Question

A solid sphere of uniform density and radius R is cut off into two pieces by a plane. The plane is at a distance R/2 from the center of sphere.

Find the center of mass of the bigger portion.

Answer 1

see the diagram.

Let the density be d.  

Consider a disc of radius r at y from center of sphere. Clearly  r² = R² - y²

dm = π r²  dy * d = π (R² - y²) d  dy
Mass of smaller part = M1.    Mass of bigger part = M2.

$M1=\pi d \int\limits_{y=\frac{R}{2}}^R {(R^2-y^2)} \, dy\\\\=\pi d[R^2y-\frac{y^3}{3}]_\frac{R}{2}^R=\frac{5}{24}\pi R^3 d\\\\C.M.\ of\ M1=\frac{\pi d}{M1} \int\limits_{y=\frac{R}{2}}^R {y*(R^2-y^2)} \, dy\\\\=\frac{\pi d}{M1}[\frac{R^2y^2}{2}-\frac{y^4}{4}]_\frac{R}{2}^R=\frac{9}{64*M1}\pi R^3 d=\frac{27R}{40}$

Let the center of mass of bigger part M2 be  y.

$M2=\frac{4}{3}\pi R^3 d-\frac{5}{24}\pi R^3 d=\frac{9}{8}\pi R^3 d\\\\combining\ two\ parts:\ \frac{27}{40}R*\frac{5}{24}\pi R^3 d+y*\frac{9}{8}\pi R^3 d=0\\\\y=-\frac{R}{8}$

Ans:     - R/8.