Practice: A 60 kg surfer is moving with 3 m/s at a certain point in a wave. Later on, he is moving with 8 m/s at a second point, 2 meters lower. Calculate the work done by the wave on the surfer.

Subjects

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Intro to Conservation of Energy | 52 mins | 0 completed | Learn |

Energy with Non-Conservative Forces | 45 mins | 0 completed | Learn |

Escape Velocity | 12 mins | 0 completed | Learn |

Conservative Forces & Inclined Planes | 50 mins | 0 completed | Learn |

Motion Along Curved Paths | 107 mins | 0 completed | Learn |

Energy in Connected Objects (Systems) | 31 mins | 0 completed | Learn |

Solving Projectile Motion Using Energy | 33 mins | 0 completed | Learn |

Springs & Elastic Potential Energy | 63 mins | 0 completed | Learn |

Force & Potential Energy | 22 mins | 0 completed | Learn |

Example #1: Energy in Curved Paths

**Transcript**

Hey guys, so in this video I want to show you how to solve problems where an object is moving along a curved path so instead of having an object moving in the horizontal and the vertical or even at the inclined plane it going to be moved in some sort of curved path and we're going to use the energy equation to do that let's check it out.

So it says here all curve path problems must be solved using energy in other words using the big energy equation something that looks like this, notice how you're going in a curved path it has to be done using energy I put here in parenthesis that there are no alternatives to really reinforce that as soon as you see a curved path you will jump immediately to the energy question, I also been an asterisk here just to make a comment that this is not really the only way you could use more advanced physics with calculus to do this but even most physics with calculus classes won't do it that way instead they will do it the energy way which is much much easier, OK? So, if you see a curved path it's automatically an energy problem. First example here what minimum speed would a block of unknown mass need to have at the bottom of a 30-meter-high smooth Hill in order to reach the top? So, what initial velocity do you need here so that you reach the top, OK? And this is a thirty meter Hill, now I'm going to use the energy equation and write it here and then we'll talk about some of the details of this question. curved path so I jumped straight into the energy equation, is a kinetic energy in the beginning? There is kinetic energy at the beginning because you have a speed if you have no speed it wouldn't be moving and you will never get to the top, there is no potential energy because you are on the ground, the work done by non-conservative force is the work done by you plus the work done by friction you're not doing anything you're just watching this box move and the hill is smooth so there is no friction so there is no work done by non-conservative forces, kinetic energy is kinetic energy at the end? And this is the trickiest of them all here because the answer is no there is no kinetic energy the reason for that is I'm asking what is the minimum speed then you need at the bottom so that you reach the top what's implied here is that you barely reach the top and your final velocity is 0, for example let's say that the velocity at the bottom here if you go...I'm just going to make up some numbers if you're moving with a 20 you get to the top with a five, well that means you can go a little bit slower let's say that going down here with an 18 gets you to the top with a 2 that means that the minimum speed is actually less than 18 you could be going even slower and still get to the top so the idea is what is the velocity here so that I just barely get here meaning the extreme case here or the boundary condition here that the final velocity is 0, cool? So that's how that works and let's keep going here there is potential energy at the end, now once you've done this this is pretty simple because I just have to expand these 1/2MV(initial) squared and this MGH final, the masses cancel and you see that the initial velocity is the square root of 2GH final, OK? And as we've discussed before this is a very common equation that we're going to see all the time, notice how the velocity depends on the height gained and not on the path taken, had you gone from here to here on a straight line it would have been the same thing you would have needed the same initial velocity to just barely get there, the fact that this is a curved path makes no difference, alright? I'm going to plug in some numbers here this is the easy part, gravity I'm going to round that to a 10 to make it easier and this is a 30 so this is going to give us the square root of 600 and the answer is 24.5 meters per second, cool? That's it for this one.

Example #2: Energy in Curved Paths

**Transcript**

Alright, so in this example we have a skater 70 kg skater dude that's riding on a half-pipe, a half pipe is the thing that skaters go back and forth on it's basically a perfect semi-circle it has a radius of 3 meters so if I kind of draw this it looks more like a circle the idea is that the distance from here to here the radius is 3 meters and we'll see why that's important in just a bit, his mass is a 70 that might cancel weÕll see and I want to calculate his speed in two different situations here, for Part A I want to know what is his speed at B? So, I want to know what is Vb? In other words, his speed down here if he starts at rest from A if Va equals zero this is motion along a curved path so we have to use the energy equation, in this case I'm giving you A and I'm asking you for B so we're going to write the energy equation from A to B like that, OK? So it's going to look this Ka, Ua work(non-conservative), Kb, Ub, the kinetic energy at point A is 0 because I have no speed it says right there, potential energy at A exists because I have a height, work non-conservative is the work done by you and we're assuming that this skater sort of is free falling and he's not actually pushing himself either way, so he's doing no work and the work done by friction we're also going to assume that it's 0, again we can assume that these two things are 0 and that the work done by you in this case is skater plus the work done by friction we can assume that they are 0 because there is no information about these so it's either 0 or you have to make up a number, right? So that wouldn't make any sense so it's zero because there's nothing referring to those as we assume, there is kinetic energy at point B because there's a speed I know that there has to be a speed because well first I'm being asked for one and second if you drop from here to here you're losing height obviously you have speed at the bottom, the potential energy at Point B is 0 because you are on the floor, let's expand this 1/2....whoops this is MGHa=1/2MVb squared, notice the masses cancel and V.B. is the square root of 2GH again, right? Very popular equation there and if you plug in these numbers if you're about to plug in the numbers you might be wondering what is the height at point A? And the height of point A is basically the whole point of this question is for me to talk about that aspect of this problem which is that the height of Point A is the same as the radius of this semi-circle so the height at point B is 0 but the height at point A is the radius, OK? So that's it because I can draw a line R for any points if I go from the very middle to all at the bottom you have R just as well, OK? So, this is square root of 2GR. Now let's plug in numbers 2, 10 and the radius here is 3 so this is the square root of 60 which is 7.7 meters per second, that's part A very straightforward let's to part B.

Part B says what is his speed at point A so that he can reach D, OK? So that he can reach D, in other words he's going from A to D and D is two meters above sea so this gap here is 2 meters. OK? 2 meters, Alright so we're going to write the energy equation I know that he gets to D that's the final point and he starts at A that's the initial point those are the two points of interest so I'm going to write the energy equation from A to D, Ka, Ua work non-conservative, Kd, UD. There is going to be kinetic energy at point A let's think about this according to conservation of energy if you start from here with no initial velocity and there's no friction to sort of dissipate your energy you should go all the way back to point C and kind of do this back and forth forever right if you start at the height of 10 here you going to go all the way down all the way up to 10 again and you keep doing this forever, in the real world that wouldn't happen because of friction so it will be sort of like 10, 9, 8 depending on how much friction you have so for you to get to D you have to actually start at A with some initial speed, OK? Otherwise you will only get up to C so hopefully makes sense so there has to be velocity at point A and therefore there is kinetic energy at point A, the potential energy at point A exists as well because we're above the ground, work non-conservative just like as before there is no work done by the skater because even though he might have been the one that pushed himself with this initial velocity at point A that happens before he has a two meters per second, or whatever velocity he has not a meters per second, two meters is the height so he gave himself that velocity if you push himself down or however he did that happen before he started moving at point A so that doesn't count it's much like how when you throw something the work that you do in throwing it doesn't count from once it leaves your hand and arm, OK? So, there is a kinetic energy there is a potential energy but there is no work by friction or by the rider and at point D you reach Point D which means you're going to do this all the way up to the top and stop so that means that the Vd equals 0 so there is no kinetic energy at point D but there is potential energy at point D, OK? The kinetic energy here is 1/2MVa squared, potential is MGHa, potential at point D MGHd notice that the masses cancel I'm looking for Va so I'm going to move everything to the other side I get the VA squared/2=GHd-GHa, OK? if you want to solve this for Va get all the letters by itself before plugging numbers it looks like this 2GHd-2GHa, OK? Again at this point right here you might of wanted to just plug in numbers already and get the answer but I want to show you something that happens here, notice I have a 2H and 2H so I can factor out those two again you wouldn't necessarily have to do this 2G and have Hd, Ha so it's Hd- Ha, so this is kind of interesting because the answer is very similar to before, I had the square root of 2GH now it's the square root of 2G and then the H here is the difference in height between D and A so it's sort of like the square to 2GH looks similar to that, alright? So, I just wanted to show you that but again once you got to this point here you're basically done it's just a matter of plugging in numbers let's put in the answers now, Va=2, gravity we're going to round that to 10, the height at point D, D is 2 meters above A so it 2+3, right? It's a 5-3 you get a 2 which is exactly this gap here should make sense the difference in height between the two and then this gives us a square of 40 which is 6.3 meters per second, OK? So here you've got to be with the 7.7 and then here you had to have left A with a 6.3, alright? That's it for this one

