Q: A rod of mass m and length L , pivoted at one of its ends is hanging vertically . A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it . The combined system now rotates with angular speed ω about the pivot . The maximum angular speed ω_{M} is achieved for x = x_{M} . Then

(a) $\omega = \frac{3 v x}{L^2 + 3 x^2}$

(b) $\omega = \frac{12 v x}{L^2 + 12 x^2}$

(c) $x_M = \frac{L}{\sqrt{3}}$

(c) $\omega_M = \frac{v}{2L} \sqrt{3}$

**Ans: (a,c,d)**

**Solution:** According to conservation of angular momentum about point of suspension ,

$\displaystyle m v x = (\frac{m l^2}{3} + m x^2 )\omega$

$\displaystyle v x = (\frac{ l^2}{3} + x^2 )\omega$

$\displaystyle \omega = \frac{3 v x}{L^2 + 3 x^2} $

For maximum ω ,

$\displaystyle \frac{d\omega}{dx} = 0 $

$\displaystyle x = \frac{L}{\sqrt{3}} $

$\displaystyle \omega = \frac{\sqrt{3} v}{2L} $