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Elastic Pendulum with Newton's equations of motion

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    A pendulum with a mass m hanging on a elastic bug rigid massless rod which may swing in the xy-plane. The pivot point is the origin of the coordinate system. The force acting on the pendulum is the sum of force of an elastic central force directed towards the origin, and gravity, which by choice of the orientation of the coordinate system, points in the positive y-direction. Friction of any kind (air, in pivot point) can be disregarded.

    Newtons's equation of motion:
    equation 1: [itex]mx''=-\frac{\partial V}{\partial x}[/itex]
    equation 2: [itex]my''=-\frac{\partial V}{\partial y}-mg[/itex]

    Gravity: [itex]g=9.81 m/s^2[/itex]
    Potential energy of the rod: [itex]V(r)=0.5*k*(r-L)^2[/itex], where [itex]r=\sqrt{x^2+y^2}[/itex]
    Length: [itex]L=1 m[/itex] (length of rod in absence of external force)
    We see from these equations that the solution does not depend separately on [itex]k[/itex] and [itex]m[/itex], but only the combination [itex]w^2=\frac{k}{m}[/itex].
    2. Relevant equations
    a)
    Here we shall on look at the pendulum moving in vertical direction ([itex]x(t)=0[/itex]). I need to derive the equation of motion in y-direction (in other words I need to show of to get equation 2).
    But I'm having problem figure out how I can get it like equation 2. More in attempt of solution.

    b) Compute the equilibrium of length [itex]L_r[/itex] for this case.

    c) Show that the solution of the equation of motion is an oscillation, up and down about the equilibrium with angular frequency [itex]w[/itex]

    3. The attempt at a solution
    a)
    So the pendulum will act like a elastic spring with a mass. The two forces acting on the pendulum in each direction. The gravity force which we can write [itex]F_g=mg[/itex]. And the force because of the potential energy which we can write [itex]F_u=-\frac{\partial V}{\partial y}[/itex].
    I know we can use Newton's second law: [itex]\sum F=ma[/itex]. And [itex]y''=a[/itex]. But I'm having trouble with the how to set it up. From the problem description I read that [itex]F_g[/itex] is positive and [itex]F_u[/itex] is negative. This gives the the sum of forces [itex]F_g-F_u=my''[/itex]. Inserting gives us: [itex]mg-(-\frac{\partial V}{\partial y})=my''[/itex]. So I get both forces positive, but it should be negative. What am I missing?

    b) I know I can find the equilibrium of length by setting [itex]my''[/itex] to 0. So I get [itex]0=-\frac{\partial V}{\partial y}-mg[/itex]. So after I derive [itex]-\frac{\partial V}{\partial y}[/itex], should I solve [itex]0=-\frac{\partial V}{\partial y}-mg[/itex] with respect to [itex]y[/itex] and insert [itex]L=1[/itex] or with the respect to [itex]L[/itex]? Which would give me a solution with something of [itex]y[/itex].

    c) I guess I need to solve this as a second order differential equation. With the equation being: itex]y''+0*y'+\frac{\partial V}{\partial y}=0[/itex].
    Is this correct? Not sure how to proceed to show the solution.

    Sorry for a long post. I really hope I have showed you that I have tried, but I really need help. And not sure if posted in the right sub-forum.
     
  2. jcsd
  3. Nov 1, 2016 #2

    ehild

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    With y pointing downward, the force of gravity is mg in equation 2. [itex]my''=-\frac{\partial V}{\partial y}+mg[/itex]
    It is not negative, but opposite to the change of the length. Fu=-k(y-L)
    NO, it should be [itex]F_g+F_u=my''[/itex]
     
  4. Nov 1, 2016 #3
    So the formula for equation in motion for y-direction ([itex]my''=-\frac{\partial V}{\partial y}-mg[/itex]) that was written in the problem description is wrong based on the choice of the direction of y?
    It should be [itex]my''=-\frac{\partial V}{\partial y}+mg[/itex] ?
    I just need to get this confirmed.

    So [itex]F_u[/itex] goes in the same direction as [itex]F_g[/itex] since both are positive? Sorry for asking stupid question. But I'm not familiar with potential energy and the force created by it. All I know is that the force created by potential energy is [itex]-\frac{\partial V}{\partial y}[/itex].
     
  5. Nov 1, 2016 #4

    ehild

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    Yes.
    my" = mg +Fu, the sum of gravity and the elastic force of the spring.
    No, the elastic force is -dV/dy. And how is the elastic potential V related to y?
    If you have difficulties with the potential, remember Hook's Law. The spring force is opposite to the change of length of the spring. If the mass stretches the spring, the length of the spring increases, y-L is positive, so the force is negative. In general, the force a spring exerts on a body attached to it is F=-k(y-L). That force is negative, if y>L and positive, if x<L.

    upload_2016-11-1_15-0-26.png
     
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