This problem involves a bullet passing through a pendulum bob, and we need to determine the minimum velocity $v$ of the bullet such that the pendulum bob barely swings through a complete vertical circle.

Problem Breakdown:

1. Bullet and Pendulum Interaction:

   • A bullet of mass $m_b = 0.00112 \, \text{kg}$ and initial speed $v$ passes through a pendulum bob of mass $m_p = 84.6 \, \text{kg}$, emerging with speed $v/2$.
   • The pendulum bob is suspended by a stiff rod of length $L = 1.08 \, \text{m}$.
   • The system experiences gravitational acceleration $g = 9.8 \, \text{m/s}^2$.

2. Condition for Barely Swinging Through a Complete Vertical Circle:

   • For the pendulum to barely complete a vertical circle, the bob must have enough energy at the highest point (when it's at the top of the circle).
   • The tension in the string should be zero at the top, meaning the kinetic energy at the top must be equal to the potential energy gained.

Steps to Solve:

1. Conservation of Linear Momentum (during collision):

The system can be analyzed using the conservation of momentum principle since no external horizontal forces act on the bullet-bob system during the collision.

$m_b v = m_b \frac{v}{2} + m_p v_p$

Where $v_p$ is the velocity of the pendulum bob just after the collision.

Solve for $v_p$: $m_b v - m_b \frac{v}{2} = m_p v_p$ $v_p = \frac{m_b v}{2 m_p}$

2. Energy Conservation (for vertical swing):

After the collision, the pendulum bob will swing upwards, converting kinetic energy into potential energy. The pendulum will barely complete a vertical circle when the velocity at the top of the swing is such that the tension in the rod is zero. The condition for this is that the centripetal force at the top equals the gravitational force.

The minimum velocity at the bottom of the swing is: $v_p^2 = 5gL$ This accounts for both the potential energy at the top and the kinetic energy required for completing the swing.

Thus, we now relate this to the initial speed $v_p$: $\left(\frac{m_b v}{2 m_p}\right)^2 = 5gL$

3. Solve for $v$:

$\frac{m_b^2 v^2}{4 m_p^2} = 5gL$ $v^2 = \frac{20 m_p^2 g L}{m_b^2}$ $v = \sqrt{\frac{20 m_p^2 g L}{m_b^2}}$

Now, plug in the values:

• $m_b = 0.00112 \, \text{kg}$
• $m_p = 84.6 \, \text{kg}$
• $g = 9.8 \, \text{m/s}^2$
• $L = 1.08 \, \text{m}$

$v = \sqrt{\frac{20 \times (84.6)^2 \times 9.8 \times 1.08}{(0.00112)^2}}$

Let me calculate this for you next.The minimum value of the bullet's velocity $v$ required for the pendulum bob to barely complete a vertical circle is approximately $1,098,986 \, \text{m/s}$.

Would you like any further details on this solution?

Related Questions:

1. How would the result change if the mass of the pendulum bob increased?
2. What if the length of the pendulum were doubled—how would that affect the minimum velocity?
3. How does the bullet's exit speed influence the final result?
4. Can we ignore air resistance in this scenario, and why?
5. What would happen if the bullet didn't pass completely through the bob but embedded itself instead?

Tip: When solving problems with pendulums, conservation laws (momentum and energy) are typically the key principles to use.