1 00:00:03,410 --> 00:00:06,680 We've defined moment of inertia of a rigid body already. 2 00:00:06,680 --> 00:00:08,540 And we often are interested in it, 3 00:00:08,540 --> 00:00:11,180 because when we rotate a body about an axis, 4 00:00:11,180 --> 00:00:14,310 say an axis that's passing through the center of mass. 5 00:00:14,310 --> 00:00:16,790 Then our kinetic energy depends in that moment of inertia 6 00:00:16,790 --> 00:00:18,230 through that axis. 7 00:00:18,230 --> 00:00:23,120 But we may also want to consider rotation about another axis. 8 00:00:23,120 --> 00:00:25,460 So suppose we consider an axis passing 9 00:00:25,460 --> 00:00:27,770 through close to the end and we rotate 10 00:00:27,770 --> 00:00:30,740 the body around this axis. 11 00:00:30,740 --> 00:00:35,090 You can see from overhead like that. 12 00:00:35,090 --> 00:00:39,140 Then we want to now consider how the mass is distributed 13 00:00:39,140 --> 00:00:42,710 about an axis that was parallel to the axis passing 14 00:00:42,710 --> 00:00:44,540 through the center of mass. 15 00:00:44,540 --> 00:00:47,090 So we'll now make a little calculation. 16 00:00:47,090 --> 00:00:49,200 But first, we want to quote the theorem. 17 00:00:49,200 --> 00:00:53,120 So let's just draw an example of our rigid body. 18 00:00:53,120 --> 00:00:55,250 And let's take the center of mass, 19 00:00:55,250 --> 00:00:57,710 and let's consider an axis that's going perpendicular 20 00:00:57,710 --> 00:00:59,000 to the center of mass. 21 00:00:59,000 --> 00:01:03,430 We can think of the object is rotating about that axis. 22 00:01:03,430 --> 00:01:05,780 And now let's consider another axis passing 23 00:01:05,780 --> 00:01:09,830 through a different point, but also parallel 24 00:01:09,830 --> 00:01:13,580 to this axis separated by a distance d. 25 00:01:13,580 --> 00:01:18,470 Then the result that we want is at the moment of inertia 26 00:01:18,470 --> 00:01:20,720 about an axis passing perpendicular 27 00:01:20,720 --> 00:01:24,050 to the plane of the object through the axis s 28 00:01:24,050 --> 00:01:27,680 is equal to the moment of inertia about an axis 29 00:01:27,680 --> 00:01:29,390 passing through the center of mass-- 30 00:01:29,390 --> 00:01:32,509 notice these are parallel axes-- plus the mass 31 00:01:32,509 --> 00:01:38,509 of the object times the distance between the two parallel axes. 32 00:01:38,509 --> 00:01:41,420 And this is a result that is very useful when 33 00:01:41,420 --> 00:01:42,979 calculating moments of inertia. 34 00:01:42,979 --> 00:01:45,020 Of course, we could calculate the moment 35 00:01:45,020 --> 00:01:46,950 about any axis we wanted. 36 00:01:46,950 --> 00:01:48,500 And we'll see that in a moment. 37 00:01:48,500 --> 00:01:54,310 For example, we know that for a rod of length L 38 00:01:54,310 --> 00:01:57,560 that the moment of inertia through the center of mass 39 00:01:57,560 --> 00:01:58,759 was 1/12. 40 00:01:58,759 --> 00:02:04,780 And let's call the length of the rod L. So we have a length L. 41 00:02:04,780 --> 00:02:08,660 And let's assume it's a uniform object. 42 00:02:08,660 --> 00:02:12,050 And we'll calculate a moment of inertia through an axis 43 00:02:12,050 --> 00:02:13,670 through the end. 44 00:02:13,670 --> 00:02:17,540 And so in this case, d is equal to L over 2. 45 00:02:17,540 --> 00:02:23,210 And so I about the end axis is 1/12 mL 46 00:02:23,210 --> 00:02:28,970 squared plus the mass times L over 2 square 47 00:02:28,970 --> 00:02:34,340 and a 12 plus a quarter is 1/3 mL squared. 48 00:02:34,340 --> 00:02:37,280 And that means that all you need to know 49 00:02:37,280 --> 00:02:39,530 is the moment through the center of mass, 50 00:02:39,530 --> 00:02:42,650 and you can calculate the moment through any other axis. 51 00:02:42,650 --> 00:02:46,410 Very useful theorem called the parallel axis theorem. 52 00:02:49,040 --> 00:02:54,200 Now as I said, we can calculate the moment s. 53 00:02:54,200 --> 00:02:58,640 And just to show you very quickly, if we pick s here, 54 00:02:58,640 --> 00:03:03,380 and we pick our little dm there, and we have a distance x, 55 00:03:03,380 --> 00:03:06,020 then the moment about s-- and let's 56 00:03:06,020 --> 00:03:09,860 say that dm is equal to the total mass divided 57 00:03:09,860 --> 00:03:14,060 by the length times some little distance dx-- 58 00:03:14,060 --> 00:03:20,450 then the moment is dm times x squared, where x we'll 59 00:03:20,450 --> 00:03:21,980 give this an integration variable 60 00:03:21,980 --> 00:03:26,000 x prime goes from 0 to x prime equals l. 61 00:03:26,000 --> 00:03:31,200 And so we have for our dm m over L dx prime. 62 00:03:31,200 --> 00:03:33,829 Remember, that's our integration variable. 63 00:03:33,829 --> 00:03:36,000 And we have x squared. 64 00:03:36,000 --> 00:03:39,110 So that's x prime squared. 65 00:03:39,110 --> 00:03:43,100 And this is just the integral of x cubed over 3. 66 00:03:43,100 --> 00:03:50,690 So we have 1/3 m over L x prime cubed evaluated from 0 to L, 67 00:03:50,690 --> 00:03:55,340 and that comes out to 1/3 mL squared. 68 00:03:55,340 --> 00:03:58,910 And that's in agreement with the parallel axis theorem. 69 00:03:58,910 --> 00:04:01,010 And that's an example of parallel axis theorems. 70 00:04:01,010 --> 00:04:06,477 You can do the same thing with the disk or other objects.