1 00:00:03,980 --> 00:00:07,080 Now let's consider the case where an object is undergoing 2 00:00:07,080 --> 00:00:12,990 circular motion, but in this motion let's 3 00:00:12,990 --> 00:00:17,237 again introduce our polar coordinate system, r-hat 4 00:00:17,237 --> 00:00:17,820 and theta-hat. 5 00:00:20,340 --> 00:00:24,480 We now want to consider the case where d theta dt is not 6 00:00:24,480 --> 00:00:26,080 constant. 7 00:00:26,080 --> 00:00:27,590 And what does that mean? 8 00:00:27,590 --> 00:00:31,380 That means that if d theta dt is positive, for instance, 9 00:00:31,380 --> 00:00:34,480 the object is speeding up in this direction, 10 00:00:34,480 --> 00:00:37,100 or if d theta dt is negative, the object 11 00:00:37,100 --> 00:00:39,320 is slowing down in this direction. 12 00:00:42,820 --> 00:00:46,430 That's one type of case. 13 00:00:46,430 --> 00:00:49,290 So in this instant, we always know 14 00:00:49,290 --> 00:00:52,620 that there is the radial component at any instant given 15 00:00:52,620 --> 00:00:56,640 by minus r d theta dt squared. 16 00:00:56,640 --> 00:00:59,460 But because it's speeding up and slowing down, 17 00:00:59,460 --> 00:01:02,340 there is now a non-zero tangential component 18 00:01:02,340 --> 00:01:04,150 to the acceleration. 19 00:01:04,150 --> 00:01:06,280 Let's see where that comes from. 20 00:01:06,280 --> 00:01:12,210 So again, if we write our velocity vector as r d theta dt 21 00:01:12,210 --> 00:01:16,910 theta-hat, this is the product of two terms. 22 00:01:16,910 --> 00:01:19,070 And because it's a product of two terms, 23 00:01:19,070 --> 00:01:21,400 we need the product rule from calculus 24 00:01:21,400 --> 00:01:24,810 in when we take a derivative. 25 00:01:24,810 --> 00:01:27,850 So the derivative will be the derivative 26 00:01:27,850 --> 00:01:32,410 of the first term times the second term 27 00:01:32,410 --> 00:01:38,910 plus the first term times the derivative of the second term. 28 00:01:38,910 --> 00:01:41,160 Now we've already analyzed this piece 29 00:01:41,160 --> 00:01:46,350 and this was precisely minus r d theta dt 30 00:01:46,350 --> 00:01:48,390 quantity squared r-hat. 31 00:01:48,390 --> 00:01:54,190 That was the always the non-zero radial acceleration. 32 00:01:54,190 --> 00:01:57,810 But now let's analyze this piece separately. 33 00:01:57,810 --> 00:02:02,190 r, for our circular motion, is a constant. 34 00:02:02,190 --> 00:02:05,810 So it's only d theta dt that is no longer constant. 35 00:02:05,810 --> 00:02:07,940 So we simply take a second derivative. 36 00:02:07,940 --> 00:02:14,290 And so we get r times d squared theta dt squared theta-hat. 37 00:02:14,290 --> 00:02:17,820 And that is our acceleration. 38 00:02:17,820 --> 00:02:20,050 Notice that it has two components. 39 00:02:20,050 --> 00:02:22,750 We'll write the first component, a theta, 40 00:02:22,750 --> 00:02:28,510 that's its tangential component, theta-hat 41 00:02:28,510 --> 00:02:31,870 plus the radial component ar. 42 00:02:31,870 --> 00:02:34,670 That's again, the component. 43 00:02:34,670 --> 00:02:37,500 And because this is a vector, r-hat 44 00:02:37,500 --> 00:02:42,670 where the a theta is now the second derivative of d theta 45 00:02:42,670 --> 00:02:44,310 squared dt squared. 46 00:02:44,310 --> 00:02:51,950 And just to remind you that ar is minus ar d theta dt squared. 47 00:02:51,950 --> 00:02:58,410 So when d squared theta dt squared is positive, 48 00:02:58,410 --> 00:03:04,080 it means d theta dt is increasing. 49 00:03:04,080 --> 00:03:07,520 And so if this object is going in this clockwise direction, 50 00:03:07,520 --> 00:03:09,010 we call that speeding up. 51 00:03:12,450 --> 00:03:15,820 In a similar fashion, it's easy to understand 52 00:03:15,820 --> 00:03:19,450 that when d squared theta dt squared is negative, 53 00:03:19,450 --> 00:03:22,540 then d theta dt is decreasing. 54 00:03:22,540 --> 00:03:24,680 And so it can be slowing down. 55 00:03:24,680 --> 00:03:27,720 Or if it slows down and stops, it 56 00:03:27,720 --> 00:03:30,240 can start to move in the other direction. 57 00:03:30,240 --> 00:03:34,260 So again, the acceleration has two components, 58 00:03:34,260 --> 00:03:37,190 a tangential component, and that depends 59 00:03:37,190 --> 00:03:39,940 on the type of circular motion we're talking about, 60 00:03:39,940 --> 00:03:42,350 whether d theta dt is constant or not. 61 00:03:42,350 --> 00:03:46,960 It always has a non-zero inward radial component 62 00:03:46,960 --> 00:03:50,300 given by the component minus r d theta dt squared, 63 00:03:50,300 --> 00:03:54,999 regardless of whether it's speeding up or slowing down.