Given an inverted pendulum system mounted on a trolley that...

Given an inverted pendulum system mounted on a trolley that can move along a horizontal path. The pendulum has a mass m and a length L, and the trolley has a mass M. Let x(t) represent the horizontal position of the trolley, 0(t) represent the angle of the pendulum with respect to the vertical axis, and the input is a force F(t) applied horizontally to the trolley. Draw the system, write the kinetic and potential energies of the system, and derive the equations of motion of the system using the Lagrange-Euler method.

Transcript

00:01 Ok we have a cart with a spring attached to it that can oscillate back and forth and there's a pendulum attached to the middle of that cart of mass m.

00:14 So we want to write out the lagrangian for this.

00:19 We have to start with the kinetic energy terms.

00:22 So for the mass m.

00:27 So the only, so i've got two generalized coordinates.

00:36 One of them is little x that tells us the position of the cart.

00:45 And it's measured from the equilibrium position of the cart which is a, so that makes it an inertial coordinate.

00:53 So for that the kinetic energy is just going to be half little m times x dot squared.

01:05 So for big m the kinetic energy is somewhat trickier because the theta, the theta generalized coordinate we want to use is not inertial.

01:19 And so we have to relate this to the inertial coordinates.

01:24 So i'm going to use x and y as the positions of the particle down here relative to an inertial frame.

01:43 Because that's what i need to calculate the kinetic energy.

01:47 So x is going to be little x plus l sine theta.

01:59 And then y will be just l cosine theta.

02:11 And so when i take big x dot we get little x dot plus l cos theta theta dot.

02:23 And y dot is l minus l sine theta theta dot.

02:33 Okay.

02:35 So when i calculate the kinetic energy for the bigger mass, mass big m, i get half big m times x dot squared capital x plus capital y dot squared.

02:55 So that's one half of big m and then i get little x dot plus l cos theta theta dot squared plus l squared sine squared theta theta dot squared.

03:16 And i can expand this and so i'm going to end up with little x dot squared plus 2l cos theta theta dot x dot.

03:32 And i get a theta dot squared times cos squared theta.

03:36 So that's just going to be sine squared plus cos sine squared is 1 l squared theta dot squared.

03:43 So there's my kinetic energy.

03:47 Okay.

03:47 Then i have two potential energy terms.

03:52 One is for the spring.

03:54 So i get one half kx squared for the spring.

04:00 And then for the gravity, usually we put u equals zero at the bottom of the swing.

04:12 So u for the gravitational potential, it's above that.

04:22 So it's l minus l cos theta.