1 00:00:03,140 --> 00:00:05,980 Now let's go back and analyze that same problem 2 00:00:05,980 --> 00:00:09,010 that we were looking at before of a wheel 3 00:00:09,010 --> 00:00:11,290 rolling down an incline plane. 4 00:00:11,290 --> 00:00:15,760 And it's rolling without slipping. 5 00:00:15,760 --> 00:00:18,220 But instead of using the energy method, 6 00:00:18,220 --> 00:00:21,340 we're now going to use the torque method. 7 00:00:21,340 --> 00:00:24,670 So let's consider the center of mass. 8 00:00:24,670 --> 00:00:26,800 And what we want to do is draw the forces. 9 00:00:26,800 --> 00:00:28,930 We have a normal force. 10 00:00:28,930 --> 00:00:30,670 We have gravity. 11 00:00:30,670 --> 00:00:34,840 And we have the friction force about the center of mass. 12 00:00:34,840 --> 00:00:37,870 Our wheel has a radius R. 13 00:00:37,870 --> 00:00:42,280 Let's choose a coordinate system i hat, j hat. 14 00:00:42,280 --> 00:00:45,160 And so we have a right-handed system k hat, 15 00:00:45,160 --> 00:00:49,270 and that will correspond to some angle theta. 16 00:00:49,270 --> 00:00:54,160 Now here we're now going to enlarge how we apply 17 00:00:54,160 --> 00:00:56,920 both translation and rotation. 18 00:00:56,920 --> 00:00:58,750 And the beauty of this problem is 19 00:00:58,750 --> 00:01:04,030 we now can decompose our motion into translational motion 20 00:01:04,030 --> 00:01:05,630 and rotational motion. 21 00:01:05,630 --> 00:01:07,660 So for the transitional motion, we'll 22 00:01:07,660 --> 00:01:11,240 apply Newton's second law. 23 00:01:11,240 --> 00:01:13,990 Now, if this is the angle phi, then that's 24 00:01:13,990 --> 00:01:15,789 the angle phi as well. 25 00:01:15,789 --> 00:01:19,120 And so our forces in the i hat direction, we 26 00:01:19,120 --> 00:01:25,240 have mg sine phi minus the friction force. 27 00:01:25,240 --> 00:01:31,180 And that's equal to the x component of the acceleration. 28 00:01:31,180 --> 00:01:35,050 Now we also can choose the center of mass 29 00:01:35,050 --> 00:01:38,320 to calculate the torque. 30 00:01:38,320 --> 00:01:40,300 And so what we're really just studying here 31 00:01:40,300 --> 00:01:44,890 is simply our old problem in the center of mass frame of fixed 32 00:01:44,890 --> 00:01:46,390 axis rotation. 33 00:01:46,390 --> 00:01:49,450 And you can see gravity is acting at the center of mass. 34 00:01:49,450 --> 00:01:52,120 So it produces no torque about the center of mass. 35 00:01:52,120 --> 00:01:56,470 The normal force is directed towards the center of mass. 36 00:01:56,470 --> 00:02:06,520 And so when we take that vector product of R cross n from cm 37 00:02:06,520 --> 00:02:08,710 to this point down here, the contact point, 38 00:02:08,710 --> 00:02:10,538 these forces are anti-parallel. 39 00:02:10,538 --> 00:02:12,970 So the normal force produces no torque. 40 00:02:12,970 --> 00:02:16,960 And the only torque that we have is from the friction force, 41 00:02:16,960 --> 00:02:18,940 and that friction torque is going 42 00:02:18,940 --> 00:02:22,750 to give us a positive angular acceleration in the k hat 43 00:02:22,750 --> 00:02:23,640 direction. 44 00:02:23,640 --> 00:02:26,270 It's at right angles with the vector R. 45 00:02:26,270 --> 00:02:33,620 So we have fs times R equals I center of mass times alpha. 46 00:02:33,620 --> 00:02:37,420 And these are our two dynamic equations. 47 00:02:37,420 --> 00:02:40,780 But remember, when the object is rolling without slipping, 48 00:02:40,780 --> 00:02:45,560 let's just remind ourselves that Vcm equals R omega. 49 00:02:45,560 --> 00:02:50,560 And if I differentiate, the Acm which is what we're calling ax, 50 00:02:50,560 --> 00:02:53,829 is equal to R alpha. 51 00:02:53,829 --> 00:02:57,520 So this ax here is the acceleration 52 00:02:57,520 --> 00:02:59,079 of the center of mass. 53 00:02:59,079 --> 00:03:02,330 And that's our third condition. 54 00:03:02,330 --> 00:03:06,820 And so now I see that I have three equations 55 00:03:06,820 --> 00:03:09,530 and my unknowns-- 56 00:03:09,530 --> 00:03:12,270 fs, ax and alpha. 57 00:03:12,270 --> 00:03:16,290 And so I'm going to solve these equations for a. 58 00:03:16,290 --> 00:03:18,130 And I'll look at these equations. 59 00:03:18,130 --> 00:03:24,280 And what I'll do is I'll just substitute for alpha ax over R. 60 00:03:24,280 --> 00:03:28,750 And then solve this equation for fs, and put it in there. 61 00:03:28,750 --> 00:03:32,500 And so I get mg sine phi. 62 00:03:32,500 --> 00:03:39,360 Now my fs is equal to Icm over R times alpha. 63 00:03:39,360 --> 00:03:45,470 But alpha is ax over R. So that's ax and an R squared. 64 00:03:45,470 --> 00:03:48,950 Notice dimensionally, I is mr squared. 65 00:03:48,950 --> 00:03:51,550 So this is just ma, the dimensions 66 00:03:51,550 --> 00:03:56,950 of force, mg dimensions of force, and that's equal to max. 67 00:03:56,950 --> 00:04:00,280 And now I can solve for ax. 68 00:04:00,280 --> 00:04:11,020 And I get mg sine phi divided by m plus Icm over R squared. 69 00:04:11,020 --> 00:04:15,790 Now ax is a constant. 70 00:04:15,790 --> 00:04:19,390 And we can, from our kinematic equations, 71 00:04:19,390 --> 00:04:28,704 if our object is moving a distance s 72 00:04:28,704 --> 00:04:34,159 as it drops height h, we know from kinematics 73 00:04:34,159 --> 00:04:35,750 and we can work this out. 74 00:04:35,750 --> 00:04:42,930 We have that Xcm is one half Acm t squared. 75 00:04:42,930 --> 00:04:53,930 And we know that the velocity Vcm equals Acm times t. 76 00:04:53,930 --> 00:04:56,510 And so when we put these two together, 77 00:04:56,510 --> 00:05:02,110 this is the distance s, we get that the velocity 78 00:05:02,110 --> 00:05:12,600 cm is equal to the square root of 2s times Acm. 79 00:05:12,600 --> 00:05:21,990 Now s is equal to h over sine phi. 80 00:05:21,990 --> 00:05:26,140 S h over s is sine phi. 81 00:05:26,140 --> 00:05:34,390 And so the Vcm equals the square root of 2h sine phi times Acm-- 82 00:05:34,390 --> 00:05:36,040 but we've solved for Acm-- 83 00:05:36,040 --> 00:05:44,290 mg sine phi over m plus Icm over R squared. 84 00:05:44,290 --> 00:05:50,290 And so we get the square root of 2mgh 85 00:05:50,290 --> 00:05:56,650 over m plus Icm divided by R squared. 86 00:05:56,650 --> 00:06:00,900 And this agrees with our energy method.