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January 3, 2018 compiled on — Wednesday January 03, 2018 at 07:41 AM

This report shows how to find the 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. It is useful to solve the same problem when possible using both methods as this will help verify the answer.

There are two degrees of freedom. The coordinate and the coordinate. Hence we need to find two equations of motion, one for each coordinate.

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.

Equation of motion along positive is

| (1) |

Equation of motion along positive is not needed since cart does not move in vertical direction. We see that to find equation for we just need to determine , since that is the only unknown in (1). will be found from the physical pendulum equation as is shown below.

We see now that the equation of motion along positive is

| (2) |

This gives us the we wanted to plug in (1). Equation (1) now becomes

Hence

| (3) |

The above is equation of motion for .

To find equation of motion for we take moments around C.G. of the rigid pendulum, using counter clock wise as positive. This gives

We know from (2). We know need to just find . This is found from resolving forces in the vertical direction for the pendulum free body diagram giving

Plugging (2) and (5) into (4) to eliminate , then (4) simplifies to

Therefore

| (6) |

The above is the required equation of motion for . Equations (3,6) are coupled and have to be solved numerically since they are nonlinear or small angle approximation can be used in order to simplify these two equations and to solve them analytically.

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

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

| (7) |

Where is K.E. of cart due to its linear motion, and is K.E. of physical pendulum due to its translation motion of its center of mass, and is K.E. of physical pendulum due to its rotational motion. Now we find

Therefore the K.E. from (7) becomes

Taking zero potential energy 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 which is given by

Hence the Lagrangian is

There are two degrees of freedom: and . The generalized forces in for are given by and the generalized force for is . Equation of motions are now found. For

Therefore

Which is the same result as Newton method found above in (3). Now we find equation of motion for

But

Hence becomes

Therefore

Which is the same ODE (6) above given by Newton’s method.