[SOLVED] Centre of mass

**Seirin** · February 25th 2009, 04:43 PM

A uniform semicircular lamina has mass M. A is the mid-point 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 mid-point of AB. Find in terms of pi, the ratio M:m

The answer given is 3pi : (3pi-8)

**skeeter** · February 25th 2009, 05:13 PM

using the theorem of Pappus ...

let $R$ = semicircle fixed radius
$r$ = axis of area rotation

$\frac{\pi R^2}{2} \cdot 2\pi r = \frac{4}{3}\pi R^3$

$r = \frac{4R}{3\pi}$

let point A be 0 ...

$x_{cm} = \frac{M\frac{4R}{3\pi} + mR}{M+m}$

$\frac{R}{2} = \frac{M\frac{4R}{3\pi} + mR}{M+m}<br />$

$\frac{1}{2} = \frac{M\frac{4}{3\pi} + m}{M+m}$

$2\left(M\frac{4}{3\pi} + m\right) = M+m$

$\frac{8M}{3\pi} + 2m = M+m$

$m = M\left(1 - \frac{8}{3\pi}\right)$

$\frac{1}{1 - \frac{8}{3\pi}} = \frac{M}{m}$

$\frac{3\pi}{3\pi - 8} = \frac{M}{m}$

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