table The first column shows the rotational variables, the second column shows the translational or linear variables, and the third column shows the relationship between the linear and translational variables.

So theta is measured in radians, omega is measured in radians per second. alpha is measured in radians per second squared, x is measured in meters, v is measured in meters per second, a is measured in meters per second squared. And the relationship between the linear and translational is shown in the third column.

You could also rearrange the equations to have Now x equal to r times theta, b equal to r times omega, a sub t equal to r times alpha. So these are all equivalent relations between translational or linear and rotational. Okay take note I'm using x here, some books will use s for the to highlight the fact that we're dealing with curves and not just straight lines, okay? So remember this relationship. If you can make a mental association between the linear and angular variables, then to an extent, module five, most of module five is just like a review of module.

two and three of module two because rotational you know the rotational variables are kind of like one one-dimensional except in the case of acceleration where you have two types the translational and the centripetal acceleration. So if you remember the three main equations we use in module two, just re-label your x as theta, your v as omega and your a as alpha and you get the equations for rotational motion. And these are the equations.

So the first equation says the linear displacement is equal to the average velocity times time. And so the angular variable or angular displacement is equal to the average angular velocity times time. So there should be like a delta here to emphasize the fact that you're referring to displacement, okay? Here as well as here. So v equals v sub o plus a t omega equals omega sub o plus alpha t.

delta x equals v o plus a t, delta theta equals omega naught t plus one half alpha t squared, and v squared equals v sub o squared plus two a x, omega squared equals omega sub o squared plus two alpha times change in theta. So exactly the same equation except we level the three variables theta. omega and alpha. And the subscripts you know a subscript of zero means initial value and the variables without subscripts means the final value.

So the problem solving strategy for rotational Kinematics is as follows. The first step is to examine the situation to determine that rotational kinematics is involved. If rotational is not involved, then you cannot use the rotational equations.

But note that you don't need to consider forces or masses because when you're doing kinematics, you're purely looking at the displacement, the velocity. the acceleration and the time. Next, identify exactly what needs to be determined in the problem.

Sometimes it's useful to sketch the situation. Third, make a list of what is given or what you can infer from the problem. Like in the previous two examples, we could infer that if something starts from rest, that means the initial velocity is zero. In the same problem, we could deduce that if something started at a certain speed and came to a stop, that means the final speed is zero. So you need to always pay attention to those keywords.

So make sure you list everything that's given and everything that you can infer from the keywords in the problem. And then the fourth step is to identify the appropriate. the equation or equations for the quantity to be determined.

Sometimes you may need to use two equations. You may need to use one equation first as a transition and then get the value from that equation and then feed it into a second equation to find the unknown. And so you're gonna apply the same strategies we use for translational.

in a motion. Then step number five, once you identify the equation, always isolate the unknown to one side and then substitute the values with the correct units and then simplify to get the final answer. Always pay attention to the units.

The SI unit for angular measure is radians. and check your answer to see if it's reasonable in terms of units. in terms of magnitude. Ask yourself, does this make sense?

If you're calculating the speed of plane or the acceleration, ask yourself if the number you got makes sense, you know, for given the, you know, the numbers in the problem. Don't just, you know, write down whatever your calculator shows you without, you know, pausing to ask yourself if it makes sense. Oftentimes you can make a simple algebraic error and by checking your answer in step number six it's easy to correct that you know that error.

So now we're going to talk about centripetal acceleration which is the acceleration that is present whenever circular motion is is taking place. So whenever an object is moving along a curve, the most be an acceleration that is directed towards the center of the curve. As you can see in this diagram, if we go make a flashback to concept of vector addition.

So this is v1 the initial velocity this is the final velocity. Take note v1 and v2 could have the same magnitude but because they're directed at different directions so they're not the same. So v2 minus v1 cannot be zero because even though they have the same magnitude they don't have the same direction. the vector minus v1 will be a vector directed in the opposite to v1. Okay so negative v1 will be directed in this direction.

