Question

A thin uniform rod AB of mass 2m and length L lies at rest horizontal on a smooth horizontal surface. A particle of mass m moving horizontally with velocity V0 strikes the rod perpendicularly at end A (see figure) and sticks to it. After the collision the angular velocity of the rod and particle system is ω. Then​

Answer 1

Answer:

Before sticking to the rod, the angular momentum of the particle is mvL and after sticking to the rod, the system will be rod + particle and hence angular momentum will be Iω, I is the moment of inertia of the system = (mL^2+mL^2/3)=4mL 2/3. Thus the angular momentum of the system will be = 4mL^2 \omega/3.

Applying conservation of angular momentum to determine the value of ω, we get 4mL^2 \omega/3 = mvL, solving we get ω=3v/4L