1 00:00:03,510 --> 00:00:05,790 Let's consider a wheel that's rolling. 2 00:00:05,790 --> 00:00:09,870 So our wheel is rolling along a surface. 3 00:00:09,870 --> 00:00:11,730 The wheel has radius r. 4 00:00:11,730 --> 00:00:15,150 And I want to consider the motion of a point P 5 00:00:15,150 --> 00:00:16,740 on the rim of the wheel. 6 00:00:16,740 --> 00:00:20,970 Now, I'll choose an origin that's fixed. 7 00:00:20,970 --> 00:00:26,910 And in fact, here we're going to talk about the reference frame 8 00:00:26,910 --> 00:00:29,155 A, which is fixed to the ground. 9 00:00:33,420 --> 00:00:36,570 And I would like to consider-- my problem is 10 00:00:36,570 --> 00:00:40,050 to find both the position of the point on the rim, 11 00:00:40,050 --> 00:00:42,120 the velocity of the point on the rim, 12 00:00:42,120 --> 00:00:44,430 and the acceleration of the point in the rim 13 00:00:44,430 --> 00:00:46,660 as a function of time. 14 00:00:46,660 --> 00:00:51,180 Now, the position vector is given from my fixed origin 15 00:00:51,180 --> 00:00:53,190 to where the point on the rim is. 16 00:00:53,190 --> 00:00:55,300 Now, this can be quite complicated. 17 00:00:55,300 --> 00:00:57,930 But what we're going to do is consider a second reference 18 00:00:57,930 --> 00:01:03,700 frame that is located at the center of mass of the wheel. 19 00:01:03,700 --> 00:01:10,680 So our second frame, we'll take reference frame 20 00:01:10,680 --> 00:01:14,530 and we'll call this frame cm, is located, 21 00:01:14,530 --> 00:01:22,440 is moving with the center of mass of the wheel. 22 00:01:27,870 --> 00:01:29,930 So in that reference frame, we'll 23 00:01:29,930 --> 00:01:34,729 denote this point by vector r cm, 24 00:01:34,729 --> 00:01:36,770 because we're in the reference frame with respect 25 00:01:36,770 --> 00:01:40,820 to the center of mass, and we're talking about the point P. 26 00:01:40,820 --> 00:01:43,770 Now, these vectors are connected. 27 00:01:43,770 --> 00:01:45,410 And we'll call this vector capital 28 00:01:45,410 --> 00:01:52,039 R. By our vector triangle, capital R equals little rP-- 29 00:01:52,039 --> 00:01:53,660 that's the position vector. 30 00:01:53,660 --> 00:01:56,509 And the fixed frame of the ground is given by capital 31 00:01:56,509 --> 00:02:01,670 R plus r cm P. 32 00:02:01,670 --> 00:02:05,090 Now, as we've seen before by differentiating, 33 00:02:05,090 --> 00:02:07,310 the velocity of the particle is given 34 00:02:07,310 --> 00:02:19,570 by V plus V cm P. This one is the velocity in frame 35 00:02:19,570 --> 00:02:22,450 fixed to the ground. 36 00:02:25,790 --> 00:02:35,845 This is the velocity of the center of mass 37 00:02:35,845 --> 00:02:43,250 with respect to ground frame. 38 00:02:43,250 --> 00:02:50,180 And finally, V cm P is the velocity 39 00:02:50,180 --> 00:02:51,510 in the center of mass frame. 40 00:02:58,260 --> 00:03:02,040 Now, in order to analyze this law, what we'll now do 41 00:03:02,040 --> 00:03:06,000 is focus on what the motion looks 42 00:03:06,000 --> 00:03:09,840 like if you're an observer moving with the center of mass. 43 00:03:09,840 --> 00:03:12,652 And that's how we'll begin our analysis.