1 00:00:03,170 --> 00:00:05,610 I'd now like to consider a very unusual problem, 2 00:00:05,610 --> 00:00:07,660 and we saw a demo of this. 3 00:00:07,660 --> 00:00:12,510 Suppose you have a yoyo, and the yoyo an inner radius 4 00:00:12,510 --> 00:00:16,140 b and an outer radius R. And the yoyo 5 00:00:16,140 --> 00:00:19,645 is rolling without slipping along the ground. 6 00:00:22,710 --> 00:00:26,880 And what we have here is we're pulling the yoyo 7 00:00:26,880 --> 00:00:29,640 with a string that's wrapped around the spool, 8 00:00:29,640 --> 00:00:37,002 and I want to find F max such that it rolls without slipping. 9 00:00:37,002 --> 00:00:38,710 If I don't pull it hard enough, the wheel 10 00:00:38,710 --> 00:00:40,630 will roll without slipping. 11 00:00:40,630 --> 00:00:44,230 And if I pull it harder, then this F max, 12 00:00:44,230 --> 00:00:47,360 the wheel will start to slip. 13 00:00:47,360 --> 00:00:49,900 So now let's analyze this problem, 14 00:00:49,900 --> 00:00:52,930 and let's use our dynamics approach. 15 00:00:52,930 --> 00:00:58,540 Let's choose an i hat, a j hat, k hat. 16 00:00:58,540 --> 00:01:01,280 As before, we'll define an angle theta. 17 00:01:01,280 --> 00:01:04,420 And now we'll apply both Newton's second law 18 00:01:04,420 --> 00:01:09,850 for linear motion and our torque about the center of mass. 19 00:01:09,850 --> 00:01:12,270 Now, we have to consider our forces. 20 00:01:12,270 --> 00:01:15,310 So I'm going to put the forces on this diagram. 21 00:01:15,310 --> 00:01:19,360 We have the normal force from the ground pointing up. 22 00:01:19,360 --> 00:01:22,000 We have the gravitational force pointing down. 23 00:01:22,000 --> 00:01:24,860 And remember, it's rolling without slipping. 24 00:01:24,860 --> 00:01:26,980 So the wheel is-- 25 00:01:26,980 --> 00:01:29,860 the contact point is instantaneously at rest. 26 00:01:29,860 --> 00:01:33,310 But because the center of mass of the wheel 27 00:01:33,310 --> 00:01:39,970 is accelerating, once again, in order to keep a equal to R 28 00:01:39,970 --> 00:01:43,300 alpha, then we need some type of torque 29 00:01:43,300 --> 00:01:45,590 that will produce a non-zero alpha. 30 00:01:45,590 --> 00:01:49,330 And that's going to come from a non-zero static friction. 31 00:01:49,330 --> 00:01:53,410 So once again, we have a case, unlike a wheel which is just 32 00:01:53,410 --> 00:01:55,120 rolling along a horizontal plane, 33 00:01:55,120 --> 00:01:57,789 because we're pulling this yoyo's string, 34 00:01:57,789 --> 00:01:59,740 the static friction is not 0. 35 00:01:59,740 --> 00:02:02,380 Static friction depends on everything else 36 00:02:02,380 --> 00:02:04,390 that's happening in the system. 37 00:02:04,390 --> 00:02:09,070 And now we can apply Newton's second law, F equals ma. 38 00:02:09,070 --> 00:02:16,045 And so in our x direction, we have F minus F equals ma. 39 00:02:16,045 --> 00:02:21,160 And if we look at the torque about the center of mass 40 00:02:21,160 --> 00:02:23,680 is Icm alpha. 41 00:02:23,680 --> 00:02:26,079 Then this is our x equation. 42 00:02:26,079 --> 00:02:32,410 Now here, both the pulling force and the static friction 43 00:02:32,410 --> 00:02:35,380 both exert torques about the center of mass. 44 00:02:35,380 --> 00:02:39,520 The static friction will exert a torque in the direction k hat, 45 00:02:39,520 --> 00:02:44,920 and the pulling force will exert a torque in the minus k hat 46 00:02:44,920 --> 00:02:46,190 direction. 47 00:02:46,190 --> 00:02:49,270 So our torques-- the normal force 48 00:02:49,270 --> 00:02:51,970 does not produce any torque about the center of mass, 49 00:02:51,970 --> 00:02:53,930 nor does gravity. 