Example #3: Energy in Curved Paths

**Transcript**

Alright so here we have a cartthat's going to go around a loop the loop of radius R, sort of a rollercoaster type problem you're going to go around through the inside of the track something like this, OK? This has a radius of R so I'm going to kind of draw some like this to indicate a radius of R and I want to derive an expression for the minimum speed that the cartneeds to have at the bottom over here you know in order to reach the top of the loop so what is the minimum velocity at the bottom so that you make it to the top of the loop if the cartis locked to the tracks, alright? So we'll talk about what that means in just second, first thing's first this is a curved path problem so we're going to use energy, now there is a distinction that I have to make here when we have curved path problems or actually more specifically when we have circular path problems not all curved paths are circular paths but if it's perfectly circular there are two types of situations if I have a problem that refers to only one point or one place then this is going to be solved using F=MA and it's going to be solved using the centripetal force version of F=MA and if there are two points it would be solved using energy, F=MA relates to a single point and energy relates to two points it's always initial to final, OK? In this case it's a circular path with two points so we're going to use the energy equation, Alright? I'm going from an initial to final so let's right that K (initial) +U (initial)+work non-conservative=K(final)+ U(final), there has to be kinetic at the beginning because if you didnÕt have a speed at the bottom you wouldn't even be moving so you definitely wouldn't get to the top, there's no potential energy at the bottom because you're on the floor, work non-conservative is the work done by you or by someone pushing the cartor maybe the engine if this thing has one but there's no reference to that so we're going to assume that that doesn't exist and we're going to assume that the work done by you or external applied force is 0and we're also going to assume that the work done by friction is 0 because there's no reference to friction here so we can just assume that. Kinetic energy final this is the tricky one and there's also something that I have to talk about with Uf, so the minimum speed so that you get to the top is 0, OK? Now the reason why the minimum speed to get to the top is 0 in this particular case is because the cartis locked to the tracks so the cartis not able to fall from the tracks so the idea is that if you're locked to the track and you would go like this and you just barely get here then you would fall, right? If you do this then you're going a bit too fast, right? So the idea is that you just barely get to the top and then you go and you make the turn right so at 0 you kind of would be gambling because you could just slide back this way but again it's the boundary condition it's the extreme situation where you split fall into the left and making it and falling to the right and not making it and that's what we have to use, the reason why this part's important if the car's locked to the tracks is because if the cart was not locked to the tracks it would actually fall before getting to the top. Imagine if you are on one of these things and you don't have seatbelts, right? And when the car's about here right before it gets to the top so let's say the top is here and you go in this or I could show here, somewhere over here you if you're going slow enough you would actually fall before even getting up there so in these problems we're going to have a difference between being locked to the tracks and then making the turn and then making the turn without things falling, right? And that's a different situation that we'll talk about later so if you're locked to the tracks you don't have to worry about falling and the velocity at the top just has to be 0, alright?

Now the other thing we have to talk about is potential energy but I'm just going to write out the equations here, so you have kinetic initial and potential final I have 1/2MV(initial) squared and MGH (final) notice the masses cancel and I'm looking for V initial and V initial is the square root of 2GH (final) again the very familiar equation square root of 2GH. What is the final height? I didn't give you the height but kind of similar to this problem up here maybe you can figure this, the height from here to here is going to be in terms of R and it's going to be two Rs and that's because I have one R this way and then one R this way, alright? Notice how this is sort of halfway this problem here was halfway through the circle and in this second problem here I am all the way at the top of the circle so one is R the other one is 2R, OK? So, the height initial is 0 but the height final is 2R, OK? So this becomes the square root of 2G2R we don't have actual numbers to plug in so we're going to try to make this a little nicer it's 4GR which is sort of enough but if you want to can step further you can get this four out of there, the square root of 4 is 2 so it would be 2GR and that is your finally answer here, Curved path a perfect circle two points so we use energy and the height at the top is 2R and the final speed at the top is 0 because the car's locked so you have to just barely make the top so that you go around, OK? So, we just set that velocity to zero that's the first one.

Now the other thing we have to talk about is potential energy but I'm just going to write out the equations here, so you have kinetic initial and potential final I have 1/2MV(initial) squared and MGH (final) notice the masses cancel and I'm looking for V initial and V initial is the square root of 2GH (final) again the very familiar equation square root of 2GH. What is the final height? I didn't give you the height but kind of similar to this problem up here maybe you can figure this, the height from here to here is going to be in terms of R and it's going to be two Rs and that's because I have one R this way and then one R this way, alright? Notice how this is sort of halfway this problem here was halfway through the circle and in this second problem here I am all the way at the top of the circle so one is R the other one is 2R, OK? So, the height initial is 0 but the height final is 2R, OK? So this becomes the square root of 2G2R we don't have actual numbers to plug in so we're going to try to make this a little nicer it's 4GR which is sort of enough but if you want to can step further you can get this four out of there, the square root of 4 is 2 so it would be 2GR and that is your finally answer here, Curved path a perfect circle two points so we use energy and the height at the top is 2R and the final speed at the top is 0 because the car's locked so you have to just barely make the top so that you go around, OK? So, we just set that velocity to zero that's the first one.

Concept #1: More Rollercoaster Problems

**Transcript**

Hey guys, so in this video I want to go over a few more roller coaster problems and really touch up on the classic types of questions you might be asked that have to do with energy let's check it out.

So, this example here it's going to be a four-part example I'm going to in the same situation ask you for four different things, it says a roller coaster cart without seat belts, without seat belts means that the person or whatever is in it could fall, right? And that's important and we'll talk about that a little bit, goes around a loop the loop of radius R so this thing here has radius R, OK? Remember this means that this is R and this is R therefore the height here at the bottom I can call this the height 2 I call this 0 and this height here height at point 3 I can call this 2R because we're at the top of the circle, OK? We want to derive an expression for all of these four different things here for the first part, I want to know what is the minimum speed that you need at the top of the loop in other words what is the minimum speed you need here, V3min=? so that the cart makes it without passengers falling? The passengers could fall because they are without seat belts and you need to do this without falling, OK? If the passengers had seatbelts your velocity at the top could be as close as 0 they wouldn't fall as long as they had seatbelts and the cart was locked to the track and then it would just go around like this but because they have no seatbelts and we want to do this without them falling this velocity here cannot be 0, so V3 is not 0, 0 is too slow you would fall in fact even before you got to the top, OK? 0 is too slow, now I want to remind you that when we have a circular path there are two ways of solving these problems two situations, first one is a situation that talks about one point only and if that's the case we're going to deal with it using the centripetal F=MA, if it's situation that deals with two points then we're going to use the energy equation F=MA always deals with one point, a single point and energy equation always deals with initial to final therefore two points, in this question which one do you think I have? Do you think I'm referring to only one point or two points? I hope you picked I hope you said one point so we're going to use F=MA and that's because I'm saying how fast do I have to be at the top so that I don't fall or passengers don't fall at the top so its own making reference to one point, so we're actually going to solve this using F=MA, the sum of all forces=MA, alright? How many forces are here? well the cart is sort of inside here right the cart's backwards so obviously you have an MG pulling down but you also have a normal force pushing down, right? Because the track is pushing or the cart is pushing against the track at the top so the track pushes down against the car when you have a car going around a roller coaster loop like this the normal force is always pushing you towards the middle, OK? And here the middle happens to be going down so there are two forces, they're both towards the middle remember when we're dealing with centripetal forces, forces towards the middle are positive, OK? so I'm going to put here +M+N=MAc. Now one thing that might be weird about using F=MA to solve this problem is that we're looking for a velocity and yet you not find a velocity here, you don't see a V anywhere there but that's because you're supposed to rewrite A as V squared/R, now I have a V that I can solve for, OK? Cool, that's what we need to get we need to solve for V here, now one of the things that or one of the key things about this question is that normal is supposed to be 0, normal at the top is 0 that's what without passengers falling gives you, it tells you that the normal equals 0, this is a little bit counterintuitive a little bit backwards because if normal is 0 normal is 0 when you're no longer touching the surface that you're pushing up against so for example in the case of passengers it's if your butt is no longer touching the seat you come off the seat your normal is 0, well wouldn't that mean that you're falling? It is actually, it does mean that you're falling and the way we're going to answer this question is by finding out at what velocity you would start to fall and that is the minimal velocity, it's almost as if you want to find out what you don't want to do so that you don't do it so you begin falling at whatever velocity you get when you are normal at the top is 0 so it's backwards you don't want to fall but the boundary condition the extreme case is when normal is 0 that's where you would start falling if that number is 10 and that means you should probably move with 11 or 12 just to be safe but the answer we would give the minimum velocity in this case would be a 10 for that example I just gave, OK? So, the normal at the top has to be 0 because if you were to fall at the limiting condition where you would fall your butt would no longer touch the seat so there would be no normal force acting on you, OK? So, this becomes 0, Alright? Now we can solve this MG=MV squared/R, the masses cancel and V becomes the square root of GR and that is the answer for part A simply the square root of GR, OK? Another way a little bit faster maybe that you could have solve this so this is the full solution but something a little bit faster you could have done I'm going to put an or here as an alternative is you could have thought of it in the following way, if you're at the top and your butt doesn't touch you're falling and if you're falling then you are free falling which means that your acceleration is just gravity, now you are sort of going in a circle except that point you would fall so I can rewrite this acceleration Ac as a V squared/R=G which then gives me V= the square root of GR, OK? And that's kind of what I had here, G=V squared/R is the same thing as saying G=Ac which should make sense if you're falling your acceleration is just gravity so this is a little bit shorter, a little bit faster to get to, so whenever you have a problem that says what is the minimum speed at the top so you don't fall you could start with F=MA and set normal to 0 or you could say well if I'm falling if I don't want to fall I have to set it up as if I did fall and in that case if I did fall then I can say that Ac=G, OK? Again, this is kind of weird if the answer here came out to be a 10, this means that at 9.999 you would certainly fall at 10.00001 you would certainly not fall in and 10 is just the minimum number in between falling and not falling, OK? Hopefully the made sense, again normal at the top is zero or Ac=G is what you would set up in a problem like this, OK?

