
[SOLVED] Centre of mass
A uniform semicircular lamina has mass M. A is the midpoint of the diameter and B is on the circumference at the other end of the axis of symmetry. A particle of mass m is attached to the lamina at B. The centre of mass of the loaded lamina is at the midpoint of AB. Find in terms of pi, the ratio M:m
The answer given is 3pi : (3pi8)

using the theorem of Pappus ...
let $\displaystyle R$ = semicircle fixed radius
$\displaystyle r$ = axis of area rotation
$\displaystyle \frac{\pi R^2}{2} \cdot 2\pi r = \frac{4}{3}\pi R^3$
$\displaystyle r = \frac{4R}{3\pi}$
let point A be 0 ...
$\displaystyle x_{cm} = \frac{M\frac{4R}{3\pi} + mR}{M+m}$
$\displaystyle \frac{R}{2} = \frac{M\frac{4R}{3\pi} + mR}{M+m}
$
$\displaystyle \frac{1}{2} = \frac{M\frac{4}{3\pi} + m}{M+m}$
$\displaystyle 2\left(M\frac{4}{3\pi} + m\right) = M+m$
$\displaystyle \frac{8M}{3\pi} + 2m = M+m$
$\displaystyle m = M\left(1  \frac{8}{3\pi}\right)$
$\displaystyle \frac{1}{1  \frac{8}{3\pi}} = \frac{M}{m}$
$\displaystyle \frac{3\pi}{3\pi  8} = \frac{M}{m}$