A uniform thin rod of length 0.55 m and mass 5.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet velocity makes an angle of 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 11.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

To find the magnitude of the bullet's velocity just before impact, we can use the principle of conservation of angular momentum.

The angular momentum of an object rotating about a fixed axis is given by the equation:

L = Iω

Where:
L is the angular momentum,
I is the moment of inertia of the object, and
ω is the angular velocity.

In this case, the bullet is fired into one end of the rod, so the moment of inertia of the system (rod + bullet) can be approximated as the moment of inertia of the rod rotating about its center of mass.

The moment of inertia of a thin rod rotating about its center of mass is given by the equation:

I = (1/12)mL^2

Where:
m is the mass of the rod, and
L is the length of the rod.

Given that the mass of the rod is 5.5 kg and the length is 0.55 m, we can calculate the moment of inertia:

I = (1/12)(5.5 kg)(0.55 m)^2
I = 0.1133 kg·m^2

Before the bullet hits the rod, the rod is at rest, so its initial angular velocity is 0 rad/s.

Now, we can use the conservation of angular momentum to find the initial angular momentum:

L_initial = L_final

Since the bullet is traveling in the same horizontal plane as the rod, the angle between the bullet's velocity and the rod is 60°. We can use this angle to determine the y-component of the initial angular momentum:

L_initial_y = (0.003 kg)(v_initial)sin(60°)

Since the rod is initially at rest, its angular momentum is 0, and we have:

L_initial_y = L_final_y

(0.003 kg)(v_initial)sin(60°) = (0.1133 kg·m^2)(11.0 rad/s)

Simplifying and solving for v_initial, we get:

v_initial = [(0.1133 kg·m^2)(11.0 rad/s)] / [(0.003 kg)sin(60°)]
v_initial ≈ 70.192 m/s

Therefore, the magnitude of the bullet's velocity just before impact is approximately 70.192 m/s.

angular momentum:

l=rmvsinQ

The rotational inertia on the rod:
I=ML^2/12

The parallel-axis theorem:
I=Irod+mr^2

Therefore:
rmvsinQ=(ML^2/12+mr^2)w
v=(5.5)(0.55)^2+(0.003)(0.275)^2(11.0)/(0.275)(0.003)sin(60.0)
v=25618.6 m/s