Practice: For the situation above, suppose the string breaks if its tension exceeds 50 N. Calculate the maximum speed that the object can attain without breaking the string.

Subjects

Sections | |||
---|---|---|---|

Uniform Circular Motion | 27 mins | 0 completed | Learn |

Centripetal Forces | 53 mins | 0 completed | Learn |

Newton's Law of Gravity | 27 mins | 0 completed | Learn |

Gravitational Forces in 2D | 26 mins | 0 completed | Learn |

Acceleration Due to Gravity | 14 mins | 0 completed | Learn |

Satellite Motion: Intro | 6 mins | 0 completed | Learn |

Satellite Motion: Speed & Period | 35 mins | 0 completed | Learn |

Geosynchronous Orbits | 15 mins | 0 completed | Learn |

Overview of Kepler's Laws | 4 mins | 0 completed | Learn |

Kepler's First Law | 10 mins | 0 completed | Learn |

Kepler's Third Law | 14 mins | 0 completed | Learn |

Concept #1: Intro to Centripetal Forces

**Transcript**

Hey guys. So in this video I'm going to introduce centripetal forces, centripetal forces are going to exist when you're moving in a circle and we're going to see that the process for solving those problems are very similar to solving other force problems that we did in the past. You draw a free body diagram you write F equals ma and so and so forth. There's only a few things that a little bit different about circular motion and centripetal force. Let's check it out. So in linear motion when you're moving in the x axis the y axis or even at an angle like this as long as it's a straight line. Right. We had forces in the x and y axis. Now we're going to have forces in the centripetal axis. And the idea is that the x or y axis can be thought of as a centripetal axis if rotation is happening in that direction. Let me show you real quick. Let's spinning a rock at the end of a string horizontally like this some spinning a rock like this. Right. So if it's here at this edge then it's pointing this it's accelerating towards the center. Right it's always accelerating towards the center. And here the X-axis is also the centripetal axis. If I'm spinning it vertically so that at the bottom it's accelerating up and at the top it's accelerating down. Remember the acceleration is centripetal. So it's towards the middle, here the y axis becomes the centripetal axis. And we're going to see that the centripetal axis takes precedence over the x and y axis. We'll see what that means a little later. So before we had sum of all forces equals ma in the X-axis and sum of all forces equals ma the y axis so forces in the y cause and acceleration y, forces in the x cause and acceleration the x. Now we're going to have same thing but in the centripetal direction, sum of all forces in the centripetal direction causes an acceleration in the centripetal direction. And you might remember that Ac equals V square over R that comes from a uniform circular motion. OK. One quick rule here that you have to know though is that when you're writing the sum of all forces in the centripetal direction, forces towards the center are positive forces away from the center or negative and tangential general forces do not get listed. I'll show you what that means in just a second. OK. So that's the convention that you have to use for centripetal forces. So let me show you three situations here. This is sort of a rollercoaster loop. This car is going to be at the bottom. This car is in the top on the top but inside of the tracks right on the rollercoaster you're always inside of the tracks. And then here the car is going to be at the rightmost point right there. OK. So what I'm going is show you the forces, so what are the forces here. I'm being pulled down by mg and I'm pushing against the track to the track pushes me back this way. Ok, forces towards the center are positive so normal is positive mg is negative. Normal is not positive because it's up it's positive because it's towards the center. OK. Now let's look here, mg obviously is always down which is positive even though it's down because it towards the center. And here you're pushing against the track this way so the track pushes you down like this as well. So they're both positive. OK. So if I were to write F equals ma here, sum of all forces equals ma first of all I would write sum of all forces are centripetal because I'm going around the circle equals mAc and the forces here will be N minus mg. And then we continue the problem. And the second one sum of our forces centripetal equals mAc would be N plus mg, why because they're both positive here ones positive the other ones negative. What about here. Well mg is always pulling down and normal is always perpendicular to the surface. Notice how in all of these normal's going to be positive because it's the force that keeps you in a circle. If I were to write this sum of forces you I have no more positive mg doesn't get listed. Why. Because you only forces that are into the circle or directly away from the circle. If you're making the 90 degree angle with that direction of towards or away from the circle which mgoes you just don't get listed so it's just normal equals mAc and say that mg here is not a centripetal force. It's not moving towards or away from the center. OK so that's the idea there. Let's do a quick example here. So a small three kilograms object so three kilograms. On top of a frictionless table. So here's a table there's no friction on it is attached to the end of a two meter string as shown. So this string here has a length or two meters. Now you might remember this if I spin on a string that has a length of two meters that is the radius of my rotation and that is a distance right. That's the distance away from. So l little r and big R are the same thing, l is the length of the rope. r is the radius of the circle that it forms are at the edge of the radius so that's my distance from the center. So basically little r equals 2m and it says here that it spins once every two seconds. That's right. So I can ride spins once. So one spin or one cycle every two seconds I can write as a fraction like this and then notice that the second is at the bottom. So this tells me. That this is frequency, frequency is 1 over 2 which tells me the period is the inverse of that, period is 2. And I wanna know the tension on the string. What is a tension in the string.

