WEBVTT
00:00:01.720 --> 00:00:07.160
A uniform lamina is bounded by the parallelogram π΄π΅πΆπ·.
00:00:07.810 --> 00:00:10.480
Where does its center of gravity lie?
00:00:11.850 --> 00:00:16.730
As the body is uniform, we know that the density is the same throughout the body.
00:00:18.040 --> 00:00:24.320
This makes finding the center of mass or center of gravity of uniform bodies straightforward.
00:00:25.020 --> 00:00:30.710
Letβs consider the parallelogram π΄π΅πΆπ· as given in this question.
00:00:31.750 --> 00:00:42.150
If we draw the two diagonals from π΄ to πΆ and from π΅ to π·, then the center of gravity is at the intersection point of the diagonals.
00:00:43.060 --> 00:00:50.600
The center of gravity will be equidistant from points π΄ and πΆ and also from points π΅ and π·.
00:00:51.520 --> 00:01:12.100
If we let the four vertices of the parallelogram have coordinates π₯ one, π¦ one; π₯ two, π¦ two; π₯ three, π¦ three; and π₯ four, π¦ four, then the center of gravity has coordinates π₯ one plus π₯ three divided by two, π¦ one plus π¦ three divided by two.
00:01:12.940 --> 00:01:15.920
This is the midpoint of π΄ and πΆ.
00:01:16.400 --> 00:01:20.600
Alternatively, we could work out the midpoint of π΅ and π·.
00:01:21.140 --> 00:01:28.770
This would have coordinates π₯ two plus π₯ four over two, π¦ two plus π¦ four over two.
00:01:29.980 --> 00:01:35.660
These formulas enable us to calculate the center of gravity for any parallelogram.