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 45o 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

We will use the principle of conservation of energy to solve the given problem.

As the slab startst to fall down, it will have an angular acceleration which will make it attain some angular velocity by the time it hits the spring. At the initial point, the block will have potential energy only, while at the final position, it will have some rotational kinetic energy and the spring will have some potential energy as it is being pressed.

We will form an equation, equating the energies at above two positions and then resolve to find the velocity.

Potential Energy (Initial) = Kinetic Energy(final) + Potential energy (Final)

That is, MgH = 0.5Iw² + 0.5Kx²

Now the inertia of the rectangular slab about the corner is given as:

I = M(a² + b²)/12 + M(a² + b²)/4 = M(a² + b²)/3

or, 15 x 9.81 x (0.32 - 0.3/22) = 0.5 x 15 x (0.09 + 0.36)w² + 0.5 x 8 x 10³ x 0.05 x0.05

or, 46.8205 = 3.375w² + 10

or, W = 3.303 rad/second

Therefore the required angular velocity is given as: W = 3.303 rad/second