Question of Physics

Question

A particle of mass m coming with velocity v₀ collides with a uniform rod of mass M and length L elastically on a horizontal table. (The rod is not pivoted and it is free to rotate about its center of mass after the collision.). The moment of inertia of the rod about its center of mass is Icm = 1/12ML².

If the particle stops after the collision, nd

a) the velocity of the center of mass of the rod, and

b) the angular velocity of the rod.

Answer 1

A particle of mass m coming with velocity v₀ collides with a uniform rod of mass M and length L elastically on a horizontal table. (The rod is not pivoted and it is free to rotate about its center of mass after the collision.). The moment of inertia of the rod about its center of mass is Icm = 1/12ML².

If the particle stops after the collision, nd

a) the velocity of the center of mass of the rod, and

b) the angular velocity of the rod.

Angular momentum of particle before collision about center of mass of rod = Mass*velocity * distance of closet approach =m*v₀*L/2 ( reference point is Center of Mass of rod)

This angular moment momentum gets transferred to the rod = I * Angular Velocity of rod

(with reference Center of Mass)

= (1/12 ) ML² * w²

here w= Angular velocity of rod end about center of mass , the angular velocity of center of mass will be zero