A 15 kg rectangular slab that is pinned in one corner is rot

A 15 kg rectangular slab that is pinned in one corner is rotated 45 degrees and released. The slab strikes a spring (k = 8 kN/m) that is unstretched and causes it to depress 0.05 m (bottom edge of the slab is horizontal at that moment). What is the angular velocity at that moment? Assume the spring compression is small and the spring does not twist or rotate.

Solution

moment of inertia is,

   I = M[a^2 + b^2] / 12 + M[a^2 + b^2] / 4 = M[a^2 + b^2] / 3 = 15[0.09+0.36] / 3 = 2.25 kg.m^2

from the law of conservation of energy, we get

    mgh = 0.5Iw^2 + 0.5kx^2

15*9.81*[0.3*sqrt(2) - [0.3/2sqrt(2)]] = 0.5(2.25)w^2 + 0.5*8000*0.05*0.05

then, the angular velocity:

     w = 5.8 rad/s