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

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.

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.

Equation of motion along positive is

| (1) |

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

| (2) |

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

| (3) |

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

The above is equation of motion for .

To find equation of motion for we need one more equation, which is the moment equation . We can take moments around either the center of mass of the pendulum, or about the pivot where the bar is hinged to the cart. If we take moments around center of mass, we would end up with extra unknown . But if we take moment around pivot, then both 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 using this method. We just need to watch out here and make sure to use parallel axis theorem to find and to remember now that the term (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.

Now, taking moments around pivot gives

The above is the required equation of motion for . 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.

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

| (6) |

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. It important here to note something. When finding K.E. of physical pendulum, even though it is pivoted at and not at its center of mass, we must use in the above, and not in the above. We should not make the mistake and write

Now we find

Therefore the K.E. in (1) 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, 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

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

But

Hence becomes

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