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Homework Help: Solving for the eqs of motion for a Double Pendulum using a Lagrangian

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Two masses [itex] m_1 [/itex] and [itex] m_2 (m_1 ≠ m_2) [/itex] are connected by a rigid rod of length [itex] d [/itex] and of negligible
    mass. An extensionless string of length[itex] l_1[/itex] is attached to [itex] m_1 [/itex] and connected to a fixed point [itex] P [/itex].
    Similarly, a string of length [itex] l_2 (l_1 ≠ l_2) [/itex] connects [itex] m_2 [/itex] and [itex] P [/itex]. Obtain the equation describing the
    motion in the plane of [itex] m_1 [/itex], [itex] m_2 [/itex], and [itex] P [/itex] and find the frequency of small oscillations around the
    equilibrium position.

    2. Relevant equations

    (1) [itex]\frac{∂L}{∂q_j}-\frac{d}{dt}\frac{∂L}{∂\dot{q_j}}+\sum_kλ_k(t)\frac{∂f_k(q_j,t)}{∂q_j} = 0 [/itex]

    (2) [itex] s=Dn-m [/itex]

    3. The attempt at a solution
    Okay so the first thing I did in this problem was I tried to figure out how many coordinates I need. Each particle can move in 2 dimensions so [itex] D=2 [/itex]. There are 2 particles so [itex] n=2 [/itex]. There are 3 constraints:(a) [itex] x_1^2+y_1^2 = l_1^2 [/itex], (b) [itex] x_2^2+y_2^2 = l_1^2 [/itex], and (c) [itex] (x_2-x_1)^2+(y_2-y_1)^2 = d^2 [/itex]

    [itex] s=(2*2)-3 = 1 [/itex] However, I'll start of with the original four coordinates and work my way to only 1:

    The Lagrangian:
    [itex] L=T-U → T=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2] [/itex] and [itex] U=g[m_1y_1+m_2y_2] [/itex]
    Then [itex] L=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2]-g[m_1y_1+m_2y_2] [/itex]

    Now I can apply 2 out of my 3 constraints (a and b):
    [itex] x_1 = l_1sin(θ_1) [/itex]
    [itex] y_1 = l_1cos(θ_1) [/itex]
    [itex] x_2 = l_2sin(θ_2) [/itex]
    [itex] y_2 = l_2cos(θ_2) [/itex]

    Applying these transformations to the Lagrangian:

    [itex] L=\frac{1}{2}[m_1l_1^2\dot{θ_1}^2+m_2l_2^2\dot{θ_2}^2]-g[m_1l_1cos(θ_1)+m_2l_2cos(θ_2)] [/itex]

    Now the angular velocity [itex] \dot{θ} [/itex] is the same for both particles because the thin massless rod of length [itex] d [/itex] holds them together. So [itex] \dot{θ_1}=\dot{θ_2} [/itex]

    Applying this to L:

    [itex] L=\frac{1}{2}[m_1l_1^2+m_2l_2^2]\dot{θ_1}^2-g[m_1l_1cos(θ_1)+m_2l_2cos(θ_2)] [/itex]

    In addition:
    [itex] \dot{θ_1}=\dot{θ_2} → θ_1=θ_2+C → f = θ_1-θ_2-C=0 [/itex]

    Okay now I can use eq. (1):

    [itex] θ_1 [/itex]: [itex] \frac{∂L}{∂θ_1} = g[m_1l_1sin(θ_1)] [/itex], [itex] \frac{d}{dt}[\frac{∂L}{∂\dot{θ_1}}] = \frac{d}{dt}[[m_1l_1^2+m_2l_2^2]\dot{θ_1}] [/itex], and [itex] λ\frac{∂f}{∂θ_1} = λ [/itex]

    (3) [itex] g[m_1l_1sin(θ_1)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}+λ=0 [/itex]

    [itex] θ_2 [/itex]: [itex] \frac{∂L}{∂θ_2} = g[m_2l_2sin(θ_2)] [/itex], [itex] \frac{d}{dt}[\frac{∂L}{∂\dot{θ_2}}] = \frac{d}{dt}[0] [/itex], and [itex] λ\frac{∂f}{∂θ_1} = -λ [/itex]

