Hamiltonian of two rotating and oscillating masses

In summary: I'll work on drawing a picture :)It is a vector: the vector from the current mass position to the point on the spring where it is attached. That is e.g. r_1-b for the first mass. The point on the spring for that particular mass is at r_1 from the origin, and the point on the spring where the second mass is attached is at r_2 from the origin, so that the distance between both points is r_2-r_1 .This is a drawing of the situation, I think. I have added a constraint that the distance between the two masses is constant. This is not stated in the problem, but it seems logical to me
  • #1
Wavefunction
99
4

Homework Statement



A massless spring of length [itex] b [/itex] and spring constant [itex] k [/itex] connects two particles of masses [itex] m_1 [/itex] and [itex] m_2 [/itex].
The system rests on a smooth table and may oscillate and rotate.

a) Determine the Lagrange's equations of motion.

b) What are the generalized momenta associated with any cyclic coordinates?

Hint: With the right choice of generalized coordinates, 3 of the 4 should be cyclic. If the force
between the masses is a central force, some of the coordinates we used earlier in the course to
describe central-force motion might be useful.


c) Determine Hamilton's equations of motion.

Homework Equations



eq(1) [itex] \frac{∂L}{∂q_j}-\frac{∂}{∂t}\frac{∂L}{∂\dot{q_j}}=0 [/itex]

eq(2) [itex] H=\sum_ip_i\dot{q_i}-L = T+U[/itex]

The Attempt at a Solution



Part A

Step 1) find the Lagrangian

[itex] T = \frac{m_1}{2}[\dot{r_1}^2+r_1^2\dot{θ_1}^2]+\frac{m_2}{2}[\dot{r_2}^2+r_2^2\dot{θ_2}^2] [/itex]

[itex] U = \frac{k}{2}[(r_1-b)^2+(r_2-b)^2] [/itex]

[itex] L= T-U = \frac{m_1}{2}[\dot{r_1}^2+r_1^2\dot{θ_1}^2]+\frac{m_2}{2}[\dot{r_2}^2+r_2^2\dot{θ_2}^2]-\frac{k}{2}[(r_1-b)^2+(r_2-b)^2] [/itex]

Step 2) Use eq(1) to determine Lagrange's eqs of motion:

[itex] \frac{∂L}{∂r_1} = m_1r_1\dot{θ_1}^2-k(r_1-b) [/itex]

[itex] \frac{d}{dt}\frac{∂L}{∂\dot{r_1}} = \frac{d}{dt}[m_1\dot{r_1}] =m_1\ddot{r_1} [/itex]

eq(3) [itex] m_1r_1\dot{θ_1}^2-k(r_1-b)-m_1\ddot{r_1}= 0 [/itex]

[itex] \frac{∂L}{∂θ_1} = 0 [/itex]

[itex] \frac{d}{dt}\frac{∂L}{∂\dot{θ_1}} = \frac{d}{dt}[m_1r_1^2\dot{θ_1}] = m_1[2r_1\dot{r_1}\dot{θ_1}+r_1^2\ddot{θ_1}] [/itex]

eq(4) [itex] -m_1[2r_1\dot{r_1}\dot{θ_1}+r_1^2\ddot{θ_1}]=0 [/itex]

Going through the same steps for [itex] m_2 [/itex]

eq(5) [itex] m_2r_2\dot{θ_2}^2-k(r_2-b)-m_1\ddot{r_2}= 0 [/itex]

eq(6) [itex] -m_2[2r_2\dot{r_2}\dot{θ_2}+r_2^2\ddot{θ_2}]=0 [/itex]

Now I can apply the given constraints: the system is only allowed to rotate on the frictionless table top and oscillate.

eq(a) [itex] θ_1 + π = θ_2 [/itex] (This basically means that spring can only extend/compress along the radial direction)

eq(b) [itex] r_1 = a^2r_2 [/itex] (Since the system can't translate radially this condition along with eq(a) ensures that the particles oscillating.)

From both of these constraints: [itex] \dot{θ_1}=\dot{θ_2} [/itex], [itex] \ddot{θ_1}=\ddot{θ_2} [/itex], [itex] \dot{r_1}=a^2\dot{r_2} [/itex], and [itex] \ddot{r_1}=a^2\ddot{r_2} [/itex].

