So, I am having a lot of troubles visualizing the following system.

The point of suspension of a plane simple pendulum of mass m and length l is constrained to move along a horizontal track and is connected to a point on the circumference of a uniform flywheel of mass M and radius a through a massless connecting rod also of length a, as shown in the figure. The flywheel rotates about a center fixed on the track. Find the hamiltonian for the combined system and determine Hamilton's equation of motion.

Here is my idea so first of all I would construct the Lagrangian and then from it I will get the hamiltonian, so for this I need to determine the kinetic energy and the potential energy.

I am having troubles constructing the cartesian coordinates of this thing which the only thing that I need I know how to do everything else.

Is the following correct

$x = l\sin(\theta) - a(\omega)t$? where I measure the angle of the regular m pendulum with respect to vertical and the other one with respect to horizontal ? and $\omega$ is how fast the flywheel rotates.

enter image description here

  • $\begingroup$ The flywheel doesn't actually rotate, right? It oscillates around its centre point? $\endgroup$ – Gert Dec 6 '15 at 23:24
  • $\begingroup$ @Gert I think the flywheel is capable of rotating - but I imagine we are interested only in small angle deviations from equilibrium. $\endgroup$ – Floris Dec 6 '15 at 23:26
  • $\begingroup$ @Floris: capable of rotating but really just oscillating, is how I see it. $\endgroup$ – Gert Dec 6 '15 at 23:28
  • $\begingroup$ Are the arms $a$ and $l$ perpendicular to each other? Easier to visualise that way. $\endgroup$ – Gert Dec 6 '15 at 23:29
  • $\begingroup$ the flywheel rotates but a doesn't is rigid so it just moves in x and y direction it doesn't like deform or anything like that. $\endgroup$ – Illustionisttt. Dec 6 '15 at 23:29

This picture might help:

enter image description here

This assumes the flywheel is fixed on the track, the point of suspension can move along the track, but there is a rod from the point of suspension to the rim of the flywheel. This means there are two equal angles $\alpha$ and $x$, the displacement of the point of suspension from the furthest possible location, is given by

$$x = 2a(1-\cos\alpha)$$

Alternatively, if you use the center of the flywheel as your origin (might be more sensible) you find that

$$x = 2a\cos\alpha$$

Since this is a homework-like question, I will for now leave the rest up to you. Can you take it from here?

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  • $\begingroup$ yeah I understand now how to do it $\endgroup$ – Illustionisttt. Dec 7 '15 at 0:42
  • $\begingroup$ I am sorry I have a question what about the y-coordinates I got my x coordinate to be as follows $x = 2acos(\alpha) - lsin(\theta)$, where I didn't assume it is fixed, where $\theta$ above is the angle the vertical makes with the pendulum above. $\endgroup$ – Illustionisttt. Dec 8 '15 at 0:59
  • $\begingroup$ does the $\alpha$ y-component matter for the potential energy of small m ? $\endgroup$ – Illustionisttt. Dec 8 '15 at 1:14
  • $\begingroup$ never mind I get it now after I did demonstration of the above with two pens it doesn't matter and the angle of $\theta$ will fix the orientation of the potential energy of mass m. $\endgroup$ – Illustionisttt. Dec 8 '15 at 1:19
  • $\begingroup$ Exactly - because the pivot point is fixed on the track them potential energy only depends on $\theta$ $\endgroup$ – Floris Dec 8 '15 at 2:32

As both the flywheel and pendulum oscillate at the same frequency (same $\omega$), calculate the kinetic energy of both in function of $\omega$ and their inertial moments. Add them up, that gives you $K$.

Calculate potential energy of the pendulum in function of $\theta$. The flywheel's potential energy is independent of $\theta$, so leave it out. That gives you $U$.

Add $