So tension is a force. So I can write F equals ma to find tension. OK. Now just to be clear here. Big T is already being used for period. So tension I'm gonna use Ft ok. Force of tension I mean just right here this is tension. And the way I'm going to do this is F equals ma, sum of all forces equals ma, now when you look at this this is a horizontal circle. So you might be tempted to write some of our forces in the x axis equals ma but it's not it's a circle. So the centripetal axis takes over the x axis takes precedence over the x axis and I write this I say centripetal F equals ma, the only force in the direction of the center is tension. So tension is pulling in towards the middle at all times and then you have sort of energy pulling you down normal pulling you up. All these two these two forces just all they do is cancel each other. And they're not in the centripetal direction right. So you're you're normal for your tension is sort of this way towards the center and you're normal's up and your mg's down. They make 90 degrees they're not either directed toward the center or away. They don't count. So the only force here is tension which will be positive because it's towards the center and, mass times acceleration. Now I have the mass I don't have the acceleration by the way. But what I can do is replace. I can go find the acceleration. Basically I'm not going to replace. It's going to go find it. So mass is three and the acceleration I need to get. So let me go off to the side here and find acceleration acceleration centripetal is V square over r. And notice I don't have a V. So I have to go get V as well. V is 2 pi rf or 2 pi r over t. All right. And if you do this if you plug in the numbers 2 pi r is 2 divided by 2. This is six point, 6.28 meters per second and if you then plug this back into acceleration you got six point twenty eight meters per second square. And you divide that by r which is 2 and the answer is nineteen point seven meters per second square. And that is what ends up going here. So my force of tension is just three times that.

My force of tension is just I have it here. Fifty nine point one Newtons so I hope I hope that makes sense. I hope you agree that it's pretty straightforward. I'm just writing and looking for force and write F equals ma in the process of doing that Ac has a special equation. If you are going around the circle which is V square over r and whenever you were writing F equals ma in the centripetal direction you should almost expect that you're going to have to rewrite your Ac as V square over that's what happens most of the time. OK. Now practice question number one here is related to Example 1 and it's it's telling that the tension the string breaks if the tension exceeds 50. So this is kind of like saying force or tension Max is 50. Now in physics whenever a problem tells you the maximum tension is 50 or the minimum speed is 30 you're just right tension equals 50 or V equals 30. You can almost ignore the fact that this is a maximum tension. Basically that's the number you're going to use the maximum or the minimum whatever extreme number they give you. That's the one you're going to use. So here tension maximum is 50 and I want to find the maximum speed. So basically tension's 50 and I want to find what is V OK and you're going to use F equals ma to do this because I'm giving you a force. So F equals ma it's in a circle so F equals ma is centripetal. So I'm going to give you that clue that hint, I want you guys to try solving this real quick and let's see if you get.

Practice: For the situation above, suppose the string breaks if its tension exceeds 50 N. Calculate the maximum speed that the object can attain without breaking the string.

Example #1: Intro to Centripetal Forces

**Transcript**

All right let's continue here so now I'm going to do another example but with vertical forces in a vertical circle. OK. So it says here are some crazy fighter pilot of mass 70 kilograms Mass equals 70 in some movie does a nosedive that is nearly circular, nearly circular it's just going to mean we're going to treat this as a circle. OK. So let's say here's your airplane what are words of fire coming out of here and you're flying this way and all of a sudden you do a nose dive nose dive looks like this. Somewhat right you go and you just kind of do. You down like that, nearly circular means that this thing forms is forms the part of the circle right. So. And I'm telling you that the speed at the bottom right here is V equals 80. You might remember you might remember that's my velocity in a circle is always tangential. So if I'm at the bottom of a circle I'm coming from here. Then the velocity is going to look like this. If you were to continue like this the velocity would have been going up but you actually went off this way, so it's always tangential to the path. So that means that this velocity at the bottom looks like this. OK. That's the velocity at the bottom and I want to know what is the centripetal acceleration.

Let's find that real quick. Ac is just I'm in a circle so I'm just gonna try V square over r and actually have these numbers so it's 80 square over three hundred. And eighty square over 300 gives me twenty one point three meters per second squared. OK. Now this is the acceleration of the airplane which is the same as the acceleration of the pilot. Just to be clear so I'm just gonna call that a they're moving together as a system. OK. Next thing is to ask for his apparent weights apparent weight means normal means normal so is a normal on the pilot's forget that point. Normal is a force and the only way you're always going to find normal is by using F equals ma. So sum of those forces misinterpret direction on the pilot is mass of the pilot times acceleration of the pilot which is a centripetal acceleration. What are the forces on him while he's down here and his velocity is this way. He obviously has mg and this might be tricky to see but the pilot is on the airplane and it's being pushed up by the seats. Right. So you're sitting there and then you have a normal force here which one of these forces are going to be greater. Well since you're going in a circle and you have to have acceleration towards the middle that's how you turn right. And that means that your normal force must be greater than your mg ok in a circle the positive force is always going to be greater than the negative force because positive towards the circle you're going that way.