So if you wanted to add, so you wanted to find delta v which is v2 minus v1 of v final minus v initial. According to the graphical tip-tail method, you will write, let's say, your v. So technically what you're doing is like you're adding minus v1 to v2. So minus v1, this is minus v1, and then v2.

tip to tail is this way and then your delta v which is your v one minus v one plus v two or v two minus v one is this vector so this is delta v this is negative v one and this is v two So delta v is the same as v2 minus v1 or minus v1 plus v2. Okay, and as you can see from this vector diagram, if I translated this triangle to the circle, you notice that delta v is directed towards the center of the circle. So your your centripetal acceleration which comes from the change in direction of the velocity vectors v1 and v2 is always directed you know towards the center of the circle and So, okay, so let's review this. So speed and velocity around a circle.

So in uniform circular motion, the velocity changes direction but the magnitude of V is constant. So when you hear the word uniform circular motion, that means the magnitude of V is the same, although the direction is changing as the particle goes around the circle. So the total distance around the circle is 2 pi r, which is the circumference of the circle.

The time required to go around a circle or to complete one cycle of motion is called the period, is denoted by uppercase T. Again, do not confuse this with tension, because tension is also represented by T. So always pay attention to the context.

So in this diagram, the circumference is 2 pi r. which is the distance. The time taken to cover the circumference is the period.

Therefore, the speed for uniform circular motion is 2 pi radius divided by t, which is the period. Therefore, for an object like the Earth, we know the period of the Earth's rotation is 24 hours. It takes the Earth 24 hours, that's one day to completely rotate around its axis. So if we know the radius of the earth, we could compute the translational speed of the earth by simply plugging in the radius for r and then plugging in 24 hours. Of course you need to convert to seconds to get the rotational speed or the tangential speed of the earth.

in meters per second. Now if you wanted to find the angular speed of the earth it would simply be changing theta divided by change in time and we know of course for one complete revolution the angle is simply two pi radians so two pi radians divided by time. So again you can see that this two pi divided by time is like your omega. So again the relation we derived earlier r omega still holds okay.

So this is the same as r times omega where omega is defined as two pi divided by the period. So if they ask you to find the angular velocity of the earth, all you need to do is do 2 pi divided by 24 hours. Convert the 24 hours to seconds and then plug it into this equation.

So is it possible for an object traveling at constant speeds to have a non-zero acceleration? You can press yes or no. There's a yes or no button. So let me see your feedback by pressing yes or no. Is it possible for an object that is traveling in a circular path with a constant speed, is it possible for it to have a non-zero acceleration?

If you think the answer is yes, you know, press yes. on the feedback buttons. So if we click on participants you see two feedback buttons.

Yes it's a green button and no it's a red circle with an x in it. So if the answer is yes click on the green circle. If the answer is no click on the red circle. Okay so Some, yes, some class members have responded and I think most of the class says yes. Okay, so this is, the current answer is yes, indeed, because even though the speed is constant, As long as the direction is changing, which is the case if something is moving in a circle, the direction changes every time because velocity is defined as the vector quantity, which is characterized by two parts, magnitude and direction.

So for any object that is traveling at constant speed around a circular path, the direction is changing all the time even though the magnitude of the speed remains constant. Okay, so this is something students find hard to digest. Because common sense tells us that if the speed is constant, then the acceleration should be zero.

But that is faulty thinking because acceleration is not defined in terms of speed. It's defined as the change in velocity divided by change in time. So speed is just one part of the velocity. It's just a magnitude. And so the magnitude could be constant if the direction changes.

then there will be an acceleration. Okay, so again, these are just reminders. The velocity vector is always tangent to the path and is perpendicular to the radius. Similarly, the centripetal acceleration vector is perpendicular to the velocity vector and points towards the center of the circle.

