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Equation of motion for pendulum with slender rod (energy method)

  1. Sep 8, 2011 #1
    hey guys, I have a question regarding how to get the potential energy for this. I can get the correct answer, but the solutions do a step that makes absolutely no sense to me, so hopefully someone leads me in the right direction :smile:

    1. The problem statement, all variables and given/known data
    Using energy method,

    derive the equation of motion for the pendulum which consists of the slender uniform rod of mass m and the bob of mass M. assume small oscillations, and neglect the radius of the bob.

    ***see attachment for diagram***
    2. Relevant equations

    [tex] I_o = \frac{1}{3}mL^2 + ML^2 [/tex] is the moment of inertia

    [tex] V = mgh [/tex]
    is the gravitational potential energy

    [tex] T = \frac{1}{2}I_o \omega^2 = \frac{1}{2}I_o \dot{\theta}^2 [/tex]
    is the kinetic energy

    [tex] \frac{d}{dt}(T+V)=0 [/tex]
    to get equation of motion

    3. The attempt at a solution

    My expression for V is this;

    [tex] V = -mg\frac{L}{2}cos(\theta)-MgLcos(\theta) [/tex]
    where I have taken the datum as being at the pivot acting postive up. This is different to the solution, which has written

    [tex] V_{solution} = mg\frac{L}{2}(1-cos(\theta))+ MgL(1-cos(\theta)) [/tex]
    I don't understand why they have 1 - cos(theta). Wouldn't you just want the cosine of the angle?

    Anyway, if I continue to the derivtive step (doing the small angle approximation), this doesn't seem to have an effect. So...
    [tex] \frac{d}{dt}(T+V)=\frac{d}{dt}\left ( \frac{1}{2}(M+\frac{m}{3})L^2\dot{\theta}^2 -mg\frac{L}{2}cos(\theta)-MgLcos(\theta)\right ) =0 [/tex]
    which yields, after alot of manipulation,
    [tex] \ddot{\theta}+\frac{3g(m+2M)}{2L(m+3M)}\theta = 0 [/tex]
    which is what the textbook says is correct.

    is the textbooks expression for the potential energy wrong in this case? or did I get lucky in getting the right answer? It doesn't make sense to me why they have 1 - cosine.

    Attached Files:

    Last edited: Sep 8, 2011
  2. jcsd
  3. Sep 8, 2011 #2


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    Homework Helper

    The change in height from when you displace it at the angle is L-Lcosθ or L(1-cosθ). Remember at the angle the length of the adjacent side is Lcosθ, but the original length was L, so change = L-Lcosθ

    Why it worked without the 1-cosθ was that you would end up differentiating 1-cosθ, which would be like this

    d/dt(1-cosθ) = d/dt(1) -d/dt(cosθ) = -d/dt(cosθ)

    so you'd still end up just differentiating cosθ wrt to t alone.
  4. Sep 9, 2011 #3
    so I need to be doing the CHANGE in height from the equilibrium position to some displaced position?
    I thought you only needed the potential AT the displaced position, ie you don't care about where it ends up only where its at when you do the energy stuff?

    I think why I was confused with this is because the textbook always has the datum at some convenient point so the potential for one position is zero.

    so in the formula
    [tex] \frac{d}{dt}(T+V)=0 [/tex]
    should it really be
    [tex] \frac{d}{dt}(\Delta T+\Delta V)=0 [/tex]
  5. Sep 9, 2011 #4


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    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You simply used a different zero point than what the solutions did. Your zero of potential energy is at the level of the pivot. In the solutions, there are two zeros, one for the rod and one for the mass. The zeros are where the mass and the center of mass are located when the pendulum is vertical. Neither method is incorrect.
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