And negative is away from the circle. So here normal will be greater than mg. But let's solve this so have normal positive plus mg negative, mass of pilot which is 70. That's just right here ma for now. So I can show that normal which is what I'm looking for it's just ma plus mg ,So it's 70 (a) is 21 point three 70g is 9.8. And that you do this you plug this all together. And the answer is 2180 Newtons is the normal force. If you were to calculate his mg, his mg would be roughly 790, his mg would be about 790. And notice how this is much bigger as we expected. Right. So that's it for this one to just have one last point to make right here at the end. Centripetal force is not a force in nature. For example gravity or the force of gravity or weight is a force of nature. Friction is a force of nature , normal's a force of nature that pushes the surface the surface pushes back. Tension is a force of nature. So and so forth centripetal force is an indication that this force in the centripetal direction so normal becomes centripetal or is categorized as a centripetal force. Tension might be a centripetal force but centripetal force itself is not a unique force. It's not a thing. It's just sort of a type of force or a indication of direction of force. OK. So you might label force as centripetal but centripetal force in itself is not anything. OK. Another force that is actually a force of nature will be categorized as a centripetal force. So I hope that makes sense. That's it for this one.

Example #2: Vertical Centripetal Forces

**Transcript**

Hey guys we're going to continue with centripetal force, now we're going to focus on vertical centripetal forces so motion in the vertical direction. A big difference here is that mg and normal Those two forces are going to play more of a role than they did in circular horizontal centripetal force. OK. Let's check it out. We're going to start with an example here. I have a Ferris wheel of radius 50 so that's right that's got a Ferris wheel draw a little circle and the radius is 50 and it takes 30 seconds to make a full cycle. 30 seconds to make a cycle that's the definition of period time to make a cycle. So can right away write that period equals 30 seconds and 80 kilogram guy rides on it. So the mass of this guy is 80 kilograms. Cool calculate his speed and centripetal acceleration. So this is straightforward. V is 2 pi r over t or 2 pi r f I have t so I'm going to go with this version of the equation. Now if you're right on the first you're always at the edge of it so your distance from the center is basically the same thing as a radius. So I can plug these numbers in here 2 pi 50 over 30. And if you do all this I have it here you get a ten point five. So nothing weird here. And for acceleration is just V square over r. Again nothing weird straightforward. Ten point five squared r is 20 and if you do this you get a 2.2, 2.2 meters per second squared. So hopefully you agree that part is straightforward. This is what's actually new here. I want to know his apparent weight at the bottom apparent weight. Remember is just what is the normal force acting on him in the bottom. So for Part B I want to know what is N bottom and for Part C I want to know what is N top the normal force and tell him when he's at the top.

So let's first draw what this looks like if he is in the bottom. So you're inside of the little cart thing for the Ferris wheel. And that's you right there. So mg pushes you down and normal pushes you up. So normal's positive mg is negative. OK. When you are at the top of the Ferris wheel. There is a little thing you're in, mg is pulling you down. So it's positive and normal push up so it's negative. So the force is basically flip in terms of science the difference between a Ferris wheel and a traditional roller coaster when you're looping is that on the ferris wheel you're always pointing up even at the top obviously and on a roller coaster you are upside down at the top.

So on the roller coaster your normal would have been down which would have been a positive. But here you're always straight up. So F equals ma is how I'm going to find forces that you always find forces sum of all forces in centripetal is F equals ma and the forces are normal positive and mg negative. This is for the bottom so the first picture here equals mAc. I'm looking for normal. So I hope you see right away normal is just ma plus mg. I have a, a is 2 so I can just plug in the numbers and I get the answer. So normal will be Eighty two point two plus eighty nine point eight. And if you plug this in you gets 960. OK. So what I want you to do now is I want you to just pause the video and try part C Try part C, I drew the diagram for you already. You should be able to just plug in F equals ma. I'm going to keep going here hopefully to try to get some more forces centripetal. And the only thing that changes is that normal is now negative and mg is now positive. So when I saw for normal I'm going to send this guy to the other side so I get this. And to get normal positive I can just multiply both sides by negative one or I can move things around however you like do it, so this is going to look like this. All the signs flip OK. So normal is mg minus ma. OK. So and mg minus ma and you do this you get 608. Newtons. That's it for this one.