So the The centripetal acceleration is directed along the radius and is therefore perpendicular to the velocity vector because the velocity vector is perpendicular to the radius. So this is the expression for the centripetal acceleration. We're not going to go into the detailed derivation.

that is not necessary for this course, but you need to remember that the centripetal acceleration a sub c is equal to v squared divided by r and v is the the tangential velocity, r is the radius of the circular path and we replace v by 2 pi r divided by t which we derived in the previous slide. Then the centripetal acceleration could be written as 4 pi squared r divided by period squared. We can simplify this expression.

Now following the concept of acceleration, we can introduce the concept of centripetal force, which is the force that results from the acceleration. Remember Newton's law. implies that whenever there's an acceleration, there must be a net force. That's the second law. F net equals m a.

So if a is not zero, which is the case for circular motion, we just established that for circular motion, there must be an acceleration directed towards the center, which we call centripetal acceleration. So that means for every circular motion there must be a force directed towards the center, which we call centripetal force. So F net in the direction towards the center equals m a c and since a c is v squared over r, so that means f c equals m v squared over r.

Now take notes, FC is not like a single force, like it's not like a normal force, tension force, friction force and so on, rather it's the net force towards the center. So FC could be the result of adding multiple forces, it could be the result of just a single force if it's... only a single force is acting towards the center.

So in terms of radius and period, the equation could be written like this. Okay and just to add we already know that v is equal to r times omega. So v squared is equal to r squared omega squared. So you could write this as m times r squared omega squared divided by r and one of the r's will cancel and so you would have m r omega squared. So this is another form of the centripetal force.

So you could use this equation. You could use this equation. You could also use this equation to find the centripetal force on an object undergoing circular motion. The difference is what is given.

So if you know V and R, then you use the first box. If you know, so here we know V and R. Now if on the other hand you know, so I'm assuming the mass is given in all cases.

Now if you knew the radius and the period, then you would use this equation. And if on the other hand you knew the radius and the angular velocity, then you would use the third equation. Okay, so you have like three different tools and you can pull out each tool depending on what is given in the problem. So let's apply this to the case of an object moving in a horizontal circle. So for an object moving in a horizontal circle as shown in this diagram, imagine something tied to a string and then caused to move in a horizontal circle.

The force of the string pulling the solid inward is known as the tension force. And this is the force responsible for the object moving in a circular path. And since this is the only force acting in the central direction, the net force in the central direction is just the tension.

And so if you wrote Newton's law, if you wrote Newton's second law for this problem, you would have... sum of forces plus the center equals m a c and the sum of forces is just the tension force equals m and the ac is v squared over r so that means for this case the tension force is equal to m v squared on r if the arcs you define v then you could simply say v squared equals R times T, multiply both sides by R, then divide by M, and then your V would equal square root of R times the tension, divide by M. On the other hand, it could ask you to find the radius.

So the radius will just be R equals, multiply both sides by R, so R times T equals MV squared, so R equals MV squared. divide by T. Okay, so the tension on the string provides the centripetal force directed towards the center and therefore you can say T equals mv squared over r.

Take note T here is the tension, it's not the period. So don't confuse this T with other T that stands for period. Okay so Here's an example. If a car with a mass m travels at constant speed around a circular curve with radius r, in what direction does the centripetal force point?

Okay, so obviously in this case the The centripetal acceleration will point towards the center of the circular curve and that corresponds to the second option. Which statement below best describes the centripetal force on the satellite that is orbiting the Earth in a uniform circular path with constant speed? The satellite is 300 miles above the Earth. So we have four options A, B, C and D.

So you can type in A if your answer is the first option and B for second option, C for the third option, D for the fourth option. okay so if you select that D, then that's the correct answer because in this case, so let's say the satellites, let's say this is the orbit of the satellites. The satellite is located somewhere here.

And this is the center of this. This orbit is the Earth. The force pulling the satellite inwards towards the Earth is the weight of the satellite or the gravitational force.