    (4) [itex] g[m_2l_2sin(θ_2)]-λ=0 [/itex]

    From eq. 4 I know [itex] λ [/itex] in terms of [itex] θ_1 [/itex]:

    Using [itex] θ_1-C=θ_2 [/itex]; [itex] λ=g[m_2l_2sin(θ_1-C)] [/itex]

    Then eq. 3 becomes:

    [itex] g[m_1l_1sin(θ_1)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}+g[m_2l_2sin(θ_1-C)]=0 [/itex]

    [itex] g[m_1l_1sin(θ_1)]+g[m_2l_2[sin(θ_1)cos(C)-sin(C)cos(θ_1)]]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0

    [itex] [gm_1l_1+gm_2l_2cos(C)]sin(θ_1)-gm_2l_2sin(C)cos(θ_1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0

    Now there is the matter of [itex] C [/itex] to handle:

    If I plug in the coordinate transforms into constraint (c) I will arrive at the law of cosines:

    [itex] l_1^2+l_2^2-2l_1l_2cos(θ_1-θ_2) = d^2 [/itex] (keep in mind here that I have defined [itex] θ_1 < 0 [/itex] and [itex] θ_2>0 [/itex])

    From the law of cosines I get:

    [itex] θ_1 = θ_2+arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}) [/itex] so I have now found what C is:

    [itex] C = arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}) [/itex]

    Now plugging in [itex] C [/itex] into the new form of eq. 3:

    [itex] [gm_1l_1+gm_2l_2cos(arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}))]sin(θ_1)-gm_2l_2sin(arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}))cos(θ_1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0

    [itex] [gm_1l_1+gm_2l_2(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2})]sin(θ_1)-gm_2l_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{-2l_1l_2})cos(θ_1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0

    Okay so now I only want to consider small oscillations about the equilibrium position [itex] \Rightarrow θ_1\ll 1 [/itex]

    Then eq. 3 becomes the Linear ODE:

    [itex] g[m_1l_1-m_2(\frac{d^2-l_1^2-l_2^2}{2l_1})](θ_1)+gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1})(1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0

    Now this equation can be put into the form [itex] \ddot{θ}+ω_0^2θ=f(t) [/itex] where [itex] f(t) [/itex] is a constant driving force.

    [itex] \ddot{θ_1}+\frac{-g[m_1l_1-m_2(\frac{d^2-l_1^2-l_2^2}{2l_1})]}{m_1l_1^2+m_2l_2^2}θ_1=\frac{gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1})}{m_1l_1^2+m_2l_2^2}

    Now then [itex] ω_0^2 \equiv \frac{-g[m_1l_1-m_2(\frac{d^2-l_1^2-l_2^2}{2l_1})]}{m_1l_1^2+m_2l_2^2} [/itex] (frequency of oscillation about the equilibrium point)

    The equilibrium condition is given as [itex] θ_1(0) = \frac{f(t)}{ω_0^2} [/itex] because in equilibrium [itex] \ddot{θ_1(t=0)}=0 [/itex] In this case [itex] f(t) = \frac{gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1})}{m_1l_1^2+m_2l_2^2}

    The general solution to eq 3. is:

    (4) [itex] θ_1(t) = Acos(ω_0t)+Bsin(ω_0t)+θ_p(t) [/itex]

    [itex] \ddot{θ_p}+ω_0^2θ_p = gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1}) → θ_p = gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2ω_0^2l_1}) [/itex]

    [itex] A = θ_1(0)-θ_p [/itex] then eq. 4 becomes [itex] θ_1(t) = [θ_1(0)-θ_p]cos(ω_0)+Bsin(ω_0t)+θ_p [/itex]

    [itex] \dot{θ_1} = Bω_0cos(ω_0t) → \dot{θ_1(0)} = Bω_0 → B=\frac{ \dot{θ_1(0)}}{ω_0} [/itex]

    So the general solution is then:

    [itex] θ_1(t) = [θ_1(0)-θ_p]cos(ω_0t)+\frac{ \dot{θ_1(0)}}{ω_0}sin(ω_0t)+θ_p [/itex]

    Thank you in advance for helping me out in checking my work(:

    Attached Files:

  2. jcsd
  3. Mar 30, 2014 #2


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    What is the difference between and "extensionless string" and a "string" ?
  4. Mar 30, 2014 #3
    In the context of the problem it is simply emphasizing the point that the particle is constrained to move on a fixed radius [itex] l_i [/itex] from point
    [itex] P [/itex]
  5. Mar 30, 2014 #4


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    Does that mean that there is no difference between a rod and a string? (in this problem)

    If that is true, then the triangle (l1 ,l2 ,d) is a rigid triangle, and then the problem admits an almost trivial solution: a simple pendulum whose you need to find the parameters.
    If that is true, you might be able to check your solution directly from this point of view.

