Physics - Center of Mass Question with Solution | TestHub

Negative-mass method about the disc centre O: the removed disc has radius R, so its mass is m = M/9 (mass proportional to area). Its centre shifts from x = 2R (the hole) to x = 4R (placed at the surface). The system = full disc (mass M, COM at O) minus m at x = 2R plus m at x = 4R:

$x_{com}=\frac{M\cdot 0-m(2R)+m(4R)}{M}=\frac{2mR}{M}$

$x_{com}=\frac{2R}{9}\{m=M/9\}$

$x_{com}=\frac{2\times 9}{9}=2\mathrm{m}$