So since this is the only force pulling the satellite inwards, if you apply the sum of forces towards the center equals mv squared. over r. So this the net force towards the center is just f of g equals mv squared over r. Okay, so and this force is mass times gravity equals mv squared over r.

You can cancel out the m. and then you just have v squared equals r times g of v equals square root of r times g. So if you knew the radius of the orbit then you could calculate the satellite speed from this equation, the radius times gravity square root.

Okay so you'll be surprised that these are some of the the simple equations. Simple calculations that have to be done to put a satellite into a certain orbit. The next problem, so a roller coaster has a mass of 5 times 10 to the 2 kilograms. When fully loaded with passengers, if the vehicle is...

if the vehicle is so low that it travels at 20 meters per second, what is the force of the track on the vehicle at A? Note G equals 9.8 meters per second squared. So at A, the first thing we need to do is to draw a free body diagram, okay? Because we're looking for a force.

And the free body diagram will look like this. The weight of the car directed downwards. the force of the track which is also known as the normal force directed upwards and the acceleration since the car is going the circular path the centripetal acceleration is v squared divided by r and we're looking for the force n so you're going to apply newton's second law f net equals m a So sum of forces in the central direction.

Remember the center of curvature is located above A. So that's why AC is pointing upwards. Okay so sum of FC equals MAC and the sum of forces we have N upwards minus mg which is downwards equals mv squared over r and so n is equal to mg plus mv squared over r or you could factor the m out g plus b squared over r and then just plug in the numbers 500 kilograms for the mass five times ten to the two and G is 9.8, V is 20 ms, R is 10 ms. Okay?

Okay. And this is the summary of the solution. And then just for completeness sake, this is Newton's universal law of gravitation.

Any two masses m and uppercase M will experience a force of attraction equal to g times m times m divided by r squared. Some books will use m1 m2 to differentiate the masses. And if you notice what we call m times g is actually this. this equation g small m uppercase m over r squared so the g that we use as 9.8 is actually this expression g is a constant the gravitational constant m sub e is the mass of the earth and then r would be the radius of the earth so picture this so this is the earth and then this is a small mass m on the earth. This is m e.

So the force of attraction from the center of the earth, which is r e, to the object m, that force is equal to g small m then m e divided by r e squared. and we just write this as m times g, okay? So the g that is 9.8 is actually the g constant times m e divided by r e squared. So if this is a good take-home problem, go to the book and look up the value of...

g, the value of the mass of the Earth is given, or you can Google it from your internet search. Also, look up the radius of the Earth, plug in those numbers, and see if it gives you a number close to 9.8 meters per second squared. Okay, so f is the...

Magnitude of the gravitational force, g is the gravitational constant, the value is given, 6.674 times 10 to the minus 11 Newton meters squared per kilogram squared. And let's go back. So this equation implies that any two objects, that means any two objects in the universe, regardless of size, experience a force, a mutual force of attraction. But because of the gravitational constant, which is of the other 10 to the minus 11, that means the force between everyday objects will be so small that you cannot even measure it using, you know, a force sensor. So even though you're experiencing a pull or push from different masses around you, the attraction is so small that you don't notice it.

Okay. But when you start thinking in terms of planets, because of the large mass of the planet, you know, the force of attraction or the gravitational force becomes important. Okay.

And then Kepler's laws. So this is again, this is for completeness sake, you need to know what is Kepler's law. Kepler's law simply relates the periods and the radius of orbit for satellites or any objects orbiting, you know, a massive planet. And we're going to run out the class today with the concept of, the concept of banking in terms of road design. So banking is just the process of raising one side of the road so that it's inclined or it's not flat, so the road is not flat.

And this is useful because without friction, the only way a car can negotiate a bend or a curve is if the car can use part of its weight. to provide a centripetal force, you know, to enable it to travel in a curved path. So consider this situation, this is a curved road. Okay, you want to negotiate this curve.

