A uniform disc of radius R has a circular portion of radius R/2 removed from it. The centre of the removed portion was initially at x = R/2. The removed piece is then placed externally with its centre at x = 3R/2, without overlapping the remaining disc. Taking the original disc centre as x = 0, the centre of mass of the complete system is at:
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Question

A uniform disc of radius R has a circular portion of radius R/2 removed from it. The centre of the removed portion was initially at x = R/2. The removed piece is then placed externally with its centre at x = 3R/2, without overlapping the remaining disc. Taking the original disc centre as x = 0, the centre of mass of the complete system is at:

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G1 Physics · Accepted Answer

Let the mass of the original disc be M. The removed circular part has mass M/4. The remaining disc has mass 3M/4. Its centre of mass is obtained by the negative-mass method: x_rem = [M(0) − (M/4)(R/2)]/(3M/4) = −R/6. Now place the removed piece of mass M/4 at x = 3R/2. The total mass again becomes M. Therefore x(CM) = [(3M/4)(−R/6) + (M/4)(3R/2)]/M = −MR/8M + 3MR/8M = R/4. Hence the correct answer is R/4.