Question

A solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface as shown below

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 $2Ma^3/3)$

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

Answer 1

Answer: (D)

Bullet provides cube an angular impulse which causes it to topple

If $I_A$ = moment of inertia of cube about point $A$ and $\omega_{C}$ = initial angular speed about $A$.
Then, initial rotational kinetic energy of cube is
$K_{i}=\frac{1}{2}I_{A}\omega_{c}^{2}$
Moment of inertia of cube about point $A$, using parallel axes theorem is
$I_{A}=I_{0}+M\left(OA\right)^{2}$
$\left[\therefore I_{0}=\frac{2}{3}Ma^{2} andOA=\sqrt{2a}\right]$
$=\frac{2Ma^{2}}{3}+M\left(\sqrt{2a}\right)^{2}=\frac{8}{3}Ma^{2}$
Note In question paper $I_0$ is misprinted as$\frac{2}{3}Ma^{3}$
$\therefore $ Initial rotational kinetic energy
$K_{i}=\frac{1}{3}\left(\frac{8}{3}Ma^{2}\right)\omega_{c}^{2}=\frac{4}{3}Ma^{2}\omega_{c}^{2}$
This rotation energy must be greater than or equal to potential energy of cube when its centre is at highest level above ground during toppling.

As, maximum height of centre of mass is a$a\sqrt{2}$, potential energy of cube in this position is
$U_{f}=Mg\left(h'\right)=Mga\sqrt{2}$
Now, by conservation of energy, we have
$U_{i}+K_{i}=U_{f}+K_{f}$
For critical (minimum) angular speed $\omega_{c},K_{f}=0$
So,
$Mg\left(\frac{2a}{2}\right)+\frac{4}{3}Ma^{2}\omega_{c^{2}}=Mg a\sqrt{2+0}$
$\Rightarrow \omega_{c}=\sqrt{\frac{3g\left(\sqrt{2-1}\right)}{4a}}$