Example #3: Vertical Centripetal Forces

**Transcript**

I know let's check this one out it says what is the minimum speed of a roller coaster cart can have at the top of a vertical loop so that the passengers don't fall so you have a roller coaster with a vertical loop so that was like that sort of and I want to know what is the minimum speed what is v min or simply what is my v At the top some here. Of a vertical loop radius ten the radius is ten in your riding at the edge of that radius so your distance to the center is always is also ten so the passengers don't fall even at the absence of restraint so talk about that in a second but first of all here's the direction of the force you will have M.G. is positive and normal is positive so that the passage wonÕt fall even at the absence of something like a seatbelt so this question is a little weird once you know what you're supposed to do is pretty simple but it's a little weird at first because. You know the way to solve this I think is a little bit backwards a little bit unintuitive but here's the idea if you were to fall if you were to fall. If you fall. This means that the normal force on you would be ZERO think about it this way normal force is the seat pushing back on you if you were falling is because you lost contact with seat So your normal force would be zero so for me to figure out the Mack the minimum speed for you not to fall I actually have to calculate the velocity I get when you normally zero so if you did fall at what velocity would that happen so that I can know then to be a little bit faster than that OK so it's back which you don't want to follow but I'm going to set it up as N equal zero so that you do fall to see what follows that starts to happen OK so it's kind of backwards you test for what you don't want to happen so you can figure out how to not. How to make it not happen sort of OK so your butt off the seat so normal is zero and now I can write a f=ma and. Again I'm writing f=ma for velocity or speed problem even though it might seem like a f=ma doesn't have a velocity in it or speeding but there's a v in here OK and if you're looking for a velocity going on a circle it's going to be f=ma problem.

So some of the forces centripetal is equal to mac at this point the forces are N Plus mg and that equals ma now what we want we want normal to be zero so actually what I get is mg Equal to ma in that cancels that means that the acceleration at that point is g

Now that should make perfect sense that I'm following freefall my accelerations is g in fact if you really wanted you could have started the question here OK you could have skipped the whole normal zero and not touch the seat and all that stuff and you could have said you know what if I'm falling my acceleration is zero OK and let's see what velocity gives you that kind of centripetal acceleration right so remember that you're always in a circle being accelerated towards the center the idea here is this this acceleration is nine point eight That means you're actually falling OK. So let's see. Acceleration equals G. Let's continue this over here from acceleration I can get V. by rewriting that acceleration is V. squared over r equals G. So v is the square root of G R. That's the solution to this OK now it's just a matter of plugging in numbers here we got a nine point eight or is Or ten.

So the minimum velocity is nine point nine meters per second OK if I go a little bit faster than that I will fall if I go a little bit slower than that I will fall. In That's the idea so this is your final answer OK nine point one I want to point out that this is the same almost exactly the same question if you had a bucket on a vertical loop so imagine you have a bucket of water bucket. And you spend this bucket in a vertical circle OK in the bucket has water inside of it so this is sort of the equivalent version version of this question and I ask how fast is a bucket have to be at the top so that the water doesn't fall right you might have done some like this to move something really quickly in the water stays inside and the answer would be exactly the same would have been well if it's for your acceleration the water's you really can treat water's sort of a one and a one point one point object or a particle and you would have gotten the same answer V. equals squared of G R where R. is the radius of rotation OK same exact setup identical question that you would have gotten now here have a practice problem have a pendulum that is made of a light two metre long rope I want to give you a little bit of this question as a hint and then I want you to try this but you release the pendulum at a certain height and it swings from side to side so let's say this is pursuing there. Here's a pendulum there's a mass at the end. And there's a length here and the pendulum is going to do this. It's going to swing from side to side all the way to its maximum sort of height on the other side and it's going to keep doing this and the length of this rope is two metres the mass of the small objects five kilograms and it says that it attains a maximum. Speed of ten meters per second the max equals ten I will tell you that the velocity at the end is zero. Because it stops. Right. And I want to know what the lowest point right here I want to draw a free body diagram for the most point. And I want you to find the acceleration at that lowest point and I want to find the tension in the rope remember tension we are going to use F t OK so using similar stuff that we've been doing for this section and let's see if you get.

Practice: A pendulum is made from a light, 2 m-long rope and a 5-kg small object. When you release the object from rest as of a certain height, it swings from side to side, attaining a maximum speed of 10 m/s. At the object’s lowest point:

(a) Draw a Free Body Diagram.

(b) Find the magnitude of its acceleration.

(c) Find the tension on the rope.

