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 = .403 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.
(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)?
Use conservation of angular momentum.
I*w = constant
I is the moment of inertia, which is
(1/12)Mrod*L^2 + 2m*R^2
R is the distance of the masses from the center of the rod.
w = 10.0 when R = 0
For (a) and (b), R = L/4
For (c), R = L/2
Let's do (a)
Angular momentum with massesw m at R=0:
= (1/12)*2.00*(0.75)^2*10
= 9.38*10^-1 kg m^2/s
(This remains constant).
When R = L/4 = 0.1875 m,
I*w = (2/12)*(0.75)^2*w + 2*(0.403)*(0.1875)^2*w
= (9.38*10^-2 + 2.83*10^-2)w
= 1.221*10^-1*w = 9.38*10^-1
w = 7.68 m/s
For (b), compare initial and final values of (1/2) I w^2
For (c), repeat the process of (a), but use R = L/2
Why did the rod start rotating? Because someone told it to "get a move on"! Now let's calculate some physics, shall we?
(a) When the masses have moved halfway to the end of the rod, we can use the principle of conservation of angular momentum. Since no external torques are acting on the system, the initial angular momentum is equal to the final angular momentum.
The initial angular momentum is given by:
L_initial = I_rod * ω_rod_initial + 2 * I_mass * ω_mass_initial,
where I_rod is the moment of inertia of the rod, I_mass is the moment of inertia of each mass, ω_rod_initial is the initial angular velocity of the rod, and ω_mass_initial is the initial angular velocity of each mass.
The final angular momentum is given by:
L_final = I_rod * ω_rod_final + 2 * I_mass * ω_mass_final,
where ω_rod_final is the final angular velocity of the rod and ω_mass_final is the final angular velocity of each mass.
Since the masses are halfway to the end, we can assume that the moments of inertia of the masses don't change. Therefore, we can ignore them for now.
Since the rod is rotating about its center, the moment of inertia of the rod is given by:
I_rod = (1/12) * M * L^2,
where M is the mass of the rod and L is its length.
Now, let's consider the conservation of angular momentum equation:
L_initial = L_final.
Since the masses start from rest, ω_mass_initial = 0, and the equation becomes:
I_rod * ω_rod_initial = I_rod * ω_rod_final.
Plugging in the values, we get:
(1/12) * M * L^2 * ω_rod_initial = (1/12) * M * L^2 * ω_rod_final.
Canceling out the common terms, we find:
ω_rod_initial = ω_rod_final.
Therefore, the rod will continue to rotate at the same speed as the masses halfway to the end. So, the rod's angular velocity will still be 10.0 rad/s.
(b) The kinetic energy of the system can be calculated using the equation:
K = (1/2) * I_rod * ω_rod^2 + 2 * (1/2) * I_mass * ω_mass^2,
where K is the kinetic energy, I_rod is the moment of inertia of the rod, I_mass is the moment of inertia of each mass, ω_rod is the angular velocity of the rod, and ω_mass is the angular velocity of each mass.
Initially, the kinetic energy is:
K_initial = (1/2) * I_rod * ω_rod_initial^2 + 2 * (1/2) * I_mass * ω_mass_initial^2.
Since the masses start from rest, ω_mass_initial = 0, and the equation simplifies to:
K_initial = (1/2) * I_rod * ω_rod_initial^2.
Finally, let's solve for the ratio of the final kinetic energy to the initial kinetic energy (K_f / K_i):
K_final / K_initial = [(1/2) * I_rod * ω_rod_final^2 + 2 * (1/2) * I_mass * ω_mass_final^2] / [(1/2) * I_rod * ω_rod_initial^2].
Since ω_rod_initial = ω_rod_final, the ratio simplifies to:
K_final / K_initial = [(1/2) * I_rod * ω_rod_final^2 + 2 * (1/2) * I_mass * ω_mass_final^2] / [(1/2) * I_rod * ω_rod_final^2].
Notice that the ω_mass terms cancel out. So, the ratio is independent of the masses' angular velocity and only depends on the angular velocity of the rod. We can substitute the expressions for the moments of inertia:
K_final / K_initial = [(1/2) * (1/12) * M * L^2 * ω_rod_final^2 + 2 * (1/2) * I_mass * ω_mass_final^2] / [(1/2) * (1/12) * M * L^2 * ω_rod_final^2].
Now, the L^2 terms cancel out, and we are left with:
K_final / K_initial = [(1/12) * M * ω_rod_final^2 + 2 * (1/2) * I_mass * ω_mass_final^2] / [(1/12) * M * ω_rod_final^2].
Simplifying further, we find:
K_final / K_initial = [1/12 + 2 * (1/2) * (I_mass / I_rod)].
Plugging in the values, we can calculate the final ratio.