Now for Part B I'm asking what is the minimum speed needed at the bottom? So, what is V (bottom is point 2 here) V2 minimum that you need at the bottom of the loop that's 0.2 so that the cart reaches the top that's point 3 with the speed found in A? So, what is V2 so that you get to the top with a V3 found in A? well what we found in A was the square root of GR What must be V2 so that I get to point 3 with a velocity of GR? And I want to go back here and ask you guys to think about this real quick, do you think I'm going to use energy or am I going to use F=MA? I said that in backwards order am I going to see F=MA or energy? Well here I clearly have 2 points, I'm saying what must this be so that I get to this with this speed? So I'm talking about two different points because I'm talking about two different points I have to use the energy equation to solve this and I'm going to write an energy equation from 2 to 3, so K2.U2.work non-conservative= K3.U3 and I warn you it going to be a little messy just because we have a lot of variables but the most important part is obviously figuring out which types of energies we have, there is kinetic energy point two because I have to obviously have a speed down here so that I get to the top, there's no potential at point 2 because we're on the floor, the work done by non-conservative forces is the work done by you or some sort of engine or external force or applied force we're going to assume that there is none of that and that once the cart starts at the top it just rolls without any kind of engine pushing it forward, it just rolls because it had some initial initial height and then it keep going due to conservation of energy, there is no friction also, most roller-coaster problems won't have friction there's no reference to friction here so we can certainly ignore it, is there a kinetic energy at point 3? There it is because of velocity at point 3 remembers it's not 0 but instead it's this, so there's kinetic energy there and there is potential energy as well. Now let's just expand this 1/2MV2 squared = 1/2MV3 squared+MGH3, notice the masses cancel I have two terms here with halves I don't like fractions and I'm going to make things simpler here by multiplying everything by 2, multiplying this 2 with this half cancels out this so we end up with V2 squared, cancels this half as well so I end up with V3 squared plus there is no half here so there's a 2GH3, let's take this over here V2 is what we're looking for I'm going to circle it, V3 I'm given it's the square root of GR so and then I have to square that right so I'm squaring V3 and V3 is the square root of GR+ 2GH3 the height at Point 3 is 2R I mentioned here because it's at the top of a perfect circle so it's twice the radius or once the diameter but we always use radius in physics so you get that there, V2 squared this is going to simplify the square of the square root just gets cancelled out like this + 4GR so obvious if you combine this is 5GR and the finally answer is that V2 is the square root of 5GR. You might remember that you very often you get something like velocity is the square root of 2GH but instead of a 2 I have a 5 and instead of a G...Actually the G's there instead of an H I have an R but the R's used to explain or to describe the height here, we're saying that the height at the top is 2R so it's pretty similar in form and that gives us an indication that this is probably right and that is in fact your final answer for this part, OK?

Let's keep going here it says find the minimum height H, OK? So that's another classic question find the minimum height H that the hill shown point 1 must have so that you get to the top with the speed you found in A? So again we want to get to the top here, the minimum speed here that we want is the square root of GR That's what we found in part A, part B said well how fast do you have to be in part in point 2 so the you get to the top of with that speed, now the next question is how high must this point be...What is the minimum height that you need here so that you do get to the top with this speed, OK? Once again, we're talking about two points so I'm asking what is the height of A so that you get to V3 with the speed of the square root of GR? Two points so I'm going to use the energy equation to do this, OK? I'm going to leave a little bit space to draw in case we have to since we can't see the drawing from here so kinetic this is a point one actually so 1 is my unknown 3 is my known so I'm going to go write the energy equation from 1 to 3, K1.U2. work non-conservative=K3.U3, OK? So, what is the minimum speed? the first question I'm going to draw this here real quick OK the diagram did indicate that the speed here is a little bit higher I mean the height here is a little bit higher than number 3, OK? So, what's the speed at number 1? The speed at point 1 it's not given to you and it will be 0 and I don't know if you've noticed but there's sort of this trend if you don't know what the numbers and there's no way to get it very often it's just 0, right? So you can think of 0 and then try to justify it with why it's 0, well here's the idea let's say this height here is 10 meters and then you leave...Let's see you leave from here and the velocity you'll need here's 5 meters per second, if you start here with an initial velocity of 0 and you get here with a velocity of 6 meters per second that means that it was more than what you needed which means this could have been a little bit lower, right? the higher this is the higher this height the faster you're going to be here, so that's one of the things that we can set it up now the other set up that's more important is the following, if this has a height of 10 meters and I have to get here with 5 meters per second, right? If the only way to do this is by having an initial velocity here of 2 because otherwise you're not going to get here with a 5 then this means that this height has to be greater, OK? It means that 10 meters is not cutting it if on top of it I also have to give you an initial push so by asking the minimum height what's kind of implied here is what minimum height do I need in such a way that this cart requires no initial velocity that just simply barely rolling from it will be enough, OK? So that implies that the initial velocity has to be 0 and therefore there is no kinetic energy in the beginning, OK? Let me clean this up a little bit so what's tricky about these questions is there's always implications that you have to be able to get to or you have to know how to do them, OK? So potential energy obviously there's a potential here because I have a certain height, work by non-conservative forces will be 0 because there's no work done by you or an engine or external of force or applied force and there's no friction either in this path there's no reference to any of that stuff, there is a kinetic energy at point 3 because there's a speed and there's a potential energy at point 3 as well because there's a height so let's expand this equation here, I have MGH1=1/2 MV3+MGH3, the mass cancel and we're looking for H1 so let's start simplifying some stuff here GH1, 1/2 I know the velocity of 3, the velocity of 3 is this here so I'm going to replace that here, now be careful realize there's a square here so it's going to be the square of the square root like that + GH3, I also have an expression for H3, we discussed this earlier this is going to be twice the radius of this thing here so 2R, OK? Now I just have to clean it up a little bit and solve for H1, GH1=1/2 the square root cancels with the 2 1/2 GR + this is 2GR, OK? Notice that the G's cancel because I have G's are everywhere...Whoops I cancelled the 2 not G and then here I have 1/2R+2R and 1/2+2=2.5 so I can just put 2.5R and that's my H1 so let me just clean that up over here H1 is 2.5R, OK? I want to show this in terms of relative to all these heights, this is 0 this is 2R right there and this has to be 2.5R so just a tiny bit taller than the height in number 3 that should make sense because the velocity of 3 is not zero So this has to be a little bit higher so that when I get all the way around and I come back to point 3 since 3 is a little bit lower than 1 I haven't consumed all the sped I still have some speed here as I should so that the passengers don't fall, OK? So that's the third one and these 3 really show up a lot in more test where your professor puts a lot of emphasis on this roller coaster problems, I would say D's probably the least likely to show up, the least common at least but this is another one of the variations you may have so let's try this, I want to know what is the force that the track exerts on the carts at the top of loop at the loops so point three? So, if you're here what is the force that the track will exert on the carts? the force that the track exert on the carts is the normal force so I want to know what is the normal force at point 3 if the cart gets there with double the speed found in A? The speed we found in A was V square root of GR so I'm asking if I get to point 3 with a V3 of double that, right? So, I want V3 to be double what I found, this is GR not 2G so if the speed is double what will the speed we found a which is this if that's the case then what will be the force that the track exerts on the carts? the force of the track exerts on the carts is normal, OK? Now am I going to use F=MA or energy? Well F=MA is for one point, energy is for two points and even though this is making a reference to answer in part A that still has to do with the speed of point 3, I'm telling you if V3 is this what will be N3? Both of these numbers or both of these variables are referred to the same point this is a single point problem with circular motion so we're going to solve this using F=MA one point which means we're going to solve this using F=MA centripetal remember when you're in the top of a circle the forces are your MG and normal, so they're both positive because they're both going.... Well in the case of a cart both forces are down, right? Because the cart spins on the inside and they're both positive because they're both towards the center MAc. Now I don't have A I have V so I'm going to replace A with V squared/R and we do this a lot in these problems, OK? So I'm looking for N so I have to replace everything else, I'm going to move MG to the other side and it's going to look at this (MV squared/R-MG, now I can still clean this up a little bit more because N3 depends on V3 and I have an expression for V3 so let's do that, MV3 the whole thing is squared and then divided by R-MG, this here will be V3 which is 2 again it's the double of the original velocity we found so 2 square root of GR, OK? I'm going to take this over here and we're almost done, M3 equals M the 2 becomes of a 4 and the squared of GR1 squared becomes just GR that's divided by R which means this R cancels with this R minus MG, N3 equals If you multiply all, it equals 4MG-MG so that's just 3MG, so the track will push on you or on the cart with 3 times the force of the weight of the carts in the situation, OK? And that's because you're going faster than the minimum if you going at the minimum speed not to make it, you're barely touching the top because you're barely making it so your normal is 0 but if you go a little bit faster you're now pushing against the top, right? Because you're going faster and the top will and push back against you, OK? So, these are the 4 types of questions you might see hopefully this made sense, very comprehensive let me know if you have any questions.