50 00:02:53,930 --> 00:02:58,150 So what we have is fs times the radius of the wheel-- 51 00:02:58,150 --> 00:03:00,910 that is the torque due to static friction-- 52 00:03:00,910 --> 00:03:06,010 minus b times the pulling force F, 53 00:03:06,010 --> 00:03:10,190 and that's equal to Icm alpha. 54 00:03:10,190 --> 00:03:13,330 So we have our two dynamic equations. 55 00:03:13,330 --> 00:03:17,440 And again, let's recopy our rolling 56 00:03:17,440 --> 00:03:20,050 without slipping condition, now expressed 57 00:03:20,050 --> 00:03:22,190 in terms of acceleration. 58 00:03:22,190 --> 00:03:24,400 So now what I'd like to do is solve 59 00:03:24,400 --> 00:03:28,030 for this force F. And the way I'll do 60 00:03:28,030 --> 00:03:32,230 it is I'll write down, from this condition, alpha 61 00:03:32,230 --> 00:03:38,160 in terms of a/R. And from this equation, I have that-- 62 00:03:38,160 --> 00:03:43,110 so alpha is a/R. And from the top equation, 63 00:03:43,110 --> 00:03:51,030 I have that a is F minus f static over mR, times R. 64 00:03:51,030 --> 00:03:54,329 So I can replace the alpha in this equation, 65 00:03:54,329 --> 00:04:04,900 and I get fsR minus bF is Icm F minus fs over mR. 66 00:04:04,900 --> 00:04:07,840 And now I just need to collect terms. 67 00:04:07,840 --> 00:04:11,430 I'll bring the static friction term to this side and my f term 68 00:04:11,430 --> 00:04:12,750 to that side. 69 00:04:12,750 --> 00:04:16,950 And so I get fs times R. And when 70 00:04:16,950 --> 00:04:19,930 I bring the static friction term over to that other side, 71 00:04:19,930 --> 00:04:22,920 I get Icm over mR. 72 00:04:22,920 --> 00:04:24,930 Let's just check dimensions. 73 00:04:24,930 --> 00:04:27,180 Icm is mR squared. 74 00:04:27,180 --> 00:04:29,210 So these terms have the dimensions of length, 75 00:04:29,210 --> 00:04:32,070 so looks like I'm OK. 76 00:04:32,070 --> 00:04:35,130 And now I bring the F to the other side, 77 00:04:35,130 --> 00:04:43,860 and I get F times b plus Icm over mR. I can now solve for F, 78 00:04:43,860 --> 00:04:50,500 and I get f static R plus Icm m/R-- 79 00:04:50,500 --> 00:04:51,990 it's a complicated answer-- 80 00:04:51,990 --> 00:04:58,350 divided by b plus Icm divided by mR. 81 00:04:58,350 --> 00:05:02,670 Now, if I want to ask the question, what 82 00:05:02,670 --> 00:05:07,050 is the maximum force I can pull in which it just slips, 83 00:05:07,050 --> 00:05:09,660 what's physically happening is the harder I pull, 84 00:05:09,660 --> 00:05:11,640 the bigger the static friction is. 85 00:05:11,640 --> 00:05:13,710 I pull harder, static friction gets bigger. 86 00:05:13,710 --> 00:05:15,780 I pull harder, static friction gets bigger. 87 00:05:15,780 --> 00:05:19,680 But static friction can only have a maximum value. 88 00:05:19,680 --> 00:05:21,550 That hasn't changed. 89 00:05:21,550 --> 00:05:24,030 The maximum value of static friction 90 00:05:24,030 --> 00:05:26,850 is the coefficient of static friction times 91 00:05:26,850 --> 00:05:28,950 a normal force, which, in this case, 92 00:05:28,950 --> 00:05:32,580 is just normal balance in gravity, mg. 93 00:05:32,580 --> 00:05:37,030 So this is the maximum value that static friction can have. 94 00:05:37,030 --> 00:05:41,250 And so now I get F max-- 95 00:05:41,250 --> 00:05:46,770 this is, again, a little complicated-- mu s mg times R 96 00:05:46,770 --> 00:06:01,250 plus Icm divided by mR divided by b plus Icm divided by mR. 97 00:06:01,250 --> 00:06:05,000 And that is the maximum force that I 98 00:06:05,000 --> 00:06:09,940 can pull this yoyo with that it still rolls without slipping.