Question

There is a cube of length c. It is in unstable equilibrium, balanced on its edge. The cube is at rest on frictionless surface A. A small shock, or impulse, causes the cube to escape the equilibrium point and topple. After the cube is toppled, the cube is under uniform gravitational field, meaning that g is constant. The cube rotates until any part of it hits the table. Before the cube hits the table, what is the cube's final angular velocity? (Important: before the cube hits the table)

The answer is sqrt[12g[sqrt(2)-1]/(5b)]

Answer #1

consider lenght of cube =c

Balanced on edge, the cube has initial potential energy
Ui=mgc/sqrt2

At the moment before it hits the ground, it has final potential
energy Uf=mgc/2

At the moment before it hits the ground, it has final angular
kinetic energy ?f=1/2 I (?_{f})^{2}.

For the question, where the edge stays fixed, that's everything. Ui=Uf+?f,

mgc/sqrt2=mgl/2+1/2 I (?_{f})^{2}.

mgc(1/sqrt2-1/2)=1/2 I (?_{f})^{2}.

?_{f=}sqrt((2mgc(sqrt2-1))/2I)

The moment of inertia for a cube rotating about an edge is
I=2/3mc^{2}

.substituting I in above we get

?_{f=} sqrt(3g(sqrt2-1)/2c)

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