A thin, uniform, metal bar, 3 m long and weighing 90 N , is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.6 m below the ceiling by a small 4-kg ball, initially traveling horizontally at 12 m/s . The ball rebounds in the opposite direction with a speed of 5 m/s. Find the angular speed of the bar just after the collision? Why linear momentum not conserved?

 w = 1.53 rad / s

Explanation:

For this exercise we will use conservation of angular momentum. Let's start by defining our system as formed for the bar and the ball, for which all the torques formed are internal and consequently the angular momentum is conserved.

Initial moment. Just before crash

         L₀ = m v₁ L₁

final moment. Right after the crash

          = I_bar w + m v₂ L₁

where L₁ is the distance of the ball L₁ = 1,6m and m its mass 4 kg, let's take as the direction of the positive initial velocity

The moment of inertia of a bar that rotates at its end is

       I_bar = ⅓ M L²

      L₀ = L_{f}

      m v₁ L₁ = (⅓ M L²) w + m v₂ L₁

      w = m L₁ (v₁ - v₂)  3 / ML²

let's calculate

velocity v₁ = 1.6 m /s and its bounce v₂ = - 5 m / s

        M = W / g

        M = 90 / 9.8

        M = 9.19 kg

        w = 4 1.6 (1.6 + 5) 3 / (9.19 3 2)

        w = 1.53 rad / s

answer: strength i guess

explanation: