The wing of a hang-glider is a uniform lamina, formed by removing from a square of side $l$ a quadrant of a circle of radius $l$, with its centre at one corner of the square. Find the distance of the centre of mass (COM) of the wing from the opposite corner.
Answer should be $0.316l$
Tried finding the COM of the circle quadrant using
$$\frac{2l\sin\alpha}{3\alpha}$$
with $\alpha = \frac{\pi}{4}$ and subtracting from $\sqrt2l$.
How would you do this?
On
$$l^2 * \frac{\sqrt2l}{2} - \frac{\pi l^2}{4} * \frac{8l}{3\pi \sqrt2} = (1-\frac{\pi}{4})l^2 * x$$
So, on simplifying $$ l^2*\frac{\sqrt2l}{2} - \frac{2l^3}{3 \sqrt2} = (1-\frac{\pi}{4})l^2 * x $$
So COM is $$ x=1.098l$$ from the corner of the quadrant of the circle.
Now from the opposite corner, the distance of the COM is $$\sqrt2 -x = 0.316l$$
$\sqrt2$ being the whole length of the diagonal.
Hint: You can treat each piece as having its entire mass concentrated at its CM and use the formula for a set of point masses:$$ \mathbf r_{CM}=\frac1M\sum m_i\mathbf r_i. $$ In this case, that gives $$ m_{\text{square}}\mathbf r_{\text{square}}=m_{\text{sector}}\mathbf r_{\text{sector}}+m_{\text{wing}}\mathbf r_{\text{wing}}.$$ By symmetry, of course, we need only work with distances along the diagonal.