Concept #2: Flat Curve

**Transcript**

Hey guys in this video I want to go over 2 very specific centripetal force problems which are going to be the flat curve and the banked curve, the flat curve is when you're turning you're driving you're turning a curve that has no, no angles it's just flat so you gonna need friction to do that. And the other one you're going to be turning this way as if you're sort of on the highway ramp or something. Or NASCAR is a good example of this where there is an angle but there is no friction. So have either friction but no angle or angle but no friction one or the other. Right. Let's get started and see how this looks. OK. So the first one we're going to start with here is a flat curve it says Find the maximum speed. So I want Vmax. That's a 800 Kilogram, car 800 kilogram car can have all going around the flat curve of radius 50. So the radius is 50 meters of the curve. You are at the edge of the curve as you're turning. So your little r is 50 without slipping without slipping if you're not slipping. That means that you have static friction. Ok that means you have static friction. If the coefficient of friction between the car and the road is point five. So I know that meu static is point five. Ok, how do I do this. Well I'm looking for a velocity in a circular problem. I'm going to use F equals ma, another hint that you would use F equals ma is that friction is a force and there's a reference to friction here. OK I want to quickly sketch what the forces look like here it's not an official free body diagram just a sketch, mg is pulling this thing down, imagine this is the the x plane and normal's pulling this thing up. Now if you go around a circle if you're going on a circle here's a top view of the car. Every time you go in and circle your velocity is tangential but you have a centripetal acceleration. The only way to have a centripetal acceleration is to have a centripetal force because F equals ma, right you need an F to get a so you need a centripetal force to get the centripetal acceleration. So there has to be a force pulling you to the center and for the flat curve you need to know that that force is your static friction. One way to think about this is just try turning without friction. Right. On an icy road or if there's oil in the road or something or or if you're going too fast or whatever. So if there's not enough friction to make a turn you don't turn in you go straight through. So you need to know that friction is what keeps you turning. Now even though the car is moving the wheels are moving. This is static friction because it's not slipping. Remember static friction is when you're moving without when you're not moving or moving without rubbing, kinetic friction is when you're running into surfaces here the wheel isn't doing that. So it's going to be static friction. OK. So the only force in the centripetal direction is static friction. It's toward the center so it's positive. So here's an example of where friction is positive so positive friction static equals mAc. Now I'm looking for velocity and hopefully you know there's a V in here Ac is Vsquare over r there's a V hiding in there so I can just replace Ac with Vsquare over r. OK. We're we're basically 80 percent done in terms of things you need to know. And I just want to tell that the rest of this is really just making this look pretty. Right. So I have to get the V to look in to be written in as simple terms as possible meaning I'm going to rewrite everything I can into simpler terms. So for example friction can be rewritten as meu static normal. In normal can be rewritten as well. So you're almost always in these kinds of derivation problems. You want to rewrite friction you want to rewrite normal until there in the simplest form. I hope you see here that normal has to equal exactly mg because there are the only two forces in the y axis. So I can rewrite this again as mg and this is mg square over r so notice that the mass is canceled. OK now I'm ready to go. There's nothing else they can replace meu with or g with or r with.

So if you move stuff around it's going to look like this. V square equals (g r meu S). Now I'm writing this in a very specific way. You might be wondering why did you write in that sequence you move the ledge around. Or why did you leave a square there. There's a reason for this which I'll show you later. So I want you to remember the equation this way and by the way you should remember this equation for a test because if it shows up as a multiple choice question if your professor does that and you remember this equation. And you know it's a flat curve you can kill this right away. OK. So that's how this works. Let's now actually calculate the number this is the least important part. But let's do that real quick. That's the square root of g which is 9.8 r which is 50 and meu which is point 5. And if you plug all of this in you get a fifteen point four. Ok at the most basic level this is how you can figure out how quickly can go around a turn. So if you work for the city and you're trying to figure out what kind of speed limit to put on this turn you'd probably put like 12 or 10 or something. Right.

Because you know people are always going to travel too fast or try to be a little travel a little faster so that's how they post Speed limits, they just you know decrease that by some percentage or whatever. And that's it. OK. So if you're going 40 on a 30 you know you're not that special. They know you're going to do that. They you know they lowered it because they knew people are always going to try to stuff later anyway. So that's easy that's the flat curve. Now the bent curve is far more complicated let's jump into that now.

