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 an angular speed ω about the pivot. The maximum angular speed ω M is achieved for x = x M . Then

Step-by-step solution

The net torque on the system (rod+bullet) will be zero about hinge point. Therefore, we can apply the principle of angular momentum conservation about hinge point. From angular momentum conservation, m v x = m L 2 3 + m x 2 ω ⇒ ω = 3 v x L 2 + 3 x 2 = 3 v L 2 x + 3 x For maximum angular velocity, d ω d x = 0 ⇒ d d x L 2 x + 3 x = 0   ⇒ - L 2 x 2 + 3 = 0   ⇒ x = L 3 ⇒ ω max = 3 v 3 L 2 L + 3 L = 3 v 2 L