Video: Identifying the Center of Mass of a Uniform Lamina Bounded by a Parallelogram

A uniform lamina is bounded by the parallelogram 𝐴𝐡𝐢𝐷. Where does its center of gravity lie?

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Video Transcript

A uniform lamina is bounded by the parallelogram 𝐴𝐡𝐢𝐷. Where does its center of gravity lie?

As the body is uniform, we know that the density is the same throughout the body. This makes finding the center of mass or center of gravity of uniform bodies straightforward. Let’s consider the parallelogram 𝐴𝐡𝐢𝐷 as given in this question.

If we draw the two diagonals from 𝐴 to 𝐢 and from 𝐡 to 𝐷, then the center of gravity is at the intersection point of the diagonals. The center of gravity will be equidistant from points 𝐴 and 𝐢 and also from points 𝐡 and 𝐷. If we let the four vertices of the parallelogram have coordinates π‘₯ one, 𝑦 one; π‘₯ two, 𝑦 two; π‘₯ three, 𝑦 three; and π‘₯ four, 𝑦 four, then the center of gravity has coordinates π‘₯ one plus π‘₯ three divided by two, 𝑦 one plus 𝑦 three divided by two. This is the midpoint of 𝐴 and 𝐢.

Alternatively, we could work out the midpoint of 𝐡 and 𝐷. This would have coordinates π‘₯ two plus π‘₯ four over two, 𝑦 two plus 𝑦 four over two. These formulas enable us to calculate the center of gravity for any parallelogram.

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