Question

A solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface as shown in the figure. The cube is free to rotate about the fixed axis $AB$. A bullet of mass $m(<<M)$ and speed $v$ is shot horizontally at the face opposite to $ABCD$ at a height $h$ above the surface to impart the cube an angular speed ${\omega }_c$ so that the cube just topples over. Then ${\omega }_c$ is (note: the moment of inertia of the cube about an axis perpendicular to the face and passing through the centre of mass is $2M{a}^3/3)$.

• A. $\sqrt{3gM/2ma}$
• B. $\sqrt{3g/4h}$
• C. $\sqrt{3g(\sqrt{2}-1)/2a}$
• D. $\sqrt{3g(\sqrt{2}-1)/4a}$

Answer 1

The correct option is D $\sqrt{3g(\sqrt{2}-1)/4a}$

Using parallel axis theorem, moment of inertia of cube about $AB$ is:

$I=\frac{2M{a}^2}{3}+2M{a}^2=\frac{8}{3}M{a}^2$

The cube just topples over if its COM can pass over AB.

Using conservation of energy,

$\frac{1}{2}I{\omega }_o^2=Mg(h_f-h_i)$

$\frac{4}{3}M{a}^2{\omega }_o^2=Mga(\sqrt{2}-1)$

${\omega }_o=\sqrt{\frac{3g(\sqrt{2}-1)}{4a}}$