Answer:
(a)
\(\bar{y}=\frac{6k+3\pi-4}{3(4k+\pi)}a.\)
(b)
\(k=\sqrt{\frac23}.\)
(a) The lamina consists of a rectangle and a quarter circle.
The rectangle has area \(ka^2\), and its centre of mass is at height \(\frac12a\) above the horizontal surface.
The quarter circle has area
\(\frac14\pi a^2.\)
Using the formula for the centre of mass of a circular sector, the distance of the centre of mass of the quarter circle from the straight side \(OA\) is
\(\frac{2a\cos(\pi/4)\sin(\pi/4)}{3(\pi/4)} =\frac{4a}{3\pi}.\)
Therefore its height above the horizontal surface is
\(a-\frac{4a}{3\pi}.\)
Let \(\bar{y}\) be the height of the centre of mass of the whole lamina above the horizontal surface. Taking moments about the horizontal surface,
\(\left(ka^2+\frac14\pi a^2\right)\bar{y} =ka^2\left(\frac12a\right)+\frac14\pi a^2\left(a-\frac{4a}{3\pi}\right).\)
Simplifying the numerator,
\(ka^2\left(\frac12a\right)+\frac14\pi a^2\left(a-\frac{4a}{3\pi}\right) =a^3\left(\frac{k}{2}+\frac{\pi}{4}-\frac13\right).\)
Thus
\(\bar{y}=a\frac{\frac{k}{2}+\frac{\pi}{4}-\frac13}{k+\frac{\pi}{4}}.\)
Multiplying numerator and denominator by \(12\),
\(\bar{y}=\frac{6k+3\pi-4}{3(4k+\pi)}a.\)
(b) On the point of toppling about \(B\), the vertical line through the centre of mass passes through \(B\). Taking moments about the vertical line \(OB\), the moment of the rectangle on one side equals the moment of the quarter circle on the other side.
The rectangle has area \(ka^2\), and its centre is a horizontal distance \(\frac12ka\) from \(OB\). The quarter circle has area \(\frac14\pi a^2\), and its centre is a horizontal distance \(\frac{4a}{3\pi}\) from \(OB\). Hence
\(ka^2\left(\frac12ka\right) =\frac14\pi a^2\left(\frac{4a}{3\pi}\right).\)
So
\(\frac12k^2a^3=\frac13a^3.\)
Therefore
\(k^2=\frac23,\)
and since \(k\gt 0\),
\(k=\sqrt{\frac23}.\)
