Bullet embedded in bar, h = 240 mm

Angular momentum of the bar and bullet about the pivot $C$ is conserved during the impact (the reaction at $C$ has no moment about $C$; the weights are non-impulsive). Take the bullet mass $m_b = 0.040$ kg, $v_0 = 600$ m/s, bar mass $M = 7$ kg, $L = 0.600$ m, $h = 0.240$ m (distance of the pivot $C$ below $A$); the bullet enters at the lower end $B$.

     A  o
        |   G = centre of bar
   h    |
     C  o-- pivot
        |
     B  o<--- bullet v0

Distances from C

• Centre of bar $G$ from $C$: $d_G = L/2 - h = 0.300 - 0.240 = 0.060$ m (below $C$)
• Bullet point $B$ from $C$: $d_B = L - h = 0.600 - 0.240 = 0.360$ m

Moment of inertia of the bar about C

$$\bar I = \frac{1}{12}ML^2 = \frac{1}{12}(7)(0.6)^2 = 0.210\ \text{kg m}^2$$ $$I_C = \bar I + Md_G^2 = 0.210 + 7(0.060)^2 = 0.2352\ \text{kg m}^2$$

(a) Angular velocity just after impact

The bullet is embedded, so its speed after impact is $\omega d_B$. Conservation of angular momentum about $C$:

$$m_bv_0d_B = I_C\omega + m_b(\omega d_B)d_B$$ $$\omega = \frac{m_bv_0d_B}{I_C + m_bd_B^2} = \frac{0.04\times600\times0.360}{0.2352 + 0.04(0.360)^2} = 35.94\ \text{rad/s}$$

(b) Impulsive reaction at C

Take the bullet direction as positive. Linear impulse-momentum for the system (bar + bullet), the impulse of the reaction being $C\,\Delta t$:

$$m_bv_0 + C\,\Delta t = M(\omega d_G) + m_b(\omega d_B)$$ $$C\,\Delta t = 7(35.94)(0.060) + 0.04(35.94)(0.360) - 0.04(600) = -8.39\ \text{N s}$$ $$C = \frac{C\,\Delta t}{\Delta t} = \frac{-8.39}{0.001} = -8387\ \text{N}$$

The negative sign means the reaction at $C$ acts opposite to the bullet's direction.

Answer: (a) $\omega = 35.94$ rad/s; (b) $C \approx 8387$ N ($\approx 8.39$ kN) opposite to the bullet velocity.