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### Cart and pendulum

The exercise of this particular examination of the cart and pendulum problem is to look at a very simple analysis of the experimental condition of a human on a moving platform. The platform then becomes the cart, and the human subject the pendulum. Furthermore, this analysis contains only one degree of freedom, that being the pendulum person's center of mass. The trajectory of the cart is specified by the experimental protocol.

## Kinematics

The kinematics of this problem are rather straightforward.  There is a pendulum that has one rotational DOF (degree of freedom) specified by the coordinate θ and a cart that has one translational DOF specified by x.  The pendulum has length L and mass m.  In addition, the joint at point J has a rotational spring K and damper b that apply torque at the pin.

We then define a coordinate system and examine the movement of important points of the model.  The position to the CoM (center of mass) can be written using the following vector equation:

$\bar{\mathbf{r}}=x-L\sin\theta\:\hat{\mathbf{i}}+L\cos\theta\:\hat{\mathbf{j}}$
Using the position vector we can determine the velocity of the CoM by taking the first derivative:

$\bar{\mathbf{v}}=\frac{d\bar{\mathbf{r}}}{dt}=\dot{x}-L\dot{\theta}\cos\theta\:\hat{\mathbf{i}}-L\dot{\theta}\sin\theta\:\hat{\mathbf{j}}$
Like-wise, acceleration is the derivative of the velocity, so the CoM acceleration can be defined as:

$\bar{\mathbf{a}}=\frac{d\bar{\mathbf{v}}}{dt}=\ddot{x}-L\ddot{\theta}\cos\theta+L\dot{\theta}^2\sin\theta\:\hat{\mathbf{i}}-L\ddot{\theta}\sin\theta-L\dot{\theta}^2\cos\theta\:\hat{\mathbf{j}}$
Note that in our kinematic relations about the CoM, we are interested in the CoM of the pendulum, and the x terms are defined quantities prescribed by how we move the cart.  This is important to note, as prescribing the motion of the cart changes how we solve this system.  We are really only interested in the motion of the pendulum.

## Kinetics

We are now interested in the forces acting on the system.  We can draw a free-body diagram for the entire system, which elucidates the action of gravity on the CoM and the ground-reaction forces that support the cart:

It may be pointed out that in this analysis the cart is not given any mass.  Since we are prescribing the motion of the cart we deal with the cart as a moving "frame" with no inherent mass.  Furthermore, this diagram doesn't give us enough information, so we move on to a diagram that focuses on the pendulum:

We now look at the external forces acting on the pendulum.  We have the forces in the X and Y directions as well as moments.  The sum of forces in the X and Y directions gives us:

$\Sigma F_x = R_x$
$\Sigma F_y = R_y - m g$
It is interesting to sum moments about the pin joint, which gives us:

$\Sigma M_J = mgL\sin\theta-M$
Where we define M as:
$M = b\dot{\theta}+k\theta$

## Dynamics

Combining what we know from the kinematics and kinetics we can now solve the equations of motion to determine important quantities about the system.  How does the CoM move when subjected to perturbations?  What are the ground-reaction forces?

In order to do this we need to connect how the external forces interact with the constrained motion of the system.  We can do this through applying conservation of linear and angular momentum.  Equating the change in linear momentum to the rectilinear external forces yields two equations:

$ma_x=R_x$

$ma_y=R_y-mg$
But by knowing these equations can we determine the motion of the pendulum?  It doesn't appear so!  However, all is not lost as we still have conservation of angular momentum up our sleeve.  First, we need to define angular momentum for the pendulum.  Since we took moments about point J, we will also want to define the angular momentum about the same point.  Angular momentum about a point is defined as:

$\bar{\mathbf{H}}_J = \bar{\mathbf{H}}_C + \bar{\mathbf{r}}_{JC} \times m\bar{\mathbf{v}}$
Note that the linear momentum term is referring to the CoM (defined above), HC is the angular momentum about the CoM and rJC is the vector:

$\bar{\mathbf{r}}_{JC} = -L\sin\theta \: \hat{\mathbf{i}} +L\cos\theta \: \hat{\mathbf{j}}$
Giving the pendulum an angular momentum of:

$\bar{\mathbf{H}}_J = mL^2\dot{\theta} -mL\dot{x}\cos\theta \: \hat{\mathbf{k}}$
The change in angular momentum about an arbitrary point is then equal to the moments about that point, defined as follows:

$\dot{\bar{\mathbf{H}}}_J + \bar{\mathbf{v}}_J \times m\bar{\mathbf{v}}= \bar{\mathbf{M}}_J$
It should be noted that the velocity vJ is written as:
$\bar{\mathbf{v}}_J = \dot{x} \hat{\mathbf{i}}$
Using the above equations to equate the change in angular momentum to the torques about point J gives the following equation:

$mL^2\ddot{\theta}-mL\ddot{x}\cos\theta=mgL\sin\theta-M$
Now we have our equation of motion that will describe how the pendulum moves. It can be rewritten with the substitution of the moment M in a more solvable form as:

$\ddot{\theta}=\frac{g\sin\theta}{L}+\frac{\ddot{x}\cos\theta}{L}-\frac{b\dot{\theta}+k\theta}{mL^2}$
Neglecting damping and looking at a linearized form of the above equation we find the following:

$\ddot{\theta} = \frac{g\theta + \ddot{x}}{L} - \frac{k}{mL^2}\theta$
Now looking back at the acceleration of the CoM in the X direction we find the following:

$a_x=\ddot{x}-L\ddot{\theta}\cos\theta+L\dot{\theta}^2\sin\theta$
And when substituting in angular acceleration we get:

$a_x=\sin\theta^2\ddot{x}-g\cos\theta\sin\theta+L\dot{\theta}^2\sin\theta+\left(b\dot{\theta}+k\theta\right)\frac{\cos\theta}{mL}$
Which when linearized and neglecting damping can be written as:

$a_x = \left( \frac{k}{mL} - g \right) \theta$
The point of this is then to show that the acceleration of the CoM in the X direction is dependent on the geometry, mass and stiffness of the pendulum system for a given perturbation.  The following applet helps show that the maximum acceleration reached by the pendulum is dependent on the stiffness of the spring:

ċ
pendulum_with_spring.m
(1k)
Jeff Bingham,
Aug 30, 2009, 8:23 PM