A classical mechanics problem by Tai Ching Kan

First, we have to find the centre of mass of the lamina with the added masses. Since the original hexagon is regular, its centre of mass is at $O$ .

Since changing the edge length of the hexagon does not change the angle we are after, we can define the hexagon to have edge lengths 1 which means that,

$OA=OB=OC=OD=OE=OF=1$

We also define that $O$ has coordinates $(0,0)$ and that $OA$ is in the direction of the positive y-axis.

We can now find the centre of mass of the lamina with the added masses.

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Taking moments about the x-axis,

$-\frac{\sqrt{3}}{2}m+0\times2m+\frac{\sqrt{3}}{4}\times3m+\frac{\sqrt{3}}{2}\times2m=(m+2m+3m+2m)x$ where $x$ is the x-coordinate of the centre of mass.

$m$ 's cancel, and after simplication,

$x=\frac{5}{32}\sqrt{3}$

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Taking moments about the y-axis,

$\frac{1}{4}\times3m+0\times m+0\times2m-\frac{1}{2}\times2m=(3m+m+2m+2m)y$ where $y$ is the y-coordinate of the centre of mass.

$y=-\frac{1}{32}$

We therefore know that the coordinates of the centre of mass (which we shall call G) is: $(\frac{5}{32}\sqrt{3},-\frac{1}{32})$ .

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In order to find theta on the diagram,

$\tan\theta=\frac{\frac{5}{32}\sqrt{3}}{1+\frac{1}{32}}$

$\theta=14.705^{\circ}$

Hence, the required angle is $60^{\circ}-\theta=60^{\circ}-14.705^{\circ}=\boxed{45.295^{\circ}}$