If the track is flat, then the only way you can negotiate this curve is if there is friction between the tires and the road. So you would need a force of friction directed towards the center of the curve to keep the car going in the curve path. Otherwise, the car will skip. of the road. Okay.

So this is the aerial view. If you look at the car from the air, that's what you see. Now, if you look from the side, then you simply see this, the normal force, that's the force of the road on the car. And then the weight of the car, which is the force of gravity.

So this is viewed from the side. Now, if we look at this. two cases, we know that from here the net force is equal to zero because the car is not moving up and down.

So that means F sub n minus F sub g equal to zero or F n equals F g which is n times g. Now here we know that the force of friction is equal to mu s times Fn, so which is mu s times mg. We already saw this in chapter four.

Now if you apply Newton's law for this case, sum of forces in the central direction equals m a c. The only force acting in the central direction is the force of friction. So that means we know the force of friction which is mu s times mg equal to mv squared over r.

You can cancel the m's and then this means that your v squared is just equal to r times mu times g. Or your v is equal to square root of... R times mu times g. So for a flat road, friction must be present for you to be able to negotiate the bend. Otherwise, the car will skate off the track.

That is why it's so hazardous to drive under icy conditions if the road doesn't have some kind of banking. That's why. You will notice most exits on most curved segments of a road, you will notice that the road is not flat.

So the engineers designing the roads must incorporate some banking into the road to provide the safety force required to safely negotiate the bend. And that ideal angle will be... will also be known as ideal banking.

Now let's look at the second case. So if friction is not present, then there must be some banking or some inclination to help the car safely negotiate a curve. So here I'm going to call the I'm going to call the angle of inclination theta. So this is the angle of inclination.

Of course, we know that the weight of the car will be directed downwards. The normal force will be directed perpendicular to the... inclination as shown to the right.

So this is the free body diagram. So this is your m times g, this is your f sub n, and if this angle is theta, the angle between the normal force and the vertical will also be theta. If you complete this triangle, you notice that this is the Y component and this is the X component.

So this is Fn, opposite is sine, so Fn sine of theta, and then Fn cosine of theta. Now to keep this curve safely on this curve, the force that... that provides the centripetal force in this case is the the Fn sine of theta. Fn cosine of theta just helps keep the car you know balanced on the road. So if we apply Newton's second law in the direction of the curve the center force which is the force directed towards the love is Fn sine of theta equals mv squared over r.

And of course Fn cosine of theta equals m times g because the car doesn't move, you know, up and down. So Fn cosine of theta minus mg equals zero or you could directly say fn cosine of theta equals m times g. So you can solve from the second equation, you could say fn equals to mg divided by cosine. And then plug this into the first equation and you have, where you have fn, you say mg divided by cosine times sine.

equal to m b squared over r. You can cancel the m's and remember that sine divided by cosine is equal to tangent from trigonometry. Okay so this means that the tangent of the angle of banking is equal to v squared over r.

So tan of theta equals v squared over r. And then from here, you can, sorry, g times the tangent. Only m cancel out, g did not cancel. So g times the tangent equals v squared over r. And So basically what we establish for this is g tangent of theta equals v squared over r.

So that means v is equal to, or you could say the tangent of the angle of banking is equal to v squared over r times g, and then the angle of banking is equal to the tan inverse of b squared over r times g. So if the axis is to calculate the angle of banking, all you need to know is the speed of the car, the radius of curvature, and then g. Alternatively, if the axis is to find v, then v will just be the square root of r times g times the tangent of the angle of banking.

Our time is up, so we'll end here today. And I'll make this recording available to the class in Blackboard by the end of the day tomorrow. So you should go back and review the concepts and then apply this to problem solving.

And next week, we're going to devote the classes to solving. you know, practice problems from chapter five, and then we'll have a review session before the test. Remember to, you know, set up your proctor appointment for the test.

Transcript for:Understanding Circular Motion and Gravitation

Transcript for:
Understanding Circular Motion and Gravitation