Rotational Kinetic Energy Of A Square

Let $a$ and $b$ be the sides of the frame and the hole, respectively; $m$ is the mass, $v$ the translational speed, and $\omega$ the angular speed. We are told that $b = 0.9\:a$ .

The moment of inertia of the frame is $I = \tfrac 16 m (a^2 + b^2) = \tfrac 16 m (1 + 0.9^2)a^2 = \frac{1.81}6 m a^2.$

The corner of the frame is at a distance $r = \tfrac12\sqrt 2 a$ from the axis of rotation. If it is instantaneously at rest at the bottom of its trajectory, then $v = r\omega = \tfrac12\sqrt 2 a\omega.$ (Compare to the condition for rolling without slipping of a wheel!)

For the rotational and translational kinetic energy we may therefore write $K_{rot} = \tfrac12I\omega^2 = \tfrac12\left(\frac{1.81}6 m a^2\right)\omega^2 = \frac{1.81}6\cdot \tfrac12 ma^2\omega^2; \\ K_{tr} = \tfrac12mv^2 = \tfrac12 m \left(\tfrac12\sqrt 2 a\omega\right)^2 = \frac12\cdot\tfrac12 ma^2\omega^2.$ Therefore $\frac{K_{rot}}{K_{tr} + K_{rot}} = \frac{1.81/6}{1/2 + 1.81/6} = \frac{1.81}{3 + 1.81} = 0.376 \approx \boxed{38\%}.$

It is just the ratio of the moment of inertia about the centre and the moment of inertia about a corner