    However, I still have some doubts about the meaning of the question.
  6. Mar 30, 2014 #5
    Yes, I suppose I could set [itex] l_1= l_2 [/itex] and [itex] m_1 = m_2 [/itex] because in that limit I should be able to calculate the reduced mass [itex] \mu [/itex] and check that the equations of motion I came up with match that of the simple pendulum. However, the point of this homework is to exercise one's skill with using Lagrangian mechanics. I really want to make sure my setup in this regard is correct more so than the equations of motion per se. Thank you for the suggestion though(:
  7. Mar 30, 2014 #6


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    I haven't gotten through all of your manipulations. But I did notice a couple of things.

    1) In getting to equations (3) and (4), you should still be treating ##\theta_1## and ##\theta_2## as independent variables in the Lagrange multiplier method.

    So, ##\frac{\partial L}{\partial \theta_2} \neq 0##. [EDIT: As discussed in the following posts, this should have been ##\frac{\partial L}{\partial \dot \theta_2} \neq 0##]

    2) The small oscillation approximation does not mean that ##\theta_1## (or ##\theta_2##) is small. When the system hangs at rest in equilibrium, ##\theta_1## has some value ##\theta_1^0## which is not necessarily small. It's the deviation of ##\theta_1## from ##\theta_1^0## that will be small.

    Note that as a check, you can obtain the frequency of oscillation by considering the system as a physical pendulum and using the well-known formula for the frequency of a physical pendulum.
    Last edited: Mar 30, 2014
  8. Mar 30, 2014 #7
    In 1) were you referring to [itex] \frac{∂L}{∂\dot{θ_2}} [/itex] because I never got that [itex] \frac{∂L}{∂θ_2}= 0 [/itex] I think I'm still ok here though because 2 of my 3 constraints were simple coordinate transformations. What about [itex]\dot{θ_1}=\dot{θ_2}[/itex]? Is that correct? I got that relation 2 different ways. The first was from the law of cosines relation and the second was treating the two particles as having a constant separation [itex] d [/itex] in the plane.

    In 2) So basically I need to reformulate the solution to 3)?
  9. Mar 30, 2014 #8
    Ahhh yes I see, wow I can't believe I overlooked that [itex] ε [/itex] the angular displacement from [itex] θ_1(0) [/itex] is what is small. Thank you for pointing that out kind sir!
  10. Mar 30, 2014 #9


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    Yes, sorry. I should have written ##\frac{∂L}{∂\dotθ_2}\neq 0##

    As I understand it, when using the Lagrange multiplier method you shouldn't use your constraint conditions until after you've derived the equations of motion from the Lagrangian (treating ##\theta_1## and ##\theta_2## as independent variables).

    Yes. But you shouldn't use it until after you have derived the correct equations of motion.
  11. Mar 30, 2014 #10
    Last edited: Mar 31, 2014
  12. Mar 31, 2014 #11


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    Things look good to me all the way through equation (7). Then, I think you should write ##\theta_1(t) = \theta_1^0 + \epsilon(t)## where ##\theta_1^0## is a constant representing the equilibrium position for ##\theta_1##. Starting with equation (8), everywhere you have ##\theta_1## you should write ##\theta_1^0##.

    You can solve for ##\theta_1^0## in terms of ##C## by using equation (7) when the system is hanging in equilibrium so that ##\ddot \theta_1 = 0##.

    Near the end you have introduced ##\alpha##. Show that ##\alpha = 0##.
  13. Mar 31, 2014 #12
  14. Mar 31, 2014 #13


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    Looks good. You can show that your expression for ##\omega## reduces to what you expect if you treat the system as a physical pendulum.
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