Question

A thing rod of mass M = 6 kg and length L = 1/4 m is hanging vertically from the pivot P as shown in figure. A small block of mass m = 2 kg travelling horizontally with speed $V^o$ strikes at the bottom of the rod and sticks to it. Find the initial velocity (in m/s) with which block should move so that the rod becomes momentarily horizontal.

Answer 1

$Conservingangularmomentumaboutthepivotm{V}_{0}L=(\frac{M{L}^2}{3}+m{L}^2)\omega \Rightarrow \omega =\frac{m{V}_0}{L(\frac{M}{3}+m)}\frac{1}{2}(\frac{M{L}^2}{3}+m{L}^2){\omega }^2=\frac{MgL}{2}+mgL\Rightarrow \omega =\sqrt{\frac{(\frac{Mg}{2}+mg)}{\frac{1}{2}(\frac{ML}{3}+mL)}}=\frac{m{V}_0}{L(\frac{M}{3}+m)}\Rightarrow V_0=\sqrt{10gl}$