Concept #3: Banked Curve

**Transcript**

All right so for the bank curve those basic set up starts with an inclined plane. OK And here's the car in this thing is only angle some turning like this and the idea is that. The incline plane sort of follows me as I'm turning right here so I'm turning here but you only have to take a snapshot of this thing in one place so you know draw like this or you can draw it like this either one I like it this way because I usually want my planes to be set up in such a way that you fall this way which is kind of to the right that makes more sense to me. Let me move this down just a bit to give us a little bit more space there's a lot of stuff that could be going on here let's right some of this information down first mass is eight hundred you're going to want a bank that means there's an angle frictionless that means there is no _. The rays of the curve is fifty and you are at the edge of that so that's your distance and the angle with the horizontal is where you want the angle always is thirty seven degrees OK And I want to know what is the. Maximum. I want to know what is the max OK So again we're going to do this using F=ma because I'm going in a circle right here is the idea you're turning here in a. Circle because the plane's going to follow you its not like the car is sort of flying near You're going to go around this ramp or the spring to curve like that and there's a radius here OK of fifty meters OK so if you were to write the sum of all forces equals to ma This would be in the centripetal direction that means they are acceleration is this way. Now here's a problem with this you might remember we talked about this earlier on in incline plane you usually tilt your X. axis right your X. Y. axis so we tilt the X. and Y. axis so it matches the plane. Right my acceleration is usually down the plane like this along my thumb so I tilted so it matches this way Now the problem is in this question My celebrations actually back in the original X. axis so this is the only case that in an inclined plane you're actually going to untilt the axis so I always talk about how an inclined plane is an exception. Type problem because you're going to tilt the axis Well this is the exception to the exception where usually you tilted but now you're not going to OK So you're back at having a traditional. X. Y. axis. Where this is going to be your positive X. and it's also your central axis and part of y is this way and this is weird for people at first because you got used to. Doing the inclined plane tilt OK So that means that now look at m G M G used to have to be decomposing to X. and Y. because it no longer found to the X. and Y. axis here it doesn't have to be decompose anymore instead instead I'm going to have to decompose normal OK because normal no longer sits in the X. and Y. So let me do that group quick I'm going to decompose normal and it's going to look like this this is my normal on the Y. axis and this is my normal along the X. axis OK you have to do this. All right back here there's no other forces there's no friction back here what are the forces in the centripetal direction well centripetal here is this way. And the only force there is Nx and Nx. is the only force equals to m v square over R. and I need to get that V. OK so now the name of the game going to be simplify everything so that v is by itself OK which means I have to. Replace an X. with something else can I get simpler terms for an X. Yes I can so first we have to decompose an X. now it could be sin or cosine and this is tricky this data I would remember that this data actually shows up here and all this data down here is in the correct place this data right here is in the wrong place OK but I can't flip it because to get this data would mean I would have to start off with this angle here and that's worse so basically you're going to have to use the wrong angle or the name going the wrong place to decompose and X. any force F. X. is usually Fcosine_. But in this case and X. is going to be Nsine_ because I have an angle against the Y. axis OK I have angle against the Y. axis so an X. will be an sign_ of an N. and NY. By the way will be Mcos_ and notice how they flip now the way that you might remember this is that of an inclined plane M.g.X. splits and when mgx splits and G.X. is an inclined plane and mG.X. is mgxsine_ you know. So X. goes with sine here I don't have to split mgx so this is on a traditional plane but here and next will be backwards so instead of mgx being backwards and N what's going to be backwards in this problem because it's the one that's on X. So lots of sort of exceptions in these kinds of questions right so anyway an X. will be Nsine_ Equals m v square r again you still can't really go ahead with this because you have to try to simplify normal into something else now if you look at the diagram here normal Y. equals mg so I'm going to have to write that to be able to simplify normal another way of thinking about this is got to solve this problem

You gonna right some of the force of the central axis and you can write some of all forces in the Y. axis so X. which is centripital in Y. that's you have two dimensions so you're going to write it on both of these and you write this in the Y. axis some of all forces in the Y. axis you get a zero because those forces cancel since I'm only accelerating this way OK and if you get a zero that means that the forces in the Y. axis simply can't so so I can write it in Y. equals M G. And if I if I expand and y I replace and Y. with what it stands for I get mcos_ equals mg so i can solve for N here and plug it into the first equation N equals mg over the cos_ and that's it right so I got stuck here because I didn't have a simpler expression for N. in this equation so what I have to do I have to go to a different equation get a simpler expression for N and plug it back so here I'm going to replace N with M.G. cosine_ and Times sine_ equals mv squared over r now something really nice happens here which is first of all your mass and so more importantly notice how your sine ends up on top of a cosine and sine over cosine is tangent so this becomes G. tangent_ equals v squared over or or V. squared equals G R tension of theta this is the form of the equation I want you to remember OK Now when I say Remember it doesn't mean that only long answer fruit response question where you have to show a word and you can just start with this unfortunately for a question like this a flat curve or a banker shows up on a test you are going to have to show this entire process of derive in this equation the kind of. The good news is that you only have to do this once in my not even show in the test and this relates this doesn't relate to anything else later so it's sort of like this one prop problem that exists sort of in a vacuum. So even if you hate this it's not going to keep coming back thankfully right one thing I will point out is that these equations look very similar This one is G R V squared equals G R new and this one's v squared equals your tangent of theta and one way to remember this is that if you're in a flat curve you need mew so it's G.R. If you on bent curve you're turning because you are. And the gravity is pulling you down the plane which helps you turn so flat curve needs friction so it has _ a bent curve needs an angle that's why it's banked so he has a theta so it's v square G.R._ in him you enjoy tension of feta because you need a theatre that's just not played by itself it's tension of the you know. G R M E U G R T And I have the silly thing to help you remember that and I usually tell my students that you can imagine like some teenager whatever trying to go as fast as possible not turn and when he's turning he's turning as fast as possible he's really angry try to go fast so he's going to go right it's really stupid to maybe help you remember and there's a girl right there right so what is the velocity when you're turning its car and Gar changing the feta OK Ghar mew because I need _ to turn tangent of theta because I need an angle right and then obviously the V. is square so these equations are very similar if you remember that those equations shows up in a multiple choice question any multiple choice question within like thirty seconds or a minute by just knowing the equations if you have to solve then you have to show this entire process OK let me just wrap this up if I were to plug this in here.

