A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be _________.

Solution

Physics sir · 4 steps

The pivot pushes back during the hit, so conserve angular momentum about the pivot, not linear momentum.

1. Pick the right law. During the collision the hinge exerts a sudden force on the rod, so linear momentum is not conserved. That force acts at the pivot and has zero torque about it, and gravity is too weak to act in that instant. So angular momentum about the pivot is conserved.

   m v L = ((M L²)/3 + m L²)ω

2. Before the hit. The particle moves horizontally at a perpendicular distance L = 1 m below the pivot. So mvL = 0.1 × 80 × 1 = 8 kg m² s⁻¹.
3. After the hit. The rod about its end has ML²/3 = (0.9 × 1²)/3 = 0.3 kg m². The stuck particle adds mL² = 0.1 kg m², so the total is 0.4 kg m².
4. So. ω = 8/0.4 = 20 rad s⁻¹, so the answer is 20.

The common slip: 26.67. Leaving out the stuck particle's mL² uses I = 0.3 kg m² and gives about 26.67, not 20.

The idea to keep. Whenever a hinged body is struck, the hinge force spoils linear momentum. Take angular momentum about the hinge, and remember to add the stuck particle's mr² to the body's moment of inertia.

Answer from the NTA final answer key. Solution by PrepWiser.