1 00:00:04,290 --> 00:00:08,280 For a particle that's moving in a circle, 2 00:00:08,280 --> 00:00:15,840 we found that when it's moving at a constant rate of d theta 3 00:00:15,840 --> 00:00:20,540 dt--- and let's recall what we meant by theta of t-- 4 00:00:20,540 --> 00:00:24,500 and here's our particle, and we introduced our polar 5 00:00:24,500 --> 00:00:29,830 coordinates r hat and theta hat, then we found that the velocity 6 00:00:29,830 --> 00:00:34,550 was r d theta dt theta hat. 7 00:00:34,550 --> 00:00:38,840 And so let's assume that this quantity is positive, 8 00:00:38,840 --> 00:00:41,830 in which case the velocity is pointing in the positive theta 9 00:00:41,830 --> 00:00:42,800 direction. 10 00:00:42,800 --> 00:00:46,660 And that means that everywhere in the circle, 11 00:00:46,660 --> 00:00:51,820 the velocity is tangential to the circle, 12 00:00:51,820 --> 00:00:55,670 and the magnitude is a constant. 13 00:00:55,670 --> 00:00:58,560 So for this case of uniform circular motion, 14 00:00:58,560 --> 00:01:01,440 we calculated that the acceleration 15 00:01:01,440 --> 00:01:10,350 was equal to minus r d theta dt quantity squared r hat, which 16 00:01:10,350 --> 00:01:15,470 means that at every point, the acceleration vector is pointing 17 00:01:15,470 --> 00:01:16,420 towards the center. 18 00:01:19,060 --> 00:01:22,990 Now we can write that acceleration vector 19 00:01:22,990 --> 00:01:30,420 as a component a of r-- r hat-- where this component is given 20 00:01:30,420 --> 00:01:32,979 by r times d theta dt squared. 21 00:01:32,979 --> 00:01:36,630 It it's always negative, because when you square this quantity, 22 00:01:36,630 --> 00:01:38,500 it's always a positive quantity. 23 00:01:38,500 --> 00:01:40,770 The minus sign, just to remember-- that 24 00:01:40,770 --> 00:01:43,770 means that the acceleration is pointing inward. 25 00:01:43,770 --> 00:01:46,420 Now how can we think about that? 26 00:01:46,420 --> 00:01:50,930 Well, if we look at the velocity vector, what's happening here 27 00:01:50,930 --> 00:01:52,940 is the velocity is not changing magnitude 28 00:01:52,940 --> 00:01:54,960 but changing direction. 29 00:01:54,960 --> 00:01:57,750 And if you compare two points-- and let's 30 00:01:57,750 --> 00:02:00,040 just pick two arbitrary points. 31 00:02:00,040 --> 00:02:04,170 So let's remove this acceleration for a moment 32 00:02:04,170 --> 00:02:10,960 and consider two arbitrary points-- say, a time t1 and t2. 33 00:02:10,960 --> 00:02:13,430 So our velocity vectors are tangent. 34 00:02:13,430 --> 00:02:16,020 The length of these vectors are the same. 35 00:02:16,020 --> 00:02:22,440 And if we move them tail to tail-- [? Vt2-- ?] 36 00:02:22,440 --> 00:02:28,570 and take the difference, delta v, where 37 00:02:28,570 --> 00:02:35,250 delta v is equal to v of t2 minus V of t1, 38 00:02:35,250 --> 00:02:38,070 then we can get an understanding why the acceleration 39 00:02:38,070 --> 00:02:40,340 is pointing inward, because recall 40 00:02:40,340 --> 00:02:42,810 that acceleration by definition is 41 00:02:42,810 --> 00:02:45,980 a limit as delta t goes to 0. 42 00:02:45,980 --> 00:02:47,870 That means as this point approaches 43 00:02:47,870 --> 00:02:53,090 that point of the change in velocity over time. 44 00:02:53,090 --> 00:02:55,730 And so when we look at this limit 45 00:02:55,730 --> 00:03:01,480 as we shrink down our time interval between t2 and t1, 46 00:03:01,480 --> 00:03:04,940 then this vector will point towards the center 47 00:03:04,940 --> 00:03:06,100 of the circle. 48 00:03:06,100 --> 00:03:09,074 And that's why the direction of a 49 00:03:09,074 --> 00:03:13,000 is in the minus r hat direction. 50 00:03:13,000 --> 00:03:16,250 Again, let's just recall that this is the case 51 00:03:16,250 --> 00:03:23,510 for we called uniform circular motion, which 52 00:03:23,510 --> 00:03:28,880 is defined by the condition that d theta dt is a constant.