A cube of side 'a' moves with a velocity 'v' | vedclass

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(A) To find the angular speed of the block after it hits point '$O$',we use the principle of conservation of angular momentum about the point '$O$'.Th

Vedclass · Accepted Answer

$\frac{3v}{4a}$

Vedclass · Answer

(A) To find the angular speed of the block after it hits point '$O$',we use the principle of conservation of angular momentum about the point '$O$'.The initial angular momentum of the block about point '$O$' is given by the product of its linear momentum and the perpendicular distance from the center of mass to the point '$O$'.$L_i = Mv 	imes \left(\frac{a}{2}ight)$The moment of inertia of the cube about the edge passing through '$O$' is calculated using the parallel axis theorem:$I_O = I_{cm} + Mr^2$Where $I_{cm} = \frac{Ma^2}{6}$ for a cube about its center,and $r$ is the distance from the center of mass to the edge '$O$',which is $\sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a}{\sqrt{2}}$.$I_O = \frac{Ma^2}{6} + M\left(\frac{a}{\sqrt{2}}ight)^2 = \frac{Ma^2}{6} + \frac{Ma^2}{2} = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$By conservation of angular momentum:$L_i = L_f$$Mv\left(\frac{a}{2}ight) = I_O \omega$$\frac{Mva}{2} = \left(\frac{2}{3}Ma^2ight)\omega$$\omega = \frac{Mva}{2} 	imes \frac{3}{2Ma^2} = \frac{3v}{4a}$

$A$ cube of side '$a$' moves with a velocity '$v$' on a horizontal surface as shown in the figure. It hits an edge at point '$O$'. What will be the angular speed of the block after it hits point '$O$'?

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