Q.A uniform solid sphere of radius R has a hole of radius R/2 drilled inside it. One end of the hole is at the centre of the sphere while the other is at the boundary. Locate the centre of mass of the remaining part of the sphere.

[!TLDR] Using the 'negative mass' method (treating the drilled hole as mass -m superposed on the full sphere), the centre of mass of the remaining solid shifts by R/14 from the original centre, on the side opposite to the hole. [!ANSW …

Full solid sphere of radius R has mass $M\propto R^3$ (say $M=kR^3$ for constant k) and centre of mass at its own centre O. The hole is a sphere of radius $\frac{R}{2}$, so its (removed) mass is $m\propto\left(\frac{R}{2}\right)^3=\frac{R^3}{8}$, i.e. $m=k\frac{R^3}{8}=\frac{M}{8}$. The hole's centre is positioned such that one end of the hole is at the sphere's centre O and the other end is at the sphere's boundary -- since the hole has radius R/2, its own centre must be at a distance $\frac{R}{2}$ from O (so that the hole spans exactly from O, at one edge of the hole, out to the boundary at radius R, the hole's far edge). Using the negative-mass method (4.13.1, Example 4.14 method II): treat the drilled-out material as a mass $-m=-\frac{M}{8}$ located at its own centre, distance $r=\frac{R}{2}$ from O, superposed on the FULL sphere (mass M at O). The centre of mass of the remaining solid is then $x_c=\frac{M(0)+\left(-\frac{M}{8}\right)\left(\frac{R}{2}\right)}{M-\frac{M}{8}}=\frac{-\frac{MR}{16}}{\frac{7M}{8}}=-\frac{R}{16}\times\frac{8}{7}=-\frac{R}{14}$. The magnitude is $\frac{R}{14}$, and the negative sign shows the remainin …

Apply the negative-mass technique: treat the hole as a mass of -M/8 (since its radius is half the full sphere's, giving 1/8 the volume) located at its own centre (distance R/2 from the main centre), a …

Misplacing the hole's own centre -- since the hole spans from the sphere's centre to its boundary, the hole's OWN centre is at R/2 from O, not at R/4 or some other v …