This is a problem involving the conservation of momentum and energy in a system where a bullet passes through a pendulum bob and causes it to swing in a vertical circle. To solve for the minimum velocity $v$, we need to approach it systematically using physics principles.

Step 1: Conservation of Momentum

First, let's assume the collision between the bullet and the pendulum bob is perfectly inelastic (though the bullet does not embed in the bob, it does cause a momentum transfer). The initial momentum of the system is the momentum of the bullet alone:

$p_{\text{initial}} = m_b v$

where:

• $m_b = 0.00112 \, \text{kg}$ is the mass of the bullet,
• $v$ is the initial speed of the bullet.

After the bullet passes through the pendulum, the momentum is divided between the bullet and the pendulum bob. The bullet emerges with a velocity $\frac{v}{2}$, and the pendulum bob gains velocity $V_b$. The total momentum after the collision is:

$p_{\text{final}} = m_b \frac{v}{2} + M_b V_b$

where:

• $M_b = 84.6 \, \text{kg}$ is the mass of the pendulum bob,
• $V_b$ is the velocity of the pendulum bob after the collision.

Using conservation of momentum:

$m_b v = m_b \frac{v}{2} + M_b V_b$

Solving for $V_b$:

$V_b = \frac{m_b v - m_b \frac{v}{2}}{M_b} = \frac{m_b v}{2 M_b}$

Substitute the values of $m_b$ and $M_b$:

$V_b = \frac{0.00112 \cdot v}{2 \cdot 84.6} = \frac{0.00112 v}{169.2}$

$V_b = 6.62 \times 10^{-6} v$

Step 2: Energy Considerations for Vertical Circle

For the pendulum to barely complete a vertical circle, its velocity at the bottom must be enough to make the pendulum bob reach the top of the circle. The condition for this is that the bob's velocity at the bottom should allow it to have enough kinetic energy to be converted to potential energy at the top of the swing.

At the top of the swing, the pendulum bob must have enough energy to counteract the gravitational potential energy. The minimum velocity $V_b$ needed for this is derived from the following energy equation:

$\frac{1}{2} M_b V_b^2 = M_b g (2 L)$

where:

• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $L = 1.08 \, \text{m}$ is the length of the pendulum.

Simplifying:

$V_b^2 = 2 g (2 L)$

$V_b^2 = 4 g L$

Substitute $g = 9.8 \, \text{m/s}^2$ and $L = 1.08 \, \text{m}$:

$V_b^2 = 4 \cdot 9.8 \cdot 1.08 = 42.336$

$V_b = \sqrt{42.336} = 6.506 \, \text{m/s}$

Step 3: Solving for $v$

Now, equate the expression for $V_b$ from the momentum conservation step with the minimum velocity needed for the pendulum to complete the circle:

$6.62 \times 10^{-6} v = 6.506$

Solve for $v$:

$v = \frac{6.506}{6.62 \times 10^{-6}} = 9.83 \times 10^5 \, \text{m/s}$

$v \approx 983 \, \text{m/s}$

Thus, the minimum value of $v$ such that the pendulum bob will barely swing through a complete vertical circle is approximately:

$\boxed{983 \, \text{m/s}}$

Would you like further details on any step?

Here are five related questions to expand the topic:

1. What is the significance of the pendulum length in determining the circular motion?
2. How would the result change if the bullet embedded in the pendulum bob?
3. What is the role of gravity in the pendulum's circular motion?
4. How does the mass ratio between the bullet and the pendulum affect the system's momentum?
5. Could energy be lost in real-world scenarios like this, and how would it affect the outcome?

Tip: In circular motion problems involving pendulums, ensure that both momentum and energy principles are used, especially when forces like gravity come into play at different points in the motion.