A cubical block of side a is moving with velo | vedclass

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(A) When the block hits the ridge at $O$,it starts rotating about an axis passing through $O$ and perpendicular to the plane of motion. Since no exter

Vedclass · Accepted Answer

$3v/4a$

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(A) When the block hits the ridge at $O$,it starts rotating about an axis passing through $O$ and perpendicular to the plane of motion. Since no external torque acts on the block about point $O$ during the impact,the angular momentum is conserved.Initial angular momentum $L_i$ about point $O$ is given by the product of linear momentum and the perpendicular distance from $O$ to the line of motion of the center of mass:$L_i = Mv 	imes (a/2) = \frac{Mva}{2}$After hitting the ridge,the block rotates about point $O$. The final angular momentum is $L_f = I_O \omega$,where $I_O$ is the moment of inertia of the cube about the axis passing through $O$.The moment of inertia of a cube about an axis passing through its center of mass $C$ is $I_C = \frac{Ma^2}{6}$.Using the parallel axis theorem,$I_O = I_C + Mr^2$,where $r$ is the distance from the center of mass to point $O$. Here,$r^2 = (a/2)^2 + (a/2)^2 = a^2/2$.Thus,$I_O = \frac{Ma^2}{6} + M(a^2/2) = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$.Equating initial and final angular momentum: $L_i = L_f$$\frac{Mva}{2} = \frac{2}{3}Ma^2 \omega$Solving for $\omega$: $\omega = \frac{Mva}{2} 	imes \frac{3}{2Ma^2} = \frac{3v}{4a}$.

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

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