Consider a uniform rod of mass M=4m and lengt | vedclass

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(B) Let the angular velocity of the system after the collision be $\omega$.The component of the velocity of mass $m$ perpendicular to the rod is $v_{\

Vedclass · Accepted Answer

$\frac{3 \sqrt{2}}{7} \frac{v}{\ell}$

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(B) Let the angular velocity of the system after the collision be $\omega$.The component of the velocity of mass $m$ perpendicular to the rod is $v_{\perp} = v \sin(	heta) = v \sin(\frac{\pi}{4}) = \frac{v}{\sqrt{2}}$.By the principle of conservation of angular momentum about the pivot (centre of the rod):$L_{initial} = L_{final}$$m v_{\perp} r = I_{total} \omega$$m \left(\frac{v}{\sqrt{2}}ight) \left(\frac{\ell}{2}ight) = \left( I_{rod} + I_{mass} ight) \omega$The moment of inertia of the rod about its centre is $I_{rod} = \frac{M \ell^2}{12} = \frac{(4m) \ell^2}{12} = \frac{m \ell^2}{3}$.The moment of inertia of the mass $m$ at distance $\frac{\ell}{2}$ is $I_{mass} = m \left(\frac{\ell}{2}ight)^2 = \frac{m \ell^2}{4}$.Substituting these values:$\frac{m v \ell}{2 \sqrt{2}} = \left( \frac{m \ell^2}{3} + \frac{m \ell^2}{4} ight) \omega$$\frac{m v \ell}{2 \sqrt{2}} = \left( \frac{4m \ell^2 + 3m \ell^2}{12} ight) \omega$$\frac{m v \ell}{2 \sqrt{2}} = \frac{7m \ell^2}{12} \omega$Solving for $\omega$:$\omega = \left( \frac{m v \ell}{2 \sqrt{2}} ight) \left( \frac{12}{7 m \ell^2} ight) = \frac{6 v}{7 \sqrt{2} \ell} = \frac{6 \sqrt{2} v}{14 \ell} = \frac{3 \sqrt{2}}{7} \frac{v}{\ell}$.

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