(c) When the masses reach the end of the rod, they will have transferred all their angular momentum to the rod. So, the final angular velocity of the rod can be found using the equation:
I_initial * ω_initial = I_final * ω_final,
where I_initial is the initial moment of inertia of the rod, ω_initial is the initial angular velocity of the rod, I_final is the final moment of inertia of the rod, and ω_final is the final angular velocity of the rod.
The initial moment of inertia of the rod is given by:
I_initial = (1/12) * M * L^2.
The final moment of inertia of the rod can be calculated by considering the masses as point masses at the ends of the rod:
I_final = (1/12) * M * L^2 + 2 * m * (L/2)^2 + 2 * m * (L/2)^2.
Here, the first term represents the moment of inertia of the rod itself, and the other two terms represent the added moments of inertia due to the masses.
Now, we can solve for the final angular velocity of the rod:
(1/12) * M * L^2 * ω_initial = [(1/12) * M * L^2 + 2 * m * (L/2)^2 + 2 * m * (L/2)^2] * ω_final.
Canceling out the common terms and solving for ω_final, we get:
ω_final = (1/12) * M * L^2 * ω_initial / [(1/12) * M * L^2 + 2 * m * (L/2)^2 + 2 * m * (L/2)^2].
Plugging in the given values, we can find the final angular velocity of the rod.
To solve this problem, we can use the principle of conservation of angular momentum and the principle of conservation of energy. Let's go step-by-step to answer each part of the question:
(a) To find out how fast the rod is rotating when the masses have moved halfway to the end of the rod, we need to analyze the conservation of angular momentum.
1. We know that the total initial angular momentum of the system is zero since the rod is rotating about its center and the masses are at the center initially.
Initial angular momentum (Li) = 0
2. When the masses move halfway to the ends, the system redistributes its mass such that the center of mass remains at the center of the rod.
3. Since there is no external torque acting on the system, the total angular momentum must be conserved.
Li = Lf
4. The angular momentum of the rod is given by L = Iω, where I is the moment of inertia of the rod and ω is the angular velocity of the rod.
5. The moment of inertia of a thin rod rotating about its center is given by I = (1/12)ML², where M is the mass of the rod and L is the length of the rod.
6. Let's calculate the initial and final angular momenta:
Initial angular momentum (Li) = 0
Final angular momentum (Lf) = I₁ω₁ + I₂ω₂, where I₁ and I₂ are the moments of inertia of the two halves of the rod, and ω₁ and ω₂ are the angular velocities of the two halves of the rod.
7. At the halfway point, the masses have moved halfway (L/2 = 37.5 cm) from the center to the ends of the rod. So, the two halves of the rod will have equal moments of inertia.
8. Now, let's calculate the final angular velocity (wf) when the masses have moved halfway.
Li = Lf
0 = 0 + 1/12ML² * ω₁ + 1/12ML² * ω₂
ω₁ + ω₂ = 0
Since the masses have moved halfway, the ratio of their velocities is directly proportional to the ratio of their distances from the center of mass. Since the ratio of the distances is 1:1, the ratio of their velocities is 1:1. This means ω₁ = ω₂. Substituting this in the equation ω₁ + ω₂ = 0, we get:
ω₁ + ω₁ = 0
2ω₁ = 0
ω₁ = 0
Therefore, the final angular velocity of the rod when the masses have moved halfway is 0 rad/s.
(b) To find the ratio of the final kinetic energy to the initial kinetic energy, we can use the principle of conservation of mechanical energy.
1. The total initial kinetic energy (Ki) is the sum of the kinetic energies of the rod and the masses.
Ki = 1/2Iω² + 1/2m(v₁)² + 1/2m(v₂)², where v₁ and v₂ are the velocities of the masses.
2. Since the masses are initially at rest, their initial velocities (v₁ and v₂) are zero.
3. Let's calculate the final kinetic energy (Kf) when the masses are halfway to the end of the rod.
Kf = 1/2Iwf² + 1/2m(vf₁)² + 1/2m(vf₂)², where wf is the final angular velocity of the rod and vf₁ and vf₂ are the final velocities of the masses.
Since the final angular velocity of the rod is 0 rad/s (as calculated in part a), the first term in the equation becomes zero.
Kf = 1/2m(vf₁)² + 1/2m(vf₂)²
Since the masses are halfway to the end of the rod, their final velocities are equal. So, vf₁ = vf₂ = vf.
Kf = 1/2m(vf)² + 1/2m(vf)²
= m(vf)²
Therefore, the ratio of the final kinetic energy to the initial kinetic energy is 1:1.
(c) To find out how fast the rod is rotating when the masses reach the end, we can again use the conservation of angular momentum.
1. The total initial angular momentum (Li) is zero, as mentioned in part a.
2. When the masses reach the end of the rod, the total angular momentum (Lf) must still be conserved.
3. Since the masses are at the end of the rod, their distances from the center of mass are equal to the length of the rod (L).
