A long thin rod of mass M=2.00 kg and length L=75.0 cm is free to rotate about its center. Two identical masses (each of mass m = .45kg) 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.

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)?

The moment of inertia of the rod rotating about its center is M*L^2/12

When the idential masses m are at radius R, the moment of intertia due to their masses is 2*m*R^2

The total angular momentum remains unchanged. Let w1 be the initial angular velocity.

A) When the masses move half way to the ends of the rod, their radius is R = L/4. The angular velocity becomes a new value w2. You can solve for it using the momentum conservation equation.

(ML^2/12)*w1 = [(ML^2/12)+2*m*(L/4)^2
]*w2

B) Use the formula for kinetic energy, (1/2)I*w^2, and compare the before and after values.
C) Repeat A), but with R = L/2 instead of L/4.

Thank you so much!

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I'd like to possibly share insights about the material and possibly help each other out in a mutually beneficial way. If you are interested, shoot me an email at William dot Cordoba at yaho dot com

Thanks
William

A) To determine the angular velocity of the rod when the masses have moved halfway to the end, we can apply the principle of conservation of angular momentum.

The equation for angular momentum is given by:
L = Iω

Where:
L is the angular momentum
I is the moment of inertia
ω is the angular velocity

In this case, the rod and the two masses are the only objects involved, and there are no external torques acting on them. Therefore, the total angular momentum is conserved.

Initially, the angular momentum of the system is given by:
Linitial = Iinitial * ωinitial

When the masses are halfway to the end, the moment of inertia has changed due to the redistribution of masses. The moment of inertia of the system is given by:
Ifinal = Irod + 2*Imass

The moment of inertia of a rod rotating about its center is given by:
Irod = (1/12) * ML^2

The moment of inertia of a point mass rotating about a point is given by:
Imass = m * r^2

Where:
M is the mass of the rod
L is the length of the rod
m is the mass of each sliding mass
r is the distance of each sliding mass from the center of rotation

For the given values:
M = 2.00 kg
L = 75.0 cm = 0.75 m
m = 0.45 kg

Since the masses are initially at the center, the distance from the center to each sliding mass is L/2 = 0.75/2 = 0.375 m.

The final angular momentum of the system is given by:
Lfinal = Ifinal * ωfinal

Since angular momentum is conserved, we can set the initial and final angular momenta equal to each other:

Linitial = Lfinal
Iinitial * ωinitial = Ifinal * ωfinal

Substituting the expressions for the moment of inertia values:
(Irod * ωinitial) = (Irod + 2*Imass) * ωfinal

Plugging in the values:
[(1/12) * (2.00 kg) * (0.75 m)^2 * 10.0 rad/s] = [(1/12) * (2.00 kg) * (0.75 m)^2 + 2 * (0.45 kg) * (0.375 m)^2] * ωfinal

Simplifying and solving for ωfinal:
ωfinal = [(1/12) * (2.00 kg) * (0.75 m)^2 * 10.0 rad/s] / [(1/12) * (2.00 kg) * (0.75 m)^2 + 2 * (0.45 kg) * (0.375 m)^2]

Calculating this expression will give you the answer to part A, the angular velocity of the rod when the masses have moved halfway to the end.

B) To find the ratio of the final kinetic energy to the initial kinetic energy (Kf/Ki), we use the conservation of angular momentum to relate the angular velocity and kinetic energy.

The equation for rotational kinetic energy is given by:
K = (1/2) * I * ω^2

Since the angular momentum is conserved, we can equate the initial and final kinetic energies:

Ki = Kf
(1/2) * Iinitial * ωinitial^2 = (1/2) * Ifinal * ωfinal^2

Using the expressions for moment of inertia and plugging in the values:
(Irod * ωinitial^2) = (Irod + 2*Imass) * ωfinal^2

We can solve for the ratio (Kf/Ki) by dividing the final kinetic energy equation by the initial kinetic energy equation:
(Kf/Ki) = [(1/2) * (Irod + 2*Imass) * ωfinal^2] / [(1/2) * Iinitial * ωinitial^2]

Simplifying the expression and substituting the given values, you can calculate the ratio of the final kinetic energy to the initial kinetic energy.

C) To find the angular velocity of the rod when the masses reach the end, we follow a similar approach to part A.

The final position of the masses at the end of the rod changes the moment of inertia of the system. The new moment of inertia is given by:
Iend = Irod + 2 * m * L^2

Using the same formulas for the moment of inertia of the rod and point masses as in part A, you can calculate Iend.

Then, using the conservation of angular momentum again, you can relate the angular velocities of the initial and final states:

(Irod * ωinitial) = (Iend * ωfinal)

Solving for ωfinal, you can substitute the calculated values to determine the angular velocity of the rod when the masses reach the end.

By following these equations and calculations, you can find the answers to parts A, B, and C of the problem.