1 00:00:03,710 --> 00:00:05,410 In problem solving, when we're working 2 00:00:05,410 --> 00:00:08,230 with problems with potential and kinetic energy, 3 00:00:08,230 --> 00:00:10,600 remember we have our main concept 4 00:00:10,600 --> 00:00:13,000 that the non-conservative work causes 5 00:00:13,000 --> 00:00:15,880 potential energy and kinetic energy to change. 6 00:00:15,880 --> 00:00:22,510 K final minus K initial plus U final minus U initial. 7 00:00:22,510 --> 00:00:26,020 Now, how are we going to apply this in problem solving? 8 00:00:26,020 --> 00:00:27,400 Well, what we'd like to introduce 9 00:00:27,400 --> 00:00:35,858 is a tool, which we're going to refer to as our energy state 10 00:00:35,858 --> 00:00:36,357 diagrams. 11 00:00:39,050 --> 00:00:44,240 So the way this works is-- let's have an example to think about. 12 00:00:44,240 --> 00:00:48,820 Suppose you have a dome. 13 00:00:48,820 --> 00:00:52,090 So here's some dome. 14 00:00:52,090 --> 00:00:55,540 We can think of this as the MIT Dome. 15 00:00:55,540 --> 00:00:58,720 And you have an object that is initially 16 00:00:58,720 --> 00:01:00,760 at the top of the dome. 17 00:01:00,760 --> 00:01:03,280 And then this object is sliding down the dome 18 00:01:03,280 --> 00:01:09,010 and at a later time it's at a point somewhere along the dome. 19 00:01:09,010 --> 00:01:11,110 Now eventually it's going to fall off the dome, 20 00:01:11,110 --> 00:01:12,970 but that's a separate question. 21 00:01:12,970 --> 00:01:17,140 So how can we use our energy ideas 22 00:01:17,140 --> 00:01:18,830 to analyze this situation? 23 00:01:18,830 --> 00:01:23,320 And the way we do it is that we choose 24 00:01:23,320 --> 00:01:31,240 first initial and final states that we're considering. 25 00:01:31,240 --> 00:01:32,740 Because our first idea in the energy 26 00:01:32,740 --> 00:01:36,400 diagram is to compare the change in kinetic energy 27 00:01:36,400 --> 00:01:38,560 between the initial and change of potential energy 28 00:01:38,560 --> 00:01:40,570 between the initial and the final states. 29 00:01:40,570 --> 00:01:45,137 So in our dome problem, we would choose our initial state. 30 00:01:48,610 --> 00:01:54,310 Let's draw our object at the top. 31 00:01:54,310 --> 00:02:00,020 And in our final state, separately, 32 00:02:00,020 --> 00:02:02,260 let's draw the object over here. 33 00:02:02,260 --> 00:02:04,990 And then we want to parametrize these states 34 00:02:04,990 --> 00:02:07,820 by some type of coordinate system. 35 00:02:07,820 --> 00:02:11,470 So here we're going to have some type of coordinate system 36 00:02:11,470 --> 00:02:12,920 that we use for both. 37 00:02:12,920 --> 00:02:16,010 Energy is a scalar, so we don't have to worry about it. 38 00:02:16,010 --> 00:02:22,930 And what I'll do is I'll define an angle theta final. 39 00:02:22,930 --> 00:02:28,390 Here from the vertical, here theta initial is 0. 40 00:02:28,390 --> 00:02:33,970 And I've now parametrized my initial and final states. 41 00:02:33,970 --> 00:02:38,090 Now, the important thing that we'll do in the energy diagram 42 00:02:38,090 --> 00:02:44,980 is to choose our 0 potential energy. 43 00:02:44,980 --> 00:02:47,470 Where are we going to choose this? 44 00:02:47,470 --> 00:02:49,860 So we could say either a surface-- we'll 45 00:02:49,860 --> 00:02:53,020 call that potential energy. 46 00:02:53,020 --> 00:02:55,150 And in this diagram, I can choose my 0 47 00:02:55,150 --> 00:02:56,800 for potential energy anywhere I want. 48 00:02:56,800 --> 00:03:00,530 But I want to draw it on my initial and my final states. 49 00:03:00,530 --> 00:03:03,910 So I'm going to choose it right here, 50 00:03:03,910 --> 00:03:08,230 and I'll denote this as u equals 0. 51 00:03:08,230 --> 00:03:14,520 And now I can now make a list-- so I want to identify. 52 00:03:14,520 --> 00:03:22,087 So step three is to identify the K and U for each state. 53 00:03:25,220 --> 00:03:29,380 So here we have K initial, it's at rest 0. 54 00:03:29,380 --> 00:03:32,540 And U initial, well, I didn't introduce a parameter 55 00:03:32,540 --> 00:03:40,400 R. U initialize is how high the gravitational potential energy. 56 00:03:40,400 --> 00:03:44,050 So in this particular case, Mg times R. 57 00:03:44,050 --> 00:03:49,390 Now over here, K final, is 1/2 M V final squared. 58 00:03:49,390 --> 00:03:55,840 And U final, we can denote it here if I plot this out-- 59 00:03:55,840 --> 00:03:58,210 and that's R. And then you can see that this 60 00:03:58,210 --> 00:04:00,860 is R cosine theta final. 61 00:04:00,860 --> 00:04:05,830 So that's MgR cosine theta final. 62 00:04:05,830 --> 00:04:11,350 Now, the fourth step is to identify W non-conservative. 63 00:04:11,350 --> 00:04:13,510 Is there any non-conservative work? 64 00:04:13,510 --> 00:04:18,483 Now, here let's assume the dome is frictionless. 65 00:04:22,790 --> 00:04:26,450 And that means that W non-conservative is 0. 66 00:04:26,450 --> 00:04:30,230 And our last step 5 is we can apply the energy 67 00:04:30,230 --> 00:04:37,970 principle, which is W non-conservative 68 00:04:37,970 --> 00:04:43,909 equals delta K-- let's write out everything explicit 69 00:04:43,909 --> 00:04:48,860 now that we've defined K final-- minus K initial plus U 70 00:04:48,860 --> 00:04:50,900 final minus U initial. 71 00:04:50,900 --> 00:04:54,980 And you see the power of these diagrams and this methodology 72 00:04:54,980 --> 00:04:58,909 is we've now defined very explicitly every single term 73 00:04:58,909 --> 00:05:00,920 that appears in the energy principle. 74 00:05:00,920 --> 00:05:07,100 And so we can write out our result that 0 K final, 75 00:05:07,100 --> 00:05:11,420 1/2 M V final squared, K initial, 0 plus U 76 00:05:11,420 --> 00:05:22,760 final, MgR cosine theta final, minus U initial, MgR, and there 77 00:05:22,760 --> 00:05:25,250 we have applied the energy principle 78 00:05:25,250 --> 00:05:29,050 using the tool of energy diagrams.