Part B

They are constants (e.g [itex] \frac{d}{dt}[m_1r_1^2\dot{θ_1}] = 0 [/itex] )

Part C

eq(7) [itex] \dot{q_i} = \frac{∂H}{∂p_i} [/itex]

eq(8) [itex] \dot{p_i} = -\frac{∂H}{∂q_i} [/itex]

The Hamiltonian is:

[itex] H = \frac{m_1}{2}[\dot{r_1}^2+r_1^2\dot{θ_1}^2]+\frac{m_2}{2}[\dot{r_2}^2+r_2^2\dot{θ_2}^2] + \frac{k}{2}[(r_1-b)^2+(r_2-b)^2] [/itex]

For [itex] r_1 [/itex]:

eq(7) yields: [itex] \dot{r_1} = \frac{p_1}{m_1} [/itex]

eq(8) yields: [itex] \dot{p_1} = m_1r_1\dot{θ_1}+k(r_1-b) [/itex]

For [itex] r_2 [/itex]

eq(7) yields: [itex] \dot{r_2} = \frac{p_2}{m_2} [/itex]

eq(8) yields: [itex] \dot{p_2} = m_2r_2\dot{θ_2}+k(r_2-b) [/itex]

For [itex] θ_1 [/itex] (L, in the following parts stands for angular momentum, NOT the Lagrangian)

eq(7) yields: [itex] \dot{θ_1} = \frac{L_1}{m_1r_1^2} [/itex]

eq(8) yields: [itex] \dot{L_1} = 0 [/itex]

For [itex] θ_2 [/itex]

eq(7) yields: [itex] \dot{θ_2} = \frac{L_2}{m_2r_2^2} [/itex]

eq(8) yields: [itex] \dot{L_2} = 0 [/itex]

Thank you in advance for checking my work(:
 
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  • #2
There is something strange with your U. It should be translation invariant: if I move both masses over the same vector, U should not change. On the other hand, ##k## does not appear, which is not credible.

Your choice of generalized coordinates is also somewhat unpractical. Would you also choose those if e.g. the spring were a stiff rod ?
 
  • #3
There is also something weird about the problem formulation: the system may oscillate and rotate, but not translate ? So what exactly is the constraint ?
 
  • #4
BvU said:
There is something strange with your U. It should be translation invariant: if I move both masses over the same vector, U should not change. On the other hand, ##k## does not appear, which is not credible.

Your choice of generalized coordinates is also somewhat unpractical. Would you also choose those if e.g. the spring were a stiff rod ?


Could you go into a little more detail as to how my U is not translation invariant? Also k appears in my equations of motion. Also I would still choose polar coordinates if it were a stiff rod because the two masses could still rotate. Is there another coordinate choices that makes more sense? Polar coordinates make the most sense to me because the two masses are rotating about a central point and oscillating in a radial direction.
 
  • #5
If I calculate U for e.g. ##r_1=10, r_2 = 20, b = 9## I get 61 k (sorry, I missed that k, so it IS present in the U expression -- disregard that comment). Do it for ##r_1=20, r_2 = 30, b = 9## I get something else, when I think it should give the same result.

The spring energy depends on how much it is stretched or compressed. That is NOT a vector like ##r_1-b, r_2 - b## .

Could you elaborate on what you mean with your choice of polar coordinates ? Where is the origin ?
Perhaps a drawing ?

And what do you mean when you say "the two masses could still rotate" ?
 
  • #6
BvU said:
If I calculate U for e.g. ##r_1=10, r_2 = 20, b = 9## I get 61 k (sorry, I missed that k, so it IS present in the U expression -- disregard that comment). Do it for ##r_1=20, r_2 = 30, b = 9## I get something else, when I think it should give the same result.

The spring energy depends on how much it is stretched or compressed. That is NOT a vector like ##r_1-b, r_2 - b## .

Could you elaborate on what you mean with your choice of polar coordinates ? Where is the origin ?
Perhaps a drawing ?

And what do you mean when you say "the two masses could still rotate" ?

As you should get a different [itex] U [/itex] for those different radial positions because your second calculation corresponds to extending the spring by 20 radial units( therefore [itex] U [/itex] should increase).