And you want to plug in all the numbers is the square root of a bunch of crap and if you want to do. Appropriately you would get that the answer is twenty nineteen point two meters per second and just to wrap up I want to build a quick little table here summarizing the flat and bent curve so these questions have to. These problems have two properties _ and theta and here's the equation that you're going to have for them which is the quick little summary on a flat curve you have mew. But you have no theta it's flat so there's no angle but you need friction on a bent curve you going to have no to me you because it's always going to be frictionless. But you're going to have a theta the equation hears v square equals Gr mew which is what you have. And here the equation is going to be v squared or tangent of theta. Because theta. Is what you have here and okay. Could you have a bent curve with friction sure but in those cases the questions are for more complicated to solve and we're not actually going to see that there's an actual piece here there most regarded says What happens if the car moves slower or faster right well if you're not going fast enough if you go down the circuit you're not going fast enough the car with a slight downward little bit right so and if you go really fast the car will start coming outside of the circle you can think of this is if you go really fast and turn what's going to happen the reason why slow down a tree so you can actually make it without just sort of flying often change right and if you're stupid slow you fall into the curve now that doesn't happen when you're driving a regular road you can go slow if you want that's because there's friction holding back but if you're playing like this and there's no friction which is the case in this particular problem you would slide down. OK So that's it for this one I have two practice problems here that are basically just going to use the equations OK I want to simulate the idea of a multiple choice question or Also the idea that once you arrive with these equations to these equations you have to know how to sort of use them for different variables so that's what these two practice problems are here for Notice that in all of these equations I have one two three variables G.'s a constant so guess what there's three different questions I can ask for an answer for V. for R. or for theta were for me or whatever that's what these questions are practicing you can start straight from those equations and try to plug in the numbers and get the answer let's let's try that.

Practice: You are designing a highway curve to allow cars to turn, without any banking, at a maximum speed of 50 m/s. The average coefficient of friction between cars and asphalt, for dry roads, is roughly 0.7. What radius would this curve have to have, for this to be possible? For the radius you just found, how much would you have to bank the same curve, in order to attain the same maximum speed, but at the absence of friction?

0 of 9 completed

Concept #1: Intro to Centripetal Forces

Practice #1: Intro to Centripetal Forces

Example #1: Intro to Centripetal Forces

Example #2: Vertical Centripetal Forces

Example #3: Vertical Centripetal Forces

Practice #2: Vertical Centripetal Forces

Concept #2: Flat Curve

Concept #3: Banked Curve

Practice #3: Flat & Banked Curves

A car travels around a curve banked at a 15° incline, with a radius of curvature of 10 m. If a 1000 kg car had a coefficient of static friction between the road and its tires of 0.5, what is the maximum speed that the car can round the curve at?

A ball of mass m = 1.30 kg is attached to a massless rope and swung in a horizontal circle of radius r = 1.70 m, as shown in the figure. The rope makes a constant angle θ with the vertical as the ball swings around the circle, and the ball makes one revolution around the circle in a time Δt = 3.00 s.
Calculate the centripetal acceleration of the ball.

A ball of mass m = 1.30 kg is attached to a massless rope and swung in a horizontal circle of radius r = 1.70 m, as shown in the figure. The rope makes a constant angle θ with the vertical as the ball swings around the circle, and the ball makes one revolution around the circle in a time Δt = 3.00 s.
Make a diagram showing all the forces acting on the ball and the net force acting on the ball. With the help of your diagram, write down equations for the horizontal and vertical forces acting on the ball and use these to determine the angle θ.

A ball of mass m = 1.30 kg is attached to a massless rope and swung in a horizontal circle of radius r = 1.70 m, as shown in the figure. The rope makes a constant angle θ with the vertical as the ball swings around the circle, and the ball makes one revolution around the circle in a time Δt = 3.00 s.
Find the tension in the rope.

A ball on a string is swung in a horizontal circle on a frictionless table. The tension in the string is To when the ball completes a rotation in 1 s. You then increase the speed so that the ball completes a rotation in 0.5 s. What is the new tension in the string?
A) 1/4 To
B) 1/2 To
C) To
D) 2 To
E) 4 To

You tie a light rope to a pail of water and you swing the pail in a vertical circle of radius 0.600 m. What minimum speed must the pail have at the highest point of its circular path if no water is to spill from it?
(a) 1.98 m/s
(b) 2.21 m/s
(c) 2.42 m/s
(d) 3.92 m/s
(e) 4.00 m/s
(f) 5.88 m/s
(g) none of the above answers

Suppose a highway curve is properly banked to eliminate friction for a speed of 45 mph. If your tires were bald and you wanted to avoid sliding on the road, you would have to drive
A) somewhat above 45 mph.
B) somewhat below 45 mph.
C) at exactly 45 mph.

A passenger on a ferris wheel moves in a vertical circle of radius R with constant speed v. What are the magnitudes of the vertical forces on the passenger at the bottom of the circle if the passenger's mass is 60 kg, the radius of the circle is 16 m, and the wheel makes one revolution in 10 s?

A car travels around a flat curve. If the radius of curvature is 150 m, the car has a mass of 1500 kg, and the car wants to be able to round the curve at 25 m/s, what is the minimum coefficient of static friction between the tires and the ground required?

