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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

The graph between angular momentum $L$ and angular velocity $\omega$ is:

$STATEMENT-1$ If there is no external torque on a body about its center of mass,then the velocity of the center of mass remains constant. because
$STATEMENT-2$ The linear momentum of an isolated system remains constant.

$A$ solid sphere and a disc of the same radius are released from the same height on an inclined plane and they reach the bottom of the plane at different times. This is due to their:

$A$ uniform rod $AB$ of mass $2 \ kg$ and length $30 \ cm$ is at rest on a smooth horizontal surface. An impulse of force $0.2 \ Ns$ is applied to end $B$. The time taken by the rod to turn through a right angle will be $\frac{\pi}{X} \ s$,where $X = \text{ . . . . . . }$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module