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A cubical block of side a is moving with a velocity v on a smooth horizontal plane as shown. It hits a ridge at point O. The angular speed of the block when it hits O is

$\begin {array} {1 1} (a)\;\frac{3v}{4a} & \quad (b)\;\frac{3v}{2a} \\ (c)\;\frac{\sqrt 3 v}{\sqrt 2 a} & \quad (d)\;zero \end {array}$

When the block hits the ridges, the impulse of the impact force produces a torque about the centre of mass of the cube.
Angular momentum conservation can be applied about 0.
$Mv \times \large\frac{a}{2}$$= I _{about0} . w I for a about an axis through centre of mass= \large\frac{Ma^2}{b} OC= \large\frac{a}{\sqrt 2} I about an axis through o and \perp to the plane of paper =\large\frac{Ma^2}{6}$$+M\bigg(a / \sqrt 2\bigg)^2$
$\qquad=\large\frac{4Ma^2}{6}$
$\qquad=\large\frac{2}{3}$$Ma^2 \therefore mV\large\frac{a}{2}=\frac{2}{3}$$ Ma^2 w$
$\Rightarrow w = \large\frac{3v}{4a}$. Therefore (A) is the answer.
edited Jun 24, 2014