Concept #2: Gravitational Energy is Relative (Pendulums)

**Transcript**

Hey guys, so in this video I want to talk about a little trick that you must know how to apply in order to solve certain types of energy question let's check it out, so this has to do with the fact that gravitational potential energy is relative, gravitational potential energy is relative so what do I mean by that? Well in energy problems it says here what matters is not an object's initial height or final height but instead the change in height so the actual heights don't matter what matters is the change in height and you might remember for example that the work done by gravity or by the weight force is negative delta U gravitational which gives a -MG Delta H and what matters here is delta H and not the actual Hs, OK? Now it says here if you know an object's change in heights then you don't really need to know the exact initial and final heights because you can set the ground level which is where H=0 to be at the lowest point in the problem, OK? And I'll show you an example of this, here in a situation like this you don't know the distance between the lowest point B over here and the ground, OK? But if you don't know this what you can do is you can say well I'm going to call this to be H=0 and since the gap between these two is 10 then those will be H=10, effectively what you're doing we can think of this as you're moving the ground up, right? But effectively what you're doing is changing the relative reference level of gravity, so it says here gravitational potential energy is relative meaning it depends on an arbitrary meaning up to you get pick where reference height and the idea is that 0 is your reference height or ground level, right? This is your reference height right here all the other heights depend on this but if I wanted to I could have made this 5 and then would have been 15 it's just that this is more complicated for no reason, OK? So you can do this in two situations, one when you don't know this here when you don't know the height between the floor and the lowest point you have to do this but you can also do this in a situation where you knew it but you just wanted to simplify things, I'm going to draw this real quick just to show you but you don't have to draw this because you need the space, let's say you knew that this was a 10 and then you knew that this was a 50 since the lowest point is right here you could actually instead of making this height here a 10, right? You have 0, 10 and 50 you could make this the 0 because it's the lowest point in your diagram and this part doesn't really matter but if this is a 0 instead of a 10 I reduced 10 from here then I have to reduce 10 from here and this will become 40 because the gap between these two here is 40 meters Ok? In this situation, I didn't have to do this I did this so that the problem was a little bit simpler, now that my height here is 0 my potential energy is 0 and I have less types of energies, OK? Here I had to do this, alright? So, this isn't just a trick to make problems easier in this case here on the left you had to do this so this problem was solvable, alright? So, like I said I hope you didn't write this I was just showing you a different situation but this is the most important case when you absolutely have to do this, OK? Just to complete here where H equals 0 so the line where H=0 is up to you, you can move that and if you do you have to adjust other things and once we've done that the pretty simple.

It's says if the roller coaster cart moves at a point A with 20 or passes that point with 20 so Va=20 calculate Vb, Ok? So we're going to use conservation of energy equation from A to B, Ka.Ua.work non-conservative=Kb.Ub, there is kinetic energy and potential energy because I have velocity and a height, there is no work non-conservative because we're going to since there's no information regarding the an external force you're not pushing that there is no engine and there's no friction we're going to assume that this doesn't exist, there is kinetic energy at point B in fact I want to know the speed at point B if B is lower than A there has to be kinetic energy there, there has to be a speed but since I raised my ground level from here to here and made that be my H equals 0 there is no potential energy at point B, right? So, I end up with just these 3 types of energies, let's expand this 1/2MVa squared+MGHa=1/2MVb squared, the masses will cancel and I'm looking for Vb so I just have to multiply the whole thing by 2 to get rid of this 1/2 here, this will be Va squared plus 2GHa=Vb squared so Vb is the square root of Va squared which is 20+2 gravity 10 and the height of A is 10, alright? So, this is going to be the square root of 600 which is 24.5 meters per second, OK? That's it so that minor trick now allows us to solve a few other problems that we couldn't do before if we didn't know this. A quick note here if you do not set H to be equals to 0 at the lowest point, point below H equals zero will have negative potential energy, what's up with that? Well let's say I'm going to redraw the situation here, let's say I got A here B here let's say this gap here between these two is 10 meters, let's say for whatever reason you wanted to make this H=0, well this point here is 10 meters below that so this has a height of H equals -10, OK? we have a negative potential, the reason I'm telling you this is so you know that it's OK to have a negative potential energy, negative potential energy just means you're below the arbitrary reference line where H=0, in fact gravitational potential energy is the only type energy that could ever be negative in terms of mechanical energies, OK? So, it could be negative and that's fine you just have to once you move the H equals you have to adjust everything accordingly. Now that we've talked about this we're going to talk about some pendulum problems because in most pendulum problems what's going to happen is you won't know the distance between the floor and the pendulum's lowest point so let me show you, here's a pendulum this thing is just a mass at the end of a string and the string is fixed here so that this thing can oscillate from left to right, OK? The points of interest are the beginning let's say it starts here at A point at the maximum, we call this the maximum amplitude or the maximum height of the left B here at the bottom is the lowest point that's important too and obviously C all the way at the other end here is important if I release this from a certain height up here at A it's going to reach the same height over here at C, same thing with the angles whatever initial angle I start here with I'm going to have the angle all the way at the other end because of conservation of energy wouldn't lose any height or any angle unless there is some sort of resistance, OK? So, the idea with these pendulum problems is that you typically won't know this gap here and because you don't know this gap you have to instead of making this your H=o you're going to make the lowest point in the pendulum your H=0, OK? So it would be the pendulum's lowest point will become your reference 0, OK? So it says our pendulum is built from a 2 Kg which is this part here so I'm going to say mass equals 2 and a rope that has a length L, length L equals 3 meters all these units are SI units we're good to go, it is attached to the ceiling and it's pulled until it is 1 meter above its lowest point so it's pulled until it's one meter above its lowest point we're going to make the lowest point H=0 therefore this H here H initial will be 1meter, OK? If the bottom is 0 and the gap between the bottom and the top is 1 then the top is 1, alright? We want to cut and then it's released its pulled and then release its implied here that it's released from rest, OK? We want to know what is the pendulum's maximum speed and the rope's tension at the bottom? Notice one last thing here says it's a light rope this just means that the mass of the rope is zero and that's how all pendulum problems will work, OK? So pendulum's max speed I'm going to call that V Max now where do you think this maximum speed happens at A, B, C or maybe somewhere else? And I hope you're thinking B because that's the answer, right? So the velocity will be 0 at point A, let me just say that V initial= Va which is 0 and Vc will be 0 as well because it stops and goes all the way to the other side and stops, right? So the maximum speed will be right at the middle of that motion which is a point B so Vmax is the same thing as Vb, this is a curved path that's why we're talking about this now and it's a curved path problem where I'm going from A to B, right? I know the initial height of A and I want to find out the speed at B so I'm going to write an energy equation from A to B, Kinetic at A, potential at A, work non-conservative, Kinetic at B, potential at B. There's no kinetic energy at point A, there is potential energy at point A because it has a height of one meter, work non-conservative will be 0 because the work done by you is 0, you're just watching this thing and then there's no friction, right? There is kinetic energy at point B because it has its speed and that's the speed we want to find out actually but there is no potential at point B because since B is the lowest point we raise the floor so that H=0 matches with B so the height of B is actually 0, right? Hb=0. Alright so very straightforwardly we have MGH on the left and MGHa and 1/2MV squared B on the right, the masses cancel and the final velocity or the velocity at the bottom will be the square root of 2GH, right? Once again, the square root of 2GH shows up irrespective of the path you take as long as you drop a height of H with no initial speed you find a velocity will always be this, so let's plug in numbers now, 2, 10 the height's 1 so I get the squared of 20 which is 4, roughly 4.5 meters per second, OK? That's it for Part A.

