A solid cube of wood of side 2a and mass M is | vedclass

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(D) The bullet provides the cube with an angular impulse,causing it to rotate about axis $AB$.Let $I_{A}$ be the moment of inertia of the cube about a

Vedclass · Accepted Answer

$\sqrt{3g(\sqrt{2}-1) / 4a}$

Vedclass · Answer

(D) The bullet provides the cube with an angular impulse,causing it to rotate about axis $AB$.Let $I_{A}$ be the moment of inertia of the cube about axis $AB$ and $\omega_{c}$ be the initial angular speed.Using the parallel axis theorem,$I_{A} = I_{CM} + M(OA)^{2}$.Given $I_{CM} = 2Ma^{2}/3$ and the distance from the center $O$ to the axis $AB$ is $OA = \sqrt{a^{2} + a^{2}} = a\sqrt{2}$.Thus,$I_{A} = \frac{2}{3}Ma^{2} + M(a\sqrt{2})^{2} = \frac{2}{3}Ma^{2} + 2Ma^{2} = \frac{8}{3}Ma^{2}$.The initial rotational kinetic energy is $K_{i} = \frac{1}{2}I_{A}\omega_{c}^{2} = \frac{1}{2}(\frac{8}{3}Ma^{2})\omega_{c}^{2} = \frac{4}{3}Ma^{2}\omega_{c}^{2}$.For the cube to just topple,its center of mass must reach the highest point above the axis $AB$,which is at a height $h' = a\sqrt{2}$ from the surface.The potential energy at this position is $U_{f} = Mgh' = Mga\sqrt{2}$.The initial potential energy with the center of mass at height $a$ is $U_{i} = Mga$.By the conservation of energy,$U_{i} + K_{i} = U_{f} + K_{f}$.For the critical case,$K_{f} = 0$,so $Mga + \frac{4}{3}Ma^{2}\omega_{c}^{2} = Mga\sqrt{2}$.$\frac{4}{3}Ma^{2}\omega_{c}^{2} = Mga(\sqrt{2} - 1)$.$\omega_{c}^{2} = \frac{3g(\sqrt{2} - 1)}{4a}$.$\omega_{c} = \sqrt{\frac{3g(\sqrt{2} - 1)}{4a}}$.

Similar Questions

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