Ball striking a pivoted rod

Given: rod $M = 3$ kg, $L = 1.2$ m, pivot O (smooth); ball $m = 0.5$ kg, $u = 10$ m/s, strikes at the free end (distance $L$ from O); $e = 0.6$.

     O (pin)
     |
     |  rod M
     |
     |
  -->o  ball m, u = 10 m/s     (impact at the lower end)

Moment of inertia of the rod about O

$$I_O = \frac{ML^2}{3} = \frac{3(1.2)^2}{3} = 1.44\ \text{kg·m}^2$$

Equations

Let $v$ be the velocity of the ball after impact (positive in the original direction) and $\omega$ the angular velocity of the rod after impact. The pin reaction is impulsive, but it passes through O, so angular momentum about O is conserved for the system of ball plus rod:

$$m u L = m v L + I_O\omega$$ $$0.5(10)(1.2) = 0.5\,v\,(1.2) + 1.44\,\omega \;\Rightarrow\; 6 = 0.6v + 1.44\omega \quad (1)$$

Restitution (relative speed of separation at the contact point, rod tip speed $= \omega L$):

$$\omega L - v = e\,u = 0.6(10) = 6 \;\Rightarrow\; v = 1.2\omega - 6 \quad (2)$$

Substituting (2) in (1): $6 = 0.6(1.2\omega - 6) + 1.44\omega = 0.72\omega - 3.6 + 1.44\omega$

$$9.6 = 2.16\,\omega \;\Rightarrow\; \omega = 4.444\ \text{rad/s}$$ $$v = 1.2(4.444) - 6 = -0.667\ \text{m/s}$$

The negative sign means the ball rebounds (moves back) at 0.667 m/s. The rod tip moves at $\omega L = 5.333$ m/s.

Kinetic energy lost

$$T_1 = \tfrac12(0.5)(10)^2 = 25.00\ \text{J} \qquad T_2 = \tfrac12(0.5)(0.667)^2 + \tfrac12(1.44)(4.444)^2 = 14.33\ \text{J}$$ $$\Delta T = 10.67\ \text{J}$$

Answer: ω = 4.44 rad/s; ball rebounds at 0.67 m/s; energy lost = 10.67 J.