There is a Part B here and let me try to make it fit here I want to know the rope's tension at the bottom, OK? So tension is a force you can't find forces using the energy equation because it does not have forces anywhere if you look through all the components of an energy question there's no force, we're going to find force like we've always found force using F=MA this is going a curved path so it's going to be sum of the forces the centripetal direction equals MAc, Alright? I'm going to over here draw, I want to know the tension at the bottom, at the bottom the block is like this and the rope is completely vertical since this thing is exactly at the bottom, the forces are MG going down and tension going up, the acceleration is always towards the center in a circular problem which means that the tension is greater than the MG, Alright? So this is going to be positive because it's towards the center this is going to be negative because it's away from the center so let's do this, I have positive T+negative MG=MAc and what we're looking for here is the tension, now to solve tension I need MG which I have the M and I have the G obviously, I have M but I need Ac I don't have AC but I can replace AC with V squared over R and that I have, so T will be M then instead of Ac I'm going to write V squared/R and then we're moving this over here plus MG, T= the mass is 2 the velocity or speed is 4.5 squared, radius what is the radius of the circle here? Where do we get that from? I hope you see that the radius comes from the length of the rope, the length of the rope gives you the radius of this partial circle it's not a complete circle but it doesn't matter it still has a radius and the radius is the distance from the middle and the edge which is L., So this is a 3 the mass is 2 and gravity we're going to go with 10. Now if you have this and you plug in all your numbers, I don't have that, this is going to be....2(23+20) and this is 40, 13.3, 33.3 newtons is the force, the mass is 2 so MG is twenty and this is 33 it's greater than 20, it's greater than MG as we expected so that is the final answer, I hope this made sense let me know if you guys have any questions.

Practice: A 60 kg surfer is moving with 3 m/s at a certain point in a wave. Later on, he is moving with 8 m/s at a second point, 2 meters lower. Calculate the work done by the wave on the surfer.

Example #4: More Pendulum Problems

**Transcript**

Hey guys, so in this video I want to show you how to solve pendulum problems when we don't know the initial height, let's check it out.

So it says here in some pendulum problems you won't be given heights and we'll need to use what I'm going to call the pendulum equation and I'll show you how to get that equation, OK? So you're going to use the energy equation with the pendulum equation, the reason why we're going to use the energy equation is because all pendulum problems are solved that way because you have motion from one point to another along a curved path, OK? So it says a pendulum is built from a 1 Kg Bob so this is the mass here M=1 that has a length of 2 meters that's the length of this rope here, length=2 the rope is light as always which means the rope is massless, OK? The pendulum is attached the ceiling and release from rest from 37 degrees with the vertical, that means that this angle over here is a 37 and that's how we're going to almost always describe the angle for pendulums and that's because the pendulum is going to move from here to here back and forth so the angle is always crossing the Y axis so instead of using the angle with the X axis as usual for pendulums you're going to use the vertical angle, right? So that's how it's always going to go. This pendulum is going to go from here to here at the same exact height if it starts at 37 degrees there at the other end here it's going to make an angle of 37 degrees as well, I'm going to call this beginning point here A the bottom here B and then the other side where it stops C, here we want to know is the pendulums maximum speed the maximum speed is at the bottom of the Vmax or Vbottom or Vb, I want to know the velocity at B and I know some information about the beginning at A, I know that the angle starts at 37 so I'm going to write an energy equation again because it's a curved path from A to B, Ka.Ua.work non-conservative=Kb.Ub, kinetic energy at point A doesn't exist because you're releasing this thing from rest, right? It's released from rest from 37 degrees there is some potential energy here because there is a height, remember in these problems because we don't know this gap here what we do is instead of calling the floor our 0 we're going to call this our 0 and you have to do that here, however I don't know this height here I know the angle and the length of the plane but I don't know this height here so we don't know Ha but it is above 0 so there is potential energy we'll get to that in a bit, work non-conservative is the work done by you you're just watching or the work done by any kind of external force plus the work done by friction there is no friction, now just to be clear there is tension pulling on this thing but the work done by tension is 0 because tension is a centripetal force here it's pulling towards the center and remember that the work done by centripetal forces is always 0, OK? And MG is not a conservative force...I'm sorry MG is not non-conservative so MG doesn't go there, right? So this is 0 because there's no work done by you or friction, kinetic energy there's is a kinetic energy at point B because we're going to get there with some speed but there's no potential as a point B because we set the height to be zero at the lowest point right so we end up with MGH=1/2MV squared, the masses cancel and Vb is going to be the square root of 2GHa, this again shows up a lot, right? The problem here is that I don't have the height of A and that's where the pendulum equation comes in, the pendulum equation I'll show you in a second will connect these 3 variables length of pendulum, initial angle of the pendulum and initial height of the pendulum as long as you have any two of these you can find a third one and I'll show you in a second in this case I have these two which allows us to find H initial which is what I need here. Alright so here's how the pendulum equation works and you may need to use this in a test so you may need to know how to do this, alright? So I'm going to draw a..... HereÕs our rope with the pendulum, this is the ceiling here's the pendulum at its initial position and here's the pendulum at the very bottom, OK? So first things first this has a length of L and this is a length of L as well because it's the same rope, right? Now this is kind of an exaggerating angle there, let me just kind of draw this a little bit more realistic and this thing is going to do this, now I need you to remember two things once you draw this, first you're going to draw a line through here one way to remember that is a lot of things in physics are going to be triangles, right? So we're going to turn a sort of semi this piece of a circle into a once I draw a line here I have a triangle right there so I need you to remember to draw this line here and then we're going to do after that is we're going to write 2 expressions for the same line, we're going to write 2....Not two equations we're going to write two expressions, two ways of describing this line here and put two green lines there and by the way this is our angle, the first one is the simpler one well if this is my H which it is and this is my L, this gap here is the whole thing minus this so this gap here's L-H and this is the first way that I can express this green line here, L-H. The second way I can express it is using a trig, right? So I can use SOH CAH TOA to express it now the simplest way here is to just realize that this side this is the opposite to that angle so this one is the adjacent in from SOH CAH TOA if you have the hypotenuse and the adjacent you're talking about cosine, right? So another way that I can express this green line here is with L cosine of theta, alright? Now these two are equivalent ways of expressing it which means I can set them equal to each other, L-H=L cosine of theta and this is the pendulum equation if your professor gives you this, awesome you can just use it but if he expects you to derive it or show how you got to it you might need to.....You would need to draw this and sort of show your work towards here, if you don't remember how to derive this equation you should still memorize it so you can use the least get most of the partial credit if you get partial credit in your tests and you know maybe you could just quickly jot down a thing like this and draw a bunch of these curly braces which at least make it seem like you were going in the right direction in terms of driving the sequence even if you don't remember all the details, alright? So again if you see some of the stuff in your homework or in class you might want to be ready for a question like this, cool? So the variables are L, H and theta and you can see if I have two of them I can find the other one, if I'm looking for H I'm going to move H to the right side and I'm in the move L cosine of theta to the other side, OK? This is how you can find H, now if you want to get fancy you could factor out L you don't have to do that that's just a little bit more work but I'll do it so you know you can do it but you could have just plugged in numbers here, the length is 2, 1-the cosine of 37, your calculator obviously has to be in degrees and when you do this you get a height of 0.4, OK? 0.4 is what's going to go right here so Vb is the square root of 2, 10.4 which means that the velocity over there will be 2.83 meters per second that is our final speed at the bottom, cool? So that's how this works let me know if you guys have any questions there is a practice problem here for you that's very similar, you're going to need to use the pendulum equation, you should maybe practice trying to go from here, right? To the equation but if you can't quite get it at least you know that you can use that equation and you're going to have to use it in this one, it's a little more complicated but if you can figure out let me know if you have any questions.

Practice: A pendulum is built from a 3 kg bob and a 4 m-long light rope. It is attached to the ceiling and pulled from its equilibrium position until it makes an angle of 53° with the vertical. It is then given an initial speed of 2 m/s directed down.

(a) Calculate the maximum speed that the pendulum will attain.

(b) Calculate the maximum angle that the pendulum will make with the vertical on the other side.

Concept #3: Energy Problems with Bumps (Part A)

**Transcript**

Hey guys, so in this video I want to show you how to solve a very particular type of problem, problems where we're going to have objects moving over humps or bumps and basically, it's going to be a box or a car that is going to be moving and then it goes over a hump like this, right? Let's check it out so I have 3 questions here or 3 parts all revolving around the same situation here which is we have a car riding over two consecutive bumps or humps consecutive means one after the other they are nearly circular, nearly circular means that even though it's not perfectly circular we're going to treat it as if it was so you can basically ignore the word nearly there and they have radii of Ra and Rb so this is Ra right here this is from the center of the circle to any of the edges and then Rb right here, OK? And right away one thing that I want to point out is that this means that this distance here is Ra as well so if you think of this as being the floor then the height at point A is the same as Ra, Ok? Same thing with Rb over here if you think of this as the floor then the height at point B is Rb and that's going to be useful a little bit later, alright? So here I want to derive an expression, derive an expression means we're going to have.... weÕre not going to have numbers we're just going to have variables everywhere, for the maximum speed so that the car can have at the top of the second bump, right? So it is Vmax b in order to not lose contact with the road, I want to remind you that lose contact in a situation like this if you lose contact it means that the normal force became 0, OK? Now in these problems the way we're going to set them up it's a little bit counterintuitive we're actually going to set it up set it so N=0, now you might be thinking you want to move in such a way that you don't lose contact you want to move in such a way so that your N is not 0, why would you set it up so that N=0? Well and that's the reason we do this is to find basically our boundary conditions or basically the line at which it would either be 0 if you are a little faster or not 0 if you're a little bit slower, basically the tipping point around which you're about to lose your contact with the road, OK? Another way to think about this is that so that you can find out at what speed you would begin to lose contact with the road so that you know to go a little slower, OK? So again remembering in these problems they are sort of counterintuitive even though losing contact means normal=0, we want to set up in that way we want to set up normal to be 0 OK? So normal at point B will be 0, that's the first point I want to make the second point I want to make really important is when you have these problems where you have an object going around the circle, there's two ways you could solve them, OK? So when you have these circular motion problems there's two ways you can solve now when I say two ways only one of these two is correct but depending on a little detail you would pick one or the other, basically if the problem refers to only one point then you're going to use the sum of all forces=M and that's what's happening with this question here, in this question part A I'm saying what is the maximum speed at B right there so that you don't lose contact with the road at point B? Everything in this problem in part A is referring to the same point, everything is reference to point B so because we're only referring to the same one point we're going to use F=MA if I was referring to two points we would use energy to solve this, what would that look like? Well something like Part B, part B says find the max radius here so that you don't lose contact here, talking about two points so we're going to use energy, OK? So one point's F=MA, two points is energy one way to remember this is that F=MA has to do with the forces centripetal forces at one point equals MA the acceleration at the same one point, the energy equation however has K initial+blah blah blah all the way to U final, the energy equation always has to do with 2 points, F=MA always has to do with one point, OK? So because we're talking about the same point here we're going to go with this, sum of all forces centripital=MA centripetal and this is at point B right here, so what are the forces at point B? There's MG pulling you down and there's normal pushing you up, normal+MG we haven't done signs yet let's talk about that now, in centripetal problem or centripital force problems forces going towards the center are going to be called positive so MG is positive and away from the center will be negative so normal is negative, OK? Now once you start writing this one of the things you might be confused about is that well why would I use F=MA If I'm looking for velocity, there is no velocity in F=MA, well actually there is remember Ac is V squared/R, so instead of writing Ac I'm going to write V squared/R and the reason for that is I want to solve for V the only way I can solve for V is obviously if it shows up in the equation so we make it show up in the equation so that we can then solve for it, last thing I got to do here is remember we said that normal was going to be 0 so that we can find at what speed you would begin to lose contact with the road, OK? If I simplify this I get MG=M(Vsquared/R), notice that the masses cancel and we're left with V=the square root of GR, OK? So just as an example if this number turned out to be.... You don't have to write this that's not in the answer but it is number turn out to be a 10 this means that at 10.001 you're going to fly off but at 9.999 you're going to make the turn because you're not going too fast, right? If you drive too fast over a speed bump what happens? You just you get shot off right so you just fly away for a little bit don't you that, Cool? And then the other thing here is just the last touch up is that this is the radius, there are two radii Ra and Rb we're talking about point B so this is the radius at point B, this is the final answer and I want to quickly point out to you that this is very similar to our loop de loop or rollercoaster problems if you've done one of those before it looked kind of like this and again if you've done it at some point I asked what is V min top? What is the minimum velocity at the top so that you don't lose touch with the tracks? Or it could also be water inside of a bucket and you were spinning it vertically what is the minimum velocity? If you did this the setup was the same, you did normal=0 and you ended up with Vmin top being the square root of GR, the answer is exactly (and I'm going to put some parenthesis here) as this the only difference is that this is the minimum velocity so you don't fall here since we're going like this it's actually the maximum velocity so that you don't shoot out of the path like that and lose contact with the road, cool? Very similar set up let's move on to part B.

Concept #4: Energy Problems with Bumps (Part B)

**Transcript**

Alright, so for part B we want to know what is the maximum radius Ra? that the first bump can have in terms of Rb, so let me talk about that real quick Ra in terms of Rb means that the answer will look something like this, Ra equals something some sort of coefficient there the will multiply it with Rb and we got to figure out what that is so that the car doesn't lose contact with the road at B, not losing contact so that the car doesn't lose contact means losing contact would mean that Nb is 0 but what we also know is it would mean that the maximum velocity we can have is as per right here maximal velocity we can have is GRb, OK? So this is our Vb, I want to know what is the maximum radius of A so that I don't lose contact? Meaning the maximum radius of A that gives me a maximum velocity at B of the square root of GRb. Now I'm going to draw two diagrams here and I want you to not draw those because it's going to take a lot of space and I'm going to erase them later just to help you kind of understand what's going on here? Let's say A was massively bigger than B, OK? And then this guy just kind of rode here with the initial velocity of 0, I just want to understand this if this is so much bigger than this you're going to get here with a really high speed and you're basically just going to get projected out this way and that's bad so that's why I want to know what is the maximum radius? Remember the radius of this thing is also the initial height here so a radius that is too big would mean that you end up with a speed it's too fast here and you'll lose touch, so that's the first point I want to make, you make sure that it's not too big. The next point that I want to make is let's say it looks something like this I'm going to do it again, let's say it look something like this and let's say VMax here is 10 as long as you're here with a 10 you're good to go, let's say then you start here with a velocity of 5, right? And then when you roll this way you get here with a velocity of 12 so this 5 gives you a 12, well that's too much so that's not going to work and then let's say that instead same height you start with a velocity of 1, right? When you start with a velocity of 1 again just making up numbers let's say you were to get here with a velocity of 9, OK? So that's good 9 is less than 10 but what that tells me is that I could actually be moving a little bit higher I could have been going a little faster or that this radius could have been a little bit higher and I still would have made it so let's do that, now this radius is a little bit higher and I'm going to start with V=1 and let's say that I get here with a V=10 that's great that's just my maximum velocity, well but once again if I had no initial velocity at all I would have been able to start from an even higher position, right? And now this thing is going to roll with a velocity of approximately 0, OK? And that's the main point I wanted to make here that since I'm not given an initial velocity at point A but I want to maximize this height here we're going to assume that the velocity at A is going to be zero if I want the maximum height then I want the minimum velocity because the more velocity I have the lower this bumb can be but I want the biggest bumb possible so I'm going to go with the lowest velocity possible, OK? So that's how we're going to know our velocity we're going to start with an initial velocity of 0, if you didn't pick up on that you would have written your energy equation and you have gotten stuck not having a velocity at A, at that point hopefully you would either figure it out or you just make it zero because it's not given to you maybe you would be guessing but that would have been correct, alright? So we don't have an initial velocity, we're going to assume it's 0 because by having no initial velocity it means that the hill could have been a little bit taller, OK? And I mentioned the energy equation we're going to use the energy equation because I'm asking what is the maximum height or what's the maximum radius here so that I get here with a certain velocity? I'm talking about 2 points so we're going to write an energy equation between those two points A and B, alright? We're going to use the energy equation from A to B, so Ka.Ua. Work non-conservative=Kb.U b, there is no kinetic energy at point A because as I just mentioning the initial velocity will be 0, there is potential energy because there's a little height, there's no work non-conservative this car is just moving by itself and there's no friction no work done by the engine we're just going through that things just riding along, right? There is going to be kinetic energy a point B because I have a specific velocity I want to point B right here and there will be a height at that point as well because there's a little bump there, OK? And if you remember the heights at point A and B are the radii, Ra and Rb are our heights, cool. So this is going to be MGHa=1/2MVb squared +MGHb, OK? So the worst part is done we figured out that Va, Vb had to be this and we figured out that we had to use the energy equation between these 2 points the rest is just plugging numbers, OK? The masses will cancel, I am looking for radius Ra let's keep going here, the initial height is Ra and that's what I'm looking for, the velocity at point B is this here the square root of GRb, this whole thing is going to be squared and This this G instead of Hb, I'm going to write Rb, the reason I wrote Ha into Ra and Hb into Rb is because I'm looking for Ra and I want it to be in terms of Rb, so I have to make those variables show up, OK? Let's see this is GRa=1/2 this is the square of a square roots so it ends up looking like this, notice that the Gs can cancel, I'm looking for Ra and this is really simple, Ra is 1/2Rb+Rb. So it's 1/2+1 so itÕs 1.5Rb, Ra=1.5Rb and that makes some sense if you want to roll from here and get here with a certain velocity so this will last velocity is 0 but this velocity is not 0 then you would imagine that this has to be a little bit bigger, right? To be able to get there with a non-zero velocity, cool. So that's the answer to this part, Ra has to be 1.5 times greater than Rb, Let's go to the next one.

