1 00:00:03,670 --> 00:00:06,340 Suppose we have a rigid body, and it's 2 00:00:06,340 --> 00:00:09,220 rotating about a fixed axis, and we 3 00:00:09,220 --> 00:00:13,450 know that the angular acceleration, alpha, is given. 4 00:00:13,450 --> 00:00:15,490 Now, it may be a function of time. 5 00:00:15,490 --> 00:00:18,220 And we'd like to find from the angular velocity 6 00:00:18,220 --> 00:00:21,710 and how much angle has rotated in some time interval. 7 00:00:21,710 --> 00:00:23,770 So the first thing we have to do is 8 00:00:23,770 --> 00:00:26,410 choose a point in the rigid body, 9 00:00:26,410 --> 00:00:28,780 and introduce a coordinate system. 10 00:00:28,780 --> 00:00:31,240 Here we have our angle theta. 11 00:00:31,240 --> 00:00:35,060 And with this point, we have our r hat and our theta hat 12 00:00:35,060 --> 00:00:36,470 direction. 13 00:00:36,470 --> 00:00:38,460 And recall that by the right-hand rule, 14 00:00:38,460 --> 00:00:44,200 the k hat is going out of the plane of the figure. 15 00:00:44,200 --> 00:00:46,300 And so we have our coordinate system. 16 00:00:46,300 --> 00:00:51,940 And we wrote alpha as the second derivative of theta 17 00:00:51,940 --> 00:00:53,950 with respect to time k hat. 18 00:00:53,950 --> 00:00:59,110 Now our goal is to find omega, which 19 00:00:59,110 --> 00:01:05,170 we'll right as the first derivative of d theta dt k hat, 20 00:01:05,170 --> 00:01:09,590 and also to find theta as a function of time. 21 00:01:09,590 --> 00:01:15,920 Now recall that our notation was that alpha z was the omega 22 00:01:15,920 --> 00:01:20,630 z dt equals d squared theta dt squared. 23 00:01:20,630 --> 00:01:25,852 And in this notation, omega z was equal to d theta dt. 24 00:01:25,852 --> 00:01:27,310 Now what we're going to do is we're 25 00:01:27,310 --> 00:01:29,710 going to integrate alpha, just like we 26 00:01:29,710 --> 00:01:34,120 did in simple circular motion kinematics 27 00:01:34,120 --> 00:01:36,039 for point-like objects. 28 00:01:36,039 --> 00:01:39,370 And so what we have here is that for some time interval, 29 00:01:39,370 --> 00:01:46,479 omega z at time t minus omega z0 is the integral of t 30 00:01:46,479 --> 00:01:51,340 prime equals 0 to t of alpha z, which 31 00:01:51,340 --> 00:01:54,970 is our component of the angular acceleration. 32 00:01:54,970 --> 00:01:58,420 We have an integration variable, dt prime. 33 00:01:58,420 --> 00:02:02,710 And recall that that implies that omega z t can 34 00:02:02,710 --> 00:02:05,080 have some initial value at time t 35 00:02:05,080 --> 00:02:09,400 equals 0 plus this integral relationship, 36 00:02:09,400 --> 00:02:14,350 which is what we want to figure out by direct integration. 37 00:02:14,350 --> 00:02:19,480 Now, this only works when alpha z is some function of time. 38 00:02:19,480 --> 00:02:24,490 In order to get the angle, theta of t, we integrate again, 39 00:02:24,490 --> 00:02:28,180 where we have theta t0 is the integral from t 40 00:02:28,180 --> 00:02:33,820 prime equal 0 to t prime equals t of this function omega z. 41 00:02:33,820 --> 00:02:39,260 Again, we have some integration variable, t prime, dt prime. 42 00:02:39,260 --> 00:02:42,850 And so we see that theta t can have 43 00:02:42,850 --> 00:02:47,829 some initial value plus this integral relationship, 44 00:02:47,829 --> 00:02:54,610 t prime equals t of omega z t prime dt prime. 45 00:02:54,610 --> 00:02:57,360 And this is how we can figure out 46 00:02:57,360 --> 00:03:03,180 how the point p has a component of angular velocity, 47 00:03:03,180 --> 00:03:07,130 and what angle the point p sweeps out 48 00:03:07,130 --> 00:03:09,324 in some time interval t.