Part 1: Calculation of the initial angular velocity

 

Conservation of angular momentum about the hinge gives 4mva/3 = I

 

I = 2Ma²/3+2Ma² (Parallel axes theorem)

  = 8Ma²/3

 

Hence _o = mv/2Ma

 

Part 2: Condition for not tipping over

 

The block now begins to rotate about side AB. If the block is to not tip over, then when the centre of mass of the block is vertically above side AB, the angular velocity is zero.

 

Since the work done by non-conservative forces at the hinge is zero, we can apply the energy conservation law as KE_i = PE_f

 

1/2 I_o² = 2 Mga

Hence 1/3 m²v²/M =  2 Mga

 

Hence the min velocity to ensure the block tips over is M/m6ga

 

I guess this is the answer that will appear in the text book. Only I am not comfortable with this business of conserving angular momentum at the beginning. Someone pls clarify how it is possible to apply it in this case.