The origin is at the center of spring. What I mean is that the two masses attached to the spring rotate ( an angular displacement ) and oscillate (along a radial direction). What coordinates do you know that involve those two things?
 
  • #7
Wavefunction said:
As you should get a different [itex] U [/itex] for those different radial positions because your second calculation corresponds to extending the spring by 20 radial units( therefore [itex] U [/itex] should increase).

The origin is at the center of spring. What I mean is that the two masses attached to the spring rotate ( an angular displacement ) and oscillate (along a radial direction). What coordinates do you know that involve those two things?

also [itex] r_i-b [\itex] is not a vector.
 
  • #8
Wavefunction said:
[itex] U = \frac{k}{2}[(r_1-b)^2+(r_2-b)^2] [/itex]

Consider an imaginary circle of radius b centered at the origin of your coordinate system. Suppose the two masses happen to be on the circle at opposite ends of a diameter. What would your expression for U reduce to? Would that be correct?

--------

The hint suggests a judicious choice of coordinates. Note that the configuration of the system can be defined in terms of the position vector of the CM, ##\small \vec{R}_{cm}##, and the vector ##\vec{r} = \vec{r}_2-\vec{r}_1## which gives the position of m2 relative to m1.

[EDIT: Although the problem only mentions oscillations and rotations, it seems to me that you are also meant to allow overall translational motion. Otherwise, you wouldn't need 4 coordinates as implied in the problem.]
 
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  • #9
The center of the spring is not a good choice; it is not nailed to the table or anything.

(if m1=m2 something like the center of the spring might be a reasonable choice ... hint hint).

For this problem, you can assume the spring can only exercise a longitudinal force along it's axis.
 
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  • #10
BvU said:
The center of the spring is not a good choice; it is not nailed to the table or anything.

(if m1=m2 something like the center of the spring might be a reasonable choice ... hint hint).

For this problem, you can assume the spring can only exercise a longitudinal force along it's axis.


We went over the problem in class so I get what both of you are saying now I need to account for the whole system moving relative to some inertial frame I'll post my recalculation when I get home later today
 
  • #11
Wavefunction said:

Homework Statement



A massless spring of length [itex] b [/itex] and spring constant [itex] k [/itex] connects two particles of masses [itex] m_1 [/itex] and [itex] m_2 [/itex].
The system rests on a smooth table and may oscillate and rotate.

a) Determine the Lagrange's equations of motion.

b) What are the generalized momenta associated with any cyclic coordinates?

Hint: With the right choice of generalized coordinates, 3 of the 4 should be cyclic. If the force
between the masses is a central force, some of the coordinates we used earlier in the course to
describe central-force motion might be useful.


c) Determine Hamilton's equations of motion.


Homework Equations



eq(1) [itex] \frac{∂L}{∂q_j}-\frac{∂}{∂t}\frac{∂L}{∂\dot{q_j}}=0 [/itex]

eq(2) [itex] H=\sum_ip_i\dot{q_i}-L = T+U[/itex]

The Attempt at a Solution



Okay so the first thing I want to do is determine the number of degrees of freedom [itex] s = DN-M [/itex] since both particles are free to move in the two dimensional plane free of constraints ( and there are 2 particles) [itex] s=(2)(2)-(0) = 4 [/itex] so I need four coordinates to describe the system.

First I want to consider the motion of the system with respect to an inertial reference frame:

Let [itex] \vec{R}[/itex] be the vector that position vector that specifies the center of mass of the system's position.

Let [itex] \vec{r}=\vec{r_2}-\vec{r_1} [/itex] where [itex] \vec{r_2} [/itex] and [itex] \vec{r_1} [/itex] specify the positions of [itex] m_2 [/itex] and [itex] m_1 [/itex] respectively. Also [itex] \vec{r} [/itex] is the radial position [itex] m_2 [/itex] relative to [itex] m_1 [/itex]

[itex] \vec{R} \equiv \frac{1}{M}[m_1\vec{r_1}+m_2\vec{r_2}] [/itex] now subbing in for [itex] \vec{r_2} → \vec{R} = \frac{1}{M}[m_1\vec{r_1}+m_2(\vec{r}+\vec{r_1})] = \vec{r_1}+\frac{m_2}{M}\vec{r} [/itex]

