Explanation

(a) The radius of gyration is the distance from the
axis of rotation at which if whole mass of the
body is supposed to be concentrated.
Here, the whole mass of the equilateral triangle
acts at point $O$. So the distance $OA$ is the radius of gyration of
this system. Now from triangle $ADB$
[image:7_image113.emf]
$x^{2} = BD^{2} + \left( \frac{x}{2} \right)^{2}$
or $BD^{2} = x^{2} - \frac{x^{2}}{4}$
or $BD^{2} = \frac{3x^{2}}{4}$
or $BD = \sqrt{3}\frac{x}{2}$
Hence, the distance, $OB = \frac{\sqrt{3}x}{2} \times \frac{2}{3}$
$\Rightarrow$ $OB = \frac{x}{\sqrt{3}}$
But, the distances $OA,\ OB\ and\ OC$ are the same.
So, $OA = \frac{x}{\sqrt{3}}$
Hence, the radius of gyration of this system is $\frac{x}{\sqrt{3}}$.