A thin rod, 0.11 m long with a mass of 4.6 kg, can rotate in a horizontal plane about a vertical axis through its center. Initially at rest, the rod is struck by a 5.0 g bullet traveling in the same plane. Viewed from above, the bullet's path makes a 60° angle with the rod. If the bullet embeds itself in the rod and their combined angular velocity is 12.0 rad/s immediately after the collision, determine the bullet's speed just before impact. Here is the corrected and concise version: A thin 0.11 m rod with a mass of 4.6 kg can rotate around a vertical axis through its center in a horizontal plane. Initially at rest, the rod is hit by a 5.0 g bullet traveling in the plane at a 60° angle to the rod. Upon impact, the bullet becomes embedded in the rod, resulting in a combined angular velocity of 12.0 rad/s. Determine the bullet's speed just before the collision.

To determine the bullet's speed just before the collision, we can use the conservation of angular momentum. The angular momentum before the collision must be equal to the angular momentum after the collision since there are no external torques acting on the system.

First, let's define the system. The rod is free to rotate about its center and the bullet is approaching at an angle of 60°. When the bullet embeds itself into the rod, the bullet-rod system will rotate around the same axis with a combined moment of inertia.

The moment of inertia of the rod rotating about its center is given by the expression for a rod rotating about its center: I_rod = (1/12) * m_rod * L^2.

Let's calculate that:

I_rod = (1/12) * 4.6 kg * (0.11 m)^2 = 0.0046475 kg·m²

The moment of inertia of the bullet at the point of impact, which is at the edge of the rod, is I_bullet = m_bullet * r^2, where r is half the length of the rod since the bullet hits the rod at the end.

r = 0.11 m / 2 = 0.055 m

So, I_bullet = 0.005 kg * (0.055 m)^2 = 1.5125e-5 kg·m²

Now, consider that the angular momentum L before the collision (L_initial) is just due to the bullet, as the rod is initially at rest:

L_initial = I_bullet * ω_bullet

where ω_bullet is the angular speed of the bullet just before impact, which we obtain through the linear speed v_bullet and the radius r:

ω_bullet = v_bullet / r

Now, ω_bullet can be substituted in the expression for L_initial:

L_initial = m_bullet * r * v_bullet

After the collision, the angular momentum L_final is due to the combined rotation of the bullet and the rod:

L_final = (I_rod + I_bullet) * ω_final

Given that L_initial = L_final and ω_final = 12 rad/s, we can set up the conservation of angular momentum equation:

m_bullet * r * v_bullet = (I_rod + I_bullet) * ω_final

Now we can solve for v_bullet:

v_bullet = ((I_rod + I_bullet) * ω_final) / (m_bullet * r)

Puting in the numbers we have:

v_bullet = ((0.0046475 kg·m² + 1.5125e-5 kg·m²) * 12 rad/s) / (0.005 kg * 0.055 m) v_bullet = (0.00466275 kg·m² * 12 rad/s) / (0.000275 kg·m) v_bullet = 0.055953 kg·m²/s / 0.000275 kg·m v_bullet = 203.4654545 m/s.

So, the bullet's speed just before the collision was approximately 203.47 m/s.