Exemple: Piston Connecting Rod

A disk of inertia Id is connected to an extremity of a bar with a distance of a of its center. This bar has length b and mass mb and in the other extremity is connected a piston of mass mp.

So, lets call (x, y) the position of the extremity of the bar connected to the disk, and (x+q, 0) the position of the piston.

Constraint functions

To solve this problem we are going to use three variables that we call x, y and q. Therefore, our variable vector X is (x, y, q).

As this problem has only one degree of freedom, that could be the angle theta of the disk, we need two constraint functions:

$\bold{g} = \begin{bmatrix} x^2+y^2-a^2 \\ y^2+q^2-b^2 \end{bmatrix}$

So, the gradient is

$\nabla \bold{g} = \begin{bmatrix} 2x & 0 \\ 2y & 2y \\ 0 & 2q \end{bmatrix}$

Energy

In this problem, we have three objects and we don't have any potential energy, only kinetic energy.

Disk's energy

If it rotates with an angular speed of w, it will have the energy of

$E_{disk} = \dfrac{1}{2}I_{d} \cdot \omega^2 = \dfrac{I_{d}}{2a^2}\cdot \left(\dot{x}^2+\dot{y}^2\right)$

Bar's energy

After some calculs, we get the energy like

$E_{bar} = \dfrac{1}{2}m_{b}\cdot \left(\dot{x}^2\ + \dfrac{1}{3}\dot{y}^2+\dfrac{1}{3}\dot{q}^2 + \dot{x}\dot{q} \right)$

Piston's energy

$E_{piston} = \dfrac{1}{2}m_{p} \cdot (\dot{x}+\dot{q})^2 = \dfrac{1}{2}m_{p}\left(\dot{x}^2 + 2\dot{x}\dot{q}+\dot{q}^2\right)$

Total energy

Summing all the energies we get

$\begin{align*} 2E & = \left(\dfrac{I_{d}}{a^2}+m_{b}+m_{p}\right)\dot{x}^2 + \left(\dfrac{I_{d}}{a^2} + \dfrac{1}{3}m_{b}\right)\dot{y}^2 + \left(\dfrac{1}{3}m_{b}+m_{p}\right)\cdot \dot{q}^2 \\ & + \left(0\right) \cdot \dot{x}\dot{y} + \left(m_{b}+2m_{p}\right) \dot{x}\dot{q} + \left(0\right)\cdot \dot{y}\dot{q} \end{align*}$

Putting it in a matricial form we get:

$2E = \begin{bmatrix} \dot{x} & \dot{y} & \dot{q} \end{bmatrix} \begin{bmatrix} \frac{I_{d}}{a^2}+m_{b}+m_{p} & 0 & \frac{1}{2}m_{b}+m_{p} \\ 0 & \frac{I_{d}}{a^2} + \frac{1}{3}m_{b} & 0 \\ \frac{1}{2}m_{b}+m_{p} & 0 & \frac{1}{3}m_{b}+m_{p} \end{bmatrix} \begin{bmatrix} \dot{x} \\ \dot{y} \\ \dot{q} \end{bmatrix} = \left[\bold{\dot{X}}\right]^{T} \left[M\right] \left[\bold{\dot{X}}\right]$

Lagrangian

As we know, we have the lagrangian operator:

$\bold{L} = \dfrac{d}{dt}\left(\dfrac{\partial E}{\partial \bold{\dot{X}}}\right) - \dfrac{\partial E}{\partial \bold{X}}$

Which, with the matricial notation of energy we get only:

$\bold{L} = \left[M\right] \cdot \left[\bold{\ddot{X}}\right]$

Lagrange multipliers

Once we get the lagrangian, and knowing that there is no external force, we have:

$\left[\bold{L}\right] + \left[\nabla \bold{g}\right] \cdot \left[\lambda\right] = \left[0\right]$

Which gives us

$\begin{bmatrix}\left[M\right] & \left[\nabla \bold{g}\right] \end{bmatrix} \cdot \begin{bmatrix} \bold{\ddot{X}}\\ \left[\lambda\right] \end{bmatrix} = \left[0\right]$

Adding the constraint functions we get

$\begin{bmatrix} \left[M\right] & \left[\nabla \bold{g}\right] \\ \left[\nabla \bold{g}\right]^{T} & \left[0\right] \end{bmatrix} \cdot \begin{bmatrix} \bold{\ddot{X}}\\ \left[\lambda\right] \end{bmatrix} = \begin{bmatrix} \left[0\right] \\ -\alpha^2 \cdot \bold{g} - 2\alpha \cdot \nabla \bold{g} \cdot \bold{\dot{X}} - \nabla \nabla \bold{g}:\bold{\dot{X}}\bold{\dot{X}} \end{bmatrix}$

Which gives, in our case

$\begin{bmatrix} M_{11}& 0 & M_{13} & 2x & 0 \\ 0 & M_{22} & 0 & 2y & 2y \\ M_{13}& 0 & M_{33} & 0 & 2q \\ 2x & 2y & 0 & 0 & 0 \\ 0 & 2y & 2q & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{q} \\ \lambda_{1} \\ \lambda_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ -\alpha^2 \left(x^2+y^2-a^2\right) - 4\alpha \left(x\dot{x}+y\dot{y}\right) - 2\left(\dot{x}^2+\dot{y}^2\right) \\ -\alpha^2 \left(y^2+q^2-b^2\right) - 4\alpha \left(y\dot{y}+q\dot{q}\right) - 2\left(\dot{y}^2+\dot{q}^2\right) \end{bmatrix}$

Solving numerically

We could use this matrix, with the initial conditions. But we are going to inverse the matrix Mexp to show that we can make the solution faster.

$\left[ M \right]^{-1} = \dfrac{1}{M_{11}M_{33}-M_{13}^2} \begin{bmatrix} M_{33} & 0 & -M_{13} \\ 0 & \dfrac{M_{11}M_{33}-M_{13}^2}{M_{22}} & 0 \\ -M_{13} & 0 & M_{11} \end{bmatrix}$

So, as show in this link, we can find the inverse of Mexp using the inverse of M and the gradient of g.

TODO

Finish and show inverse of Mexp

Provide feedback

Saved searches

Use saved searches to filter your results more quickly