JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Rotational Mechanics

Released from rest, the rod turns about its pivot and energy conservation balances the fall of its centre of mass against its rotational kinetic energy, using $I=\frac{m{\ell }^2}{3}$. The centre of mass begins a height $\frac{\ell }{2}\sin 30^{\circ }$ above the table, so $mg\frac{\ell }{2}\sin 30^{\circ }=\frac{1}{2}\cdot \frac{m{\ell }^2}{3}{\omega }^2$. With $\ell =1$ m and $g=10$, this reduces to ${\omega }^2=\frac{3g}{2\ell }=15$, so the angular speed at impact is $\sqrt{15}{\mathrm{s}}^{-1}$ and $n=15$.