4. Let's calculate the final angular velocity (wf) when the masses reach the end using the conservation of angular momentum:
Li = Lf
0 = 0 + 1/12ML² * ω₁ + 1/12ML² * ω₂
ω₁ + ω₂ = 0
Since the masses are at the ends, their distances from the center of mass are greater than zero. As a result, their velocities must be greater than zero. This means ω₁ and ω₂ cannot both be zero.
Since we know that the sum of their velocities must be zero, one of them must be positive while the other is negative.
Let's assume ω₁ is positive (counterclockwise rotation) and ω₂ is negative (clockwise rotation).
ω₁ + ω₂ = 0
ω₁ - |ω₂| = 0 (since ω₂ is negative)
ω₁ = |ω₂|
ω₁ = |ω₂| implies that the magnitude of the positive angular velocity is equal to the magnitude of the negative angular velocity.
5. The angular velocity (ω) is the ratio of the tangential velocity (v) to the radius (r). Since the radius is the length of the rod (L/2), and the tangential velocities (v₁ and v₂) of the masses are equal, we can write:
v₁ / (L/2) = v₂ / (L/2)
v₁ = v₂
Therefore, the magnitudes of the velocities of the masses at the ends are equal.
6. Since the magnitudes of the velocities of the masses at the ends are equal, the magnitudes of their angular velocities must also be equal.
ω₁ = |ω₂| implies ω₁ = ω₂
Since the magnitudes of their angular velocities are equal and their signs are opposite, the magnitude of the angular velocity (wf) of the rod when the masses reach the end is equal to their individual magnitudes and in the same direction.
wf = |ω₁| = |ω₂|
Therefore, the angular velocity of the rod when the masses reach the end is equal in magnitude to the angular velocities of the masses and in the same direction.
To summarize:
(a) The rod is rotating at 0 rad/s when the masses have moved halfway to the end of the rod.
(b) The ratio of the final kinetic energy to the initial kinetic energy is 1:1 when the masses are halfway to the end of the rod.
(c) The rod is rotating with an angular velocity equal in magnitude and direction to the angular velocities of the masses when the masses reach the end.
To solve this problem, we need to apply the law of conservation of angular momentum. The total angular momentum of the system remains constant, assuming no external torques acting on it. Let's go through each part of the problem step by step:
(a) When the masses have moved halfway to the end of the rod, we can consider the initial and final states of the system. Initially, the rod is rotating at 10.0 rad/s, and the two masses are at the center of the rod (point of rotation). Finally, the two masses have moved halfway towards the end of the rod.
Using the conservation of angular momentum, we can write:
(I_rod * ω_i) + (2 * I_masses * 0) = (I_rod * ω_f) + (2 * I_masses * v_rel)
Here, I_rod represents the moment of inertia of the rod about its center, I_masses represents the moment of inertia of each mass, ω_i is the initial angular velocity of the rod, ω_f is the final angular velocity of the rod, and v_rel is the relative velocity of the masses with respect to the rod.
Since the two masses move halfway towards the end of the rod, v_rel becomes L/2 multiplied by the final angular velocity of the rod (v_rel = (L/2) * ω_f).
Simplifying the equation, we get:
(I_rod * ω_i) = (I_rod * ω_f) + (m * L^2 * ω_f / 4)
To find ω_f, we can solve for it:
(I_rod * ω_i) - (I_rod * ω_f) = (m * L^2 * ω_f / 4)
(I_rod - (m * L^2 / 4)) * ω_f = (I_rod * ω_i)
ω_f = (I_rod * ω_i) / (I_rod - (m * L^2 / 4))
Now, substitute the given values:
Mass of rod, M = 2.00 kg
Length of rod, L = 75.0 cm = 0.75 m
Mass of each mass, m = 0.403 kg
Initial angular velocity of rod, ω_i = 10.0 rad/s
The moment of inertia of a rod rotating about its center is (1/12) * M * L^2, so the moment of inertia of the rod, I_rod = (1/12) * M * L^2.
Plug in the values to calculate ω_f.
(b) When the masses are halfway to the end of the rod, the ratio of the final kinetic energy to the initial kinetic energy (Kf/Ki) can be calculated using the conservation of mechanical energy.
The initial kinetic energy (Ki) is given by (1/2) * I_rod * ω_i^2, where I_rod is the moment of inertia of the rod and ω_i is the initial angular velocity of the rod.
The final kinetic energy (Kf) is given by (1/2) * [I_rod + 2 * (1/2) * m * (L/2)^2] * ω_f^2, where I_rod is the moment of inertia of the rod, m is the mass of each mass, L is the length of the rod, and ω_f is the final angular velocity of the rod.
Substitute the given values and calculate the ratio Kf/Ki.
(c) When the masses reach the end of the rod, the final angular velocity of the rod, ω_f, can be calculated by again using the conservation of angular momentum. Use the same equation as in part (a) and substitute the given values to find ω_f.