So [itex] \vec{r_1} = \vec{R}-\frac{m_2}{M}\vec{r} [/itex] and [itex] \vec{r_2} = \vec{R}+\frac{m_1}{M}\vec{r} [/itex]

Also [itex] \vec{R} = (X,Y) [/itex] with transforms [itex] x_2=x = rcos(θ) [/itex] and [itex] y_2=y=rsin(θ) [/itex]

So [itex] T=\frac{M}{2}[\dot{X}^2+\dot{Y}^2]+\frac{μ}{2}[(\dot{r}cos(θ)-r\dot{θ}sin(θ))^2+(\dot{r}sin(θ)+r\dot{θ}cos(θ))^2] = \frac{M}{2}[\dot{X}^2+\dot{Y}^2]+\frac{μ}{2}[(\dot{r}^2+r^2\dot{θ}^2] [/itex] where [itex] μ \equiv \frac{m_1m_2}{m_1+m_2} [/itex]

[itex] U=\frac{k}{2}(r-b)^2 [/itex]

Since [itex] U [/itex] has no dependence on any generalized velocity and the coordinate transforms did not explicitly involve time. [itex] \Rightarrow H=T+U [/itex]

Using Hamilton's equations: [itex] \dot{p_j} = -\frac{∂H}{∂q_j} [/itex] and [itex] \dot{q_j} = \frac{∂H}{∂p_j} [/itex] and [itex] p_j = \frac{∂L}{∂\dot{q_j}} [/itex]

So [itex] p_{X} = M\dot{X} [/itex] , [itex] p_{Y} = M\dot{Y} [/itex] , [itex] p_r = μ\dot{r} [/itex], [itex] p_θ = μr^2\dot{θ} [/itex]

then [itex] H [/itex] becomes:

[itex] H =\frac{1}{2M}[p_X^2+p_Y^2]+\frac{1}{2μ}[p_r^2]+\frac{1}{2μr^2}[p_θ^2] +\frac{k}{2}(r-b)^2 [/itex]

For [itex] X [/itex]: [itex] \dot{X} = \frac{p_X}{M} [/itex] and [itex] \dot{p_X}=0 [/itex]

For [itex] Y [/itex]: [itex] \dot{X} = \frac{p_Y}{M} [/itex] and [itex] \dot{p_Y}=0 [/itex]

For [itex] r [/itex]: [itex] \dot{r} = \frac{p_r}{μ} [/itex] and [itex] \dot{p_r}=-k(r-b) [/itex]

For [itex] θ [/itex]: [itex] \dot{θ} = \frac{p_θ}{μr^2} [/itex] and [itex] \dot{p_θ}=0 [/itex]

Okay so my take-away(s) from this problem are that unless the particles are constrained I should consider all possible motions with respect to an inertial reference frame. I should also check my Hamilton's equation to make sure they make sense because looking at my old ones compared to the ones I just got I was way off. Thank you again to everyone who helped me!
 

FAQ: Hamiltonian of two rotating and oscillating masses

What is the Hamiltonian of two rotating and oscillating masses?

The Hamiltonian of two rotating and oscillating masses is a mathematical expression that describes the total energy of a system consisting of two masses that are both rotating and oscillating.

How is the Hamiltonian of two rotating and oscillating masses calculated?

The Hamiltonian is calculated by summing the kinetic and potential energies of the two masses. The kinetic energy is the energy of motion, while the potential energy is the energy stored in the position of the masses.

What factors affect the Hamiltonian of two rotating and oscillating masses?

The Hamiltonian is affected by the masses of the two objects, their velocities, the distance between them, and the strength of the forces between them. It is also influenced by external factors such as friction and air resistance.

What is the significance of the Hamiltonian in physics?

The Hamiltonian is an important concept in physics, particularly in the study of classical mechanics and quantum mechanics. It is used to describe the dynamics of a system and can be used to predict the behavior of the system over time.

How does the Hamiltonian change over time for a system of two rotating and oscillating masses?

The Hamiltonian can change over time if there are external forces acting on the system, such as friction or air resistance. However, in an isolated system where there are no external forces, the Hamiltonian remains constant, indicating that the total energy of the system is conserved.

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