A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis passing through D and perpendicular to face ABCD. A bullet of mass m and speed v is shot at a height of 4a//3 as shown in the figure. The bullet becomes embedded in the cube. Find the minimum value of v required to topple the cube. Assume mgt gtM.

A solid cube of wood of side 2a and mass M is resting on a horizontal

1.	A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis passing through D and perpendicular to face ABCD. A bullet of mass m and speed v is shot at a height of 4a//3 as shown in the figure. The bullet becomes embedded in the cube. Find the minimum value of v required to topple the cube. Assume mgt gtM.
Answer» SOLUTION :Let `omega` be the angular velocity of cube just after collison about an axis passing through `D`. From conservation of the angular mometum about an axis passing through `D`. `mv((4A)/3)=I_(D).omega` Here `I_(D)=I_(com)+mr^(2)` where `r=` perpedicular distance of axis of ROTATION passing through `D` from centre of mass (com) of the cube `=sqrt(2)a` or `(4mva)/3=[M((4^(2)+4a^(2))/12)+M(2a^(2))]omega` or `(4mva)/3=8/3Ma^(2).omega` `omega=1/2 (mv)/(Ma)`........i The cube will topple if its `COM` is just able to reach in a vertical height `h_(2)` as shown in figure. Hence applying conservation of mechanical energy `1/2I_(D)omega^(2)=Mg(h_(2)-h_(1))` (`:'I_(D)=8/3Ma^(2))` or `1/2 (8/3Ma^(2)).{1/2(mv)/(Ma)}^(2)=Mg(sqrt(2-1)a` or `1/2(m/M)^(2)v^(2)=ag(sqrt(2)-1)` or `v=M/m[3ag(sqrt(2)-1)]^(1/2)`