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Finding equations of motion for pendulum on moving cart

Nasser M. Abbasi

September 18, 2017 compiled on — Monday September 18, 2017 at 10:42 PM [public]

Contents

1 Introduction
2 Newton’s Method
 2.1 FBD for cart
 2.2 FBD for pendulum
3 Lagrange method

1 Introduction

This report shows how to find equations of motion of a rigid bar pendulum (physical pendulum) on a moving cart as shown in the following diagram using both Newton’s method and the energy (Lagrangian) method.


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Figure 1: Pendulum on moving cart

2 Newton’s Method

In the above, the angular motion is such that it is positive in counter clock wise, which is the standard. The first step is to make free body diagram (FBD). One for the cart and one for the physical pendulum and equate each FBD to the kinematics diagrams in order to write down the equations of motion.

2.1 FBD for cart


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Figure 2: Pendulum on moving cart

Equation of motion along positive x  is

− kx − c˙x+ F +  Px = M ¨x
(1)

Equation of motion along positive y  is not needed. But it is given here for illustration

Py + N − M g = M  ¨y = 0
(2)

We see that to find equation for ¨x  we just need to determine Px  , since that is the only unknown in (1). Px  will be found from the physical pendulum equation as is shown below.

2.2 FBD for pendulum


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Figure 3: Pendulum on moving cart

We see now that the equation of motion along positive x  is

− Px = m ¨x− m  L¨θ sinθ − m L-˙θ2cosθ
               2          2
(3)

This gives us the Px  we wanted to plug in (1) in order to find equation of motion for ¨x  . Equation (1) becomes

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The above is equation of motion for ¨x  .

To find equation of motion for ¨θ we need one more equation, which is the moment equation Iθ¨= τ  . We can take moments around either the center of mass of the pendulum, or about the pivot o  where the bar is hinged to the cart. If we take moments around center of mass, we would end up with extra unknown Py  . But if we take moment around pivot, then both Px,Py  do not appear, which is good. So we will do that. In this case also, we see that equation (2) above was not needed, since we do not need to find Py  using this method. We just need to watch out here and make sure to use parallel axis theorem to find Io = Icg + m (L-)2
              2   and to remember now that the term m ¨x  (which is due to the transfer acceleration from the cart), will produce a reaction force  acting on the pendulum in opposite direction, will also now contribute a positive torque. The following diagram illustrates this.


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Figure 4: Pendulum on moving cart

Now, taking moments around pivot gives

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The above is the required equation of motion for θ¨ . Equations (4,5) are coupled and have to be solved numerically since they are nonlinear. Small angle approximation can be applied if needed to simplify these two equations and to solve them analytically.

3 Lagrange method

The first step in using Lagrange method is to make a velocity diagram to each object. This is shown below


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Figure 5: Pendulum on moving cart

From the velocity diagram above we see that the kinetic energy of the system is

T = 1-M ˙x2 + 1mv2 +  1Icg ˙θ2
    2        2       2
(6)

Where 1M x˙2
2   is K.E. of cart due to its linear motion, and 1mv2
2   is K.E. of physical pendulum due to its translation motion of its center of mass, and 1   ˙2
2Ic.gθ   is K.E. of physical pendulum due to its rotational. It important here to note something. When finding K.E. of physical pendulum, even though it is pivoted at o  and not at its center of mass, we must use 12 Icg ˙θ2   in the above, and not 12Io˙θ2   in the above.  We should not make the mistake and write

         1       1       1
Twrong = -M  ˙x2 +--mv2 + --Ioθ˙2
         2       2       2

Now we find v

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Therefore the K.E. in (1) becomes

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Taking zero potential energy V  as the horizontal level where the pendulum is attach to the cart, then P.E. comes from only spring extension and change of vertical position of center of mass of pendulum, given by

V = mg L-sinθ + 1kx2
        2       2

Hence the Lagrangian Γ  is

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There are two degrees of freedom: x  and θ  . The generalized forces in for x  are given by Q  =  F − c˙x
  x  and the generalized force for θ  is Q  = 0
 θ  . Equation of motions are now found. For x

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Which is the same result as Newton method found above in (4). Now we find equation of motion for θ

d-∂-Γ   ∂Γ-
dt ∂ ˙θ − ∂θ = 0

But

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Hence  d-∂Γ − ∂Γ = 0
dt∂˙θ   ∂θ  becomes

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Which is the same ODE (5) above given by Newton’s method.