If the total mass of object mass M then the mass removed part will be 4M .

XCM=M−4MM⋅2a−4M(43a)=43M2Ma−163Ma=125a
YCM=M−4MM⋅2b−4M(43b)=43M2Mb−163Mb=125b
A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be:

Held on 8 Apr 2019 · Verified 6 Jul 2026.
125a,125b
35a,35b
32a,32b
43a,43b
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