A 600-kg car is going around a banked curve with a radius of 110 m at a speed of 24.5 m/s. What is the appropriate banking angle so that the car stays on its path without the assistance of friction?[A] 29.1°[B] 13.5°[C] 33.8°[D] 56.2°[E] 60.9°

A passenger on a Ferris wheel moves in a vertical circle of radius R with constant speed v. Assuming that the seat remains upright during the motion, derive an expression for the magnitude of the upward force the seat exerts on the passenger at the bottom of the circle if the passenger's mass is m.

A ball of mass m = 1.30 kg is attached to a massless rope and swung in a horizontal circle of radius r = 1.70 m, as shown in the figure. The rope makes a constant angle θ with the vertical as the ball swings around the circle, and the ball makes one revolution around the circle in a time Δt = 3.00 s.Determine the ball's speed as it moves around the circle.

A small rock is tied to a light string set it motion in a horizontal circle with constant speed. The string is 8.00 m long and makes a constant angle of 36.9° with the vertical direction. The tension in the string is 6.20 N.What is the radius r of the circular path of the rock?

A small rock with mass 2.00 kg slides on the inside of a circular track that has radius 0.600 m. When the block is at highest point of its path (point A), its speed is 3.00 m/s. What is the downward normal force that the track exerts on the block when it is at point A during its motion?A) 10.4 NB) 15.0 NC) 19.6 ND) 30.0 NE) 49.6 NF) None of the above answers

A new park attraction consists of a vertical wheel of radius R, spinning at constant angular velocity, with cars for passengers. The passengers feel weightless at the top of the wheel. Assume that the typical mass of a passenger is m.
a) What is the magnitude of the force exerted on a passenger by the car at the top of the wheel?
b) What is the speed of the car?
c) What is the magnitude of the force exerted on a passenger by the car at the bottom of the wheel?
d) What is the magnitude of the force exerted on a passenger by the car when they are 90° from the bottom, going up?
Write your results in terms of R, m, and g. Check the units/dimensions for each answer.

A small block is sliding on the inside of a circular track that has radius R = 2.0 m. Point A is at the top of the block's path. What is the minimum speed v that the block must have at point A so the block continues in its circular path an doesn't fall from the track at point A.A) 3.1 m/sB) 4.4 m/sC) 5.4 m/sD) 6.3 m/sF) None of the above answers

(a) Which direction best approximates the direction of when the object is at position 1?A. straight upB. downward to the leftC. downward to the rightD. straight down(b) Which direction best approximates the direction of when the object is at position 2?A. straight upB. upward to the rightC. straight downD. downward to the left(c) Which direction best approximates the direction of when the object is at position 3?A. upward to the rightB. to the rightC. straight downD. downward to the right

The figure (figure 1) shows two balls of equal mass moving in vertical circles Is the tension in string A greater than less than, or equal to the tension in string B if the balls travel over the top of the circle with equal angular velocity?a. The tension in string A is greater than the tension in string Bb. The tension in string A is less than the tension in string Bc. The tension in string A is equal to the tension in string B

How will the desired centripetal force be set?1. By measuring the velocity, radius, and mass of the test mass.2. By choosing a value for the hanging mass.3. By measuring the acceleration and mass of the test mass.4. The centripetal force will not be set in this experiment.

A binary star system consists of two stars of masses m1 and m2. The stars, which gravitationally attract each other, revolve around the center of mass of the system. The star with mass m1 has a centripetal acceleration of magnitude a1.Note that you do not need to understand universal gravitation to solve this problem.Find a2, the magnitude of the centripetal acceleration of the star with mass m2.Express the acceleration in terms of quantities given in the problem introduction.

Six roller-coaster carts pass over the same semicircular"bump." (Intro 1 figure) The mass M of each cart (including passenger) and the normal force n of the track on the cart at the top of each bump are given in the figures. Rank the speeds of the different carts as each passes over the top of the bump.Rank from largest to smallest. Torank items as equivalent, overlap them.a) 200N, 400kgb) 400N, 100kgc) 300N, 300kgd) 800N, 100kge) 800N, 800kgf) 400N, 200kg

1. A particle traveling around a circle at constant speed will experience an acceleration. 2. The test-mass is referred to as m and it hangs from the test-mass riser. 3. A particle travels 17 times around a 15-cm radius circle in 30 seconds. What is the average speed (in m/s) of the particle?4. What measurements will be made to determine the magnitude of the test-mass centripetal acceleration?(1) The mass of the test-mass.(2) The velocity of the test-mass.(3) The radius of the circular path.(4) The mass of the hanging mass.(5) The spring constant.(6) The period of the orbital motion.5. Before rotating the platform, the hanging mass is disconnected from the test mass and removed from the platform.6. A particle in uniform circular motion requires a net force acting in what direction? 7. The centripetal force acting on a particle is given by F = mv2/r. If the centripetal force and mass are kept constant, increasing the radius of the particle's circular path will mean that the particle's velocity must increase.

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **55 hours** of Physics videos that follow the topics **your textbook** covers.