Concept #5: Energy Problems with Bumps (Part C)

**Transcript**

Alright so here we want to know what is the force that the track will exert on the cart (So the force that the track exerts on the cart is the normal force always) at the top of the second bump B? so I want to know what is normal at point B If Ra is 20 percent greater than Rb? 20 percent greater means that Ra will be 1.2 Rb, OK? So, Ra is 20 percent greater than Rb is this part here and the part that I'm asking for the force is this part here, OK? Normal is a force I want to know the force at point B so I can write the sum of all forces equals MAc this is at point B, OK? if I'm looking for force it has to be with F=MA not energy, cool? So, let's draw...I'm going to draw a little B here, point B you are right there and at that point you have MG pulling down on you and normal pulling up, MG will be positive normal and normal will be negative, positive towards the center negative away from the center so I have +MG-Normal=MAc. Now in a lot of these problems we're going to rewrite Ac as V squared/R because I'm not going to find Ac I'm going to have V squared instead but even if you left it as Ac for now it's OK we're looking for normal so let's move some stuff around, MG I'm going to move this to stuff here to the left so that normal is by itself on the right side, normal is a negative so it's going to show up here as a positive and this guy shows up as MV squared/R, now if you try to solve this you know the masses actually you don't know the mass so that's a problem, you know gravity we're talking about point B so this is Rb but you don't know the velocity so you don't know M, M or V, you know R and G so obviously you're stuck, right? And you have to try to figure out a way at this point this is just basic physics hustle, a way to find either M or V, in the beginning of the problem if you notice you were never given the mass if you're not given the mass it means that it's going to cancel somehow so M is going to cancel somehow and you have to have a little faith that it's going to come through and it's going to happen that it's going to cancel so instead of focusing on M we're going to focus on V, If I'm at point B this V Here is the velocity at point B, I don't have that but I can find that, right? I can find it, I told you that the radius at point A is 1.2 times the radius of point B, I'm given some information about point A so maybe we can find the velocity at point B and if I want to find the velocity of point B by using some information at point A, that's two points so that's energy so I have to stop here and go over here to try to find the velocity at point B by using energy from A to B, the energy from points A to B, so Ka.UA.Work non-conservative=Kb+Ub. Remember we're assuming that the initial velocity is 0 because this cart just sort of rolls from the top of point A, there is a potential energy there, work non-conservative is 0 no engines you're not pushing no friction, kinetic at point B is not going to be 0 there's some speed there and there's also potential in there so try it out, MGHa=1/MV squared B+ MGHb, the masses will cancel and remember out of all this we're looking for Vb, OK? We're looking for Vb, one of the things we can do right away is multiply both sides by 2 to get rid of that half but actually I'm not going to do that because you might not have thought of doing that so that's fine, gravity is 10 it's actually probably easier not to do that, gravity is 10 the heights at point A is.... I'm sorry we're not plugging numbers gravity is G, the height at point A is Ra and then 1/2Vb squared+GRb, If you look at this I got some Gs over here and I have an Ra and an Rb and I can solve for Vb but instead of having Ra and Rb I can simplify things here because there's a relationship between those two, instead of Ra I can write 1.2 Rb, so instead of Ra I'm going to write 1.2Rb and that simplifies things because now I have 2Rbs instead of one Ra and one Rb which means I can combine the Rbs to make this a little bit simpler, so G or 1.2GRb-GRb=1/2Vb squared, I'm running out of space here let me disappear, so I'm just going to sort of skip one step here or at least not write it out, this is 0.2GRb and then I have to multiply both sides by 2 so it's going to be 0.4GRb, hopefully you see that this is a 0.2 GRb right there, I multiply by 2.4 and then I got to take the square root of that so the VB is this, you don't have to simplify try to get that 0.4 out of there because when you plug it in here Vb squared anyway so this stuff will go back to being 0.4GRb, so this is to be an MG-M/Rb, you can still see that and then I have this guy here squared so 0.4GRb squared, this squared cancels with the squared and we're back to just having this here, little short on space here I'm sorry that this was so tight, N=MG-M/Rb(0.4GRb), notice what happens here, the Rbs will cancel and then I have N=MG-0.4MG, right? And this will become MG-0.4Mg is 0.6MG, OK? 0.6MG is the force we're going to get I was actually expecting that Ms would cancel I mentioned earlier but actually the Ms don't cancel and that's OK because this is just letting us know that the normal force will be 60 percent of the weight, right? If you're driving horizontally the normal force equals exactly the weight this is just telling us that there will be a normal force, it's a little bit less than our weight which is what happens when you go over a hump, right? you feel a little bit lighter temporarily as you go up like this but it's certainly not 0 still got a long way to go from 0.6MG to 0MG, OK? So that's the final answer there just to recap real quick we're looking for the normal force, right? Given this situation here and as we were solving for it we realized we didn't have the velocity but we had enough information to find the velocity using energy so this question is very tricky this last part because you had to use both F=MA centripetal to find a normal force because it's a force so you do F=MA and then you have to use energy to find velocity from this point to this point the velocity down here to be able to plug it back into the F=MA equation, right? So, tricky got stuck have to go figure out another question to get out of it, alright? Let me know if you guys have any questions.

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Example #1: Energy in Curved Paths

Example #2: Energy in Curved Paths

Example #3: Energy in Curved Paths

Concept #1: More Rollercoaster Problems

Concept #2: Gravitational Energy is Relative (Pendulums)

Practice #1: More Curved Path Problems

Example #4: More Pendulum Problems

Practice #2: More Pendulum Problems

Concept #3: Energy Problems with Bumps (Part A)

Concept #4: Energy Problems with Bumps (Part B)

Concept #5: Energy Problems with Bumps (Part C)

A pendulum is released from rest at an angle of 40 o. If the pendulum is made of a light, 1 m string supporting a 2.5 kg bowling ball, what is the speed of the bowling ball at its lowest point in the motion? What is the tension in the string at the lowest point?

A 5 kg object undergoes uniform circular motion. If the object has a tangential speed of 15 m/s in a 0.7 m orbit and undergoes a 1/4 revolution, how much work was done on the object by the centripetal force?

A 500 kg car carrying 200 kg of passengers on a rollercoaster encounters a stop on the track shown in the figure below. How high does the car have to start in order to traverse the loop without any passengers being in danger of falling out if their harness broke? Note that the car starts from rest at the top of the slope and that there is no friction between the car and the track.

A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius exttip{R}{R}.From what minimum height h must the block start to make it around the loop without falling off?
Give your answer as a multiple of R.

