find the position of centre of mass of the uniform planner sheet with respect to origin
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The centre of mass is a point where all the mass is assumed to be located.
In the above figure, the sheet is divided into 3 segments. The (X, Y) coordinates of the 3 segments can be taken as (A/2, 3A/2) for segment 1, for segment 2 it is (A/2, A/2), and for segment 3 are (3A/2, A/2).
Each segment will be having a mass of M/3.
Therefore,
X CM = (1/M) [A/2(M/3) + A/2(M/3) + 3A/2(M/3)] = 5A/6
Y CM = (1/M) [3A/2(M/3) + A/2(M/3) + A/2(M/3)] = 5A/6
In the above figure, the sheet is divided into 3 segments. The (X, Y) coordinates of the 3 segments can be taken as (A/2, 3A/2) for segment 1, for segment 2 it is (A/2, A/2), and for segment 3 are (3A/2, A/2).
Each segment will be having a mass of M/3.
Therefore,
X CM = (1/M) [A/2(M/3) + A/2(M/3) + 3A/2(M/3)] = 5A/6
Y CM = (1/M) [3A/2(M/3) + A/2(M/3) + A/2(M/3)] = 5A/6
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