A long thin rod of mass M = 2.00 kg and length L = 75.0 cm is free to rotate about its

center as shown. Two identical masses (each of mass m = .421 kg) slide
without friction along the rod. The two masses begin at the rod's point of rotation when
the rod is rotating at 10.0 rad/s.

[weight]------(counterclockwise(mid))-------[weight]

(a) When they have moved halfway to the end of the rod, how fast (rad/s) is the rod
rotating?

(b) When the masses are halfway to the end of the rod, what is the ratio of the �final
kinetic energy to the initial kinetic energy (Kf =Ki)?

(c) When they reach the end, how fast is the rod rotating (rad/s)?

Disregard. I found a post with it

To solve this problem, we will use the principle of conservation of angular momentum. According to this principle, the total angular momentum remains constant unless acted upon by an external torque.

(a) To find the final angular velocity of the rod when the masses have moved halfway to the end, we can use the conservation of angular momentum. Initially, the system has an angular momentum given by:

L_initial = I_rod * ω_rod_initial + I_mass * ω_mass_initial + I_mass * ω_mass_initial

where I_rod is the moment of inertia of the rod, ω_rod_initial is the initial angular velocity of the rod, I_mass is the moment of inertia of each mass (assuming they are point masses), and ω_mass_initial is the initial angular velocity of each mass.

At this point, the two masses have moved halfway to the end of the rod. Assuming the rod rotates without any external torque and the masses do not leave the rod, the moment of inertia of the masses does not change. However, the moment of inertia of the rod changes since the masses have moved away from the center of rotation.

The moment of inertia of a rod rotating about its center is given by:

I_rod = (1/12) * M * L^2

where M is the mass of the rod and L is its length.

The moment of inertia of each mass about the center of the rod can be approximated as:

I_mass = m * (L/2)^2

where m is the mass of each mass and L is the length of the rod.

We are given M = 2.00 kg, L = 75.0 cm = 0.75 m, m = 0.421 kg, and ω_rod_initial = 10.0 rad/s. Plugging in these values, we can calculate the initial angular momentum.

L_initial = (1/12) * M * L^2 * ω_rod_initial + 2 * (0.421 kg) * (0.375 m)^2 * 10.0 rad/s + 2 * (0.421 kg) * (0.375 m)^2 * 10.0 rad/s

(b) To find the ratio of the final kinetic energy (Kf) to the initial kinetic energy (Ki) when the masses are halfway to the end of the rod, we can use the conservation of angular momentum. Since kinetic energy is related to angular velocity by the equation:

K = (1/2) * I * ω^2

where K is the kinetic energy, I is the moment of inertia, and ω is the angular velocity, we can find the kinetic energy at each point and calculate the ratio.

The initial kinetic energy can be calculated using the initial angular velocity of the rod and masses:

Ki = (1/2) * I_rod * ω_rod_initial^2 + (1/2) * I_mass * ω_mass_initial^2 + (1/2) * I_mass * ω_mass_initial^2

Similarly, the final kinetic energy can be calculated using the final angular velocity of the rod and masses:

Kf = (1/2) * I_rod * ω_rod_final^2 + (1/2) * I_mass * ω_mass_final^2 + (1/2) * I_mass * ω_mass_final^2

(c) To find the final angular velocity of the rod when the masses reach the end, we can again use the conservation of angular momentum. Since no external torques act on the system, the total angular momentum remains constant.

Using the same approach as in part (a), we can calculate the final angular velocity of the rod when the masses reach the end. The initial angular momentum equals the final angular momentum.

L_initial = I_rod * ω_rod_initial + I_mass * ω_mass_initial + I_mass * ω_mass_initial

L_final = I_rod * ω_rod_final + I_mass * ω_mass_final + I_mass * ω_mass_final

By setting L_initial equal to L_final, we can solve for ω_rod_final.