A car in an amusement park ride rolls without friction around a track
. The car starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)?If the car starts at height
h=
4.10
R
and the radius is
exttip{R_{
m 1}}{R_1} = 15.0 m , compute the speed of the passengers when the car is at point C, which is at the end of a horizontal diameter.Compute the radial acceleration of the passengers when the car is at point C, which is at the end of a horizontal diameter.Compute the tangential acceleration of the passengers when the car is at point C, which is at the end of a horizontal diameter.

A roller-coaster car shown in the figure is pulled up to point 1 where it is released from rest. Assuming no friction, calculate the speed at point 2.Assuming no friction, calculate the speed at point 3.Assuming no friction, calculate the speed at point 4.

Suppose the roller-coaster car in the figure passes point 1 with a speed of 3.10 m/s . If the average force of friction is equal to 0.23 of its weight, with what speed will it reach point 2? The distance traveled is 55.0 m .

The small mass m sliding without friction along the looped track shown in the figure is to remain on the track at all times, even at the very top of the loop of radius r.If the actual release height is 4
h,
calculate the normal force exerted by the track at the bottom of the loop.If the actual release height is 4
h,
calculate the normal force exerted by the track at the top of the loop.If the actual release height is 6
h,
calculate the normal force exerted by the track after the block exits the loop onto the flat section.Determine the minimum release height h.

A 350 kg roller coaster starts from rest at point exttip{A}{A} and slides down the frictionless loop-the-loop shown in the accompanying figure.How fast is this roller coaster moving at point exttip{B}{B}?How hard does it press against the track at point exttip{B}{B}?

You are testing a new amusement park roller coaster with an empty car with a mass of
120 kg. One part of the track is a vertical loop with a radius of
12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of
8.00 m/s.
You may want to review (Pages 203 - 212).For related problemsolving
tips and strategies, you may
want to view a Video Tutor Solution of A vertical circle with friction.
As the car rolls from point A to point B, how much work is done by friction?

A pendulum is formed from a small ball of mass exttip{m}{m} on a string of length exttip{L}{L}. As the figure shows, a peg is height
exttip{h}{h}
=L/3
above the pendulums lowest point. From what minimum angle exttip{ heta }{theta} must the pendulum be released in order for the ball to go over the top of the peg without the string going slack?

A
95.0-kg
mail bag hangs by a vertical rope
3.9 m
long. A postal worker then displaces the bag to a position
2.9 m
sideways from its original position, always keeping the rope taut.What horizontal force is necessary to hold the bag in the new position?As the bag is moved to this position, how much work is done by the rope?As the bag is moved to this position, how much work is done by the worker?

A pendulum is made by tying a 510 g ball to a
47.0 cm -long string. The pendulum is pulled 24.0 to one side, then released.
You may want to review (Pages 234 - 238).What is the balls speed at the lowest point of its trajectory?To what angle does the pendulum swing on the other side?

A 20 kg child is on a swing that hangs from 3.2-m-long chains.You may want to review (Pages 234 - 238).For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Car rolling down a hill.What is her maximum speed if she swings out to a 46 angle?

Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of 45 with the vertical, steps off his tree limb, and swings down and then up to Janes open arms. When he arrives, his vine makes an angle of 30 ^circ with the vertical.Calculate Tarzans speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.

A ball is attached to a horizontal cord of length l whose other end is fixed .If the ball is released, what will be its speed at the lowest point of its path?A peg is located a distance h directly below the point of attachment of the cord. If h = 0.80l, what will be the speed of the ball when it reaches the top of its circular path about the peg?

A 0.325 kg potato is tied to a string with length 2.40 m , and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released.What is the speed of the potato at the lowest point of its motion?What is the tension in the string at this point?

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 with the vertical. Air resistance is negligible.What is the speed of the rock when the string passes through the vertical position?What is the tension in the string when it makes an angle of 45 45 with the vertical?What is the tension in the string as it passes through the vertical?

A 25.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 43.0 from the vertical and releases her from rest.What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?How fast will she be moving at the bottom of the swing?How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

A sled starts from rest at the top of the frictionless, hemispherical, snow-covered hill shown in the figure. Find an expression for the sleds speed when it is at angle exttip{phi }{phi}.Use Newtons laws to find the maximum speed the sled can have at angle exttip{phi }{phi} without leaving the surface.At what angle exttip{phi _{
m max}}{phi_max} does the sled "fly off" the hill?

A 56.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 3.10 m .What speed does he need at the bottom?

A small rock of mass m is attached to a strong string and whirled in a vertical circle of radius R. When the rock is at the lowest point in its path, the tension in the string is five times the weight of the rock. At this point the speed of the rock isA) √2gRB) √3gRC) 2√gRD) 3√gRE) √5gRF) √6gRG) None of the above answers

A 240 kg roller coaster car starts from rest at point A and slides down the frictionless loop-the-loop shown in the accompanying figure. Point A is 25.0 m above the ground and point B is 12.0 m above the ground. The height of other points on the track are shown in the diagram.How fast is this roller coaster moving at point B?

A 240 kg roller coaster car starts from rest at point A and slides down the frictionless loop-the-loop shown in the accompanying figure. Point A is 25.0 m above the ground and point B is 12.0 m above the ground. The height of other points on the track are shown in the diagram.What is the magnitude of the normal force that the track exerts on it at point B?

A 5.00 m long light rope is tied to the ceiling. A steel ball with mass 2.00 kg is attached to the lower end of the rope. The ball is pulled to one side and released, and swings back and forth as a pendulum. As the ball passes through its lowest point, with the rope vertical, its speed is 6.00 m/s. As the ball swings through this point, what is the tension in the rope?(a) 34.0 N(b) 26.8 N(c) 19.6 N(d) 14.4 N(e) 12.4 N(f) 5.2 N(g) none of the above answers

One end of a 6.00 m long rope is tied to the ceiling. A small rock with mass 0.500 kg is tied to the other end of the rope. The rock is released from rest with the rope horizontal. What is the tension in the rope when the rock is swinging through its lowest point, where the rope is vertical?A) zeroB) 4.9 NC) 9.8 ND) 14.7 NE) 19.6 NF) None of the above answers

A roller-coaster car shown in Figure P8.72 is released from rest from a height h and then moves freely with negligible friction. The roller-coaster track includes a circular loop of radius R in a vertical plane.(a) First suppose the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless. Find the required height h of the release point above the bottom of the loop in terms of R.(b) Now assume the release point is at or above the minimum required height. Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the car’s weight. The normal force on each rider follows the same rule. Such a large normal force is dangerous and very uncomfortable for the riders. Roller coasters are therefore not built with circular loops in vertical planes. Figure P6.17 (page 170) shows an actual design.

A block of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure P8.6. Determine(a) the block’s speed at points B and C and(b) the net work done by the gravitational force on the block as it moves from point A to point C.

A pendulum, comprising a light string of length L and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. P8.68).(a) Show that if the sphere is released from a height below that of the peg, it will return to this height after the string strikes the peg.(b) Show that if the pendulum is released from rest at the horizontal position (θ = 90°) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5.

Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with person-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force , on a vine having length L and initially making an angle θ with the vertical (Fig. P8.81). Take D = 50.0 m, F = 110 N, L = 40.0 m, and θ = 50.0°.(a) With what minimum speed must Jane begin her swing to just make it to the other side?(b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume Tarzan has a mass of 80.0 kg.

A ball of mass m = 300 g is connected by a strong string of length L = 80.0 cm to a pivot and held in place with the string vertical. A wind exerts constant force F to the right on the ball as shown in Figure P8.82. The ball is released from rest. The wind makes it swing up to attain maximum height H above its starting point before it swings down again.(a) Find H as a function of F. Evaluate H for(b) F = 1.00 N and(c) F = 10.0 N. How does H behave(d) as F approaches zero and(e) as F approaches infinity?(f) Now consider the equilibrium height of the ball with the wind blowing. Determine it as a function of F. Evaluate the equilibrium height for(g) F = 10 N and(h) F going to infinity.

A 400-N child is in a swing that is attached to a pair of ropes 2.00 m long. Find the gravitational potential energy of the child–Earth system relative to the child’s lowest position when (a) the ropes are horizontal, (b) the ropes make a 30.0° angle with the vertical, and (c) the child is at the bottom of the circular arc.

A light, rigid rod is 77.0 cm long. Its top end is pivoted on a frictionless, horizontal axle. The rod hangs straight down at rest with a small, massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?

A ball whirls around in a vertical circle at the end of a string. The other end of the string is fixed at the center of the circle. Assuming the total energy of the ball– Earth system remains constant, show that the tension in the string at the bottom is greater than the tension at the top by six times the ball’s weight.

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