Solution:
Conserving angular momentum about pivot: \(L_i = mv\frac{\ell}{2}\). Total moment of inertia is \(I = \frac{1}{3}m\ell^2 + m(\ell/2)^2 = \frac{7}{12}m\ell^2\). Equating \(I\omega = L_i\) yields \(\omega = \frac{6v}{7\ell}\).
A uniform rod of mass \(m\) and length \(\ell\) is pivoted about one end and hung vertically. Another mass \(m\) hits it perpendicular to its length with a velocity \(v\) at its midpoint and sticks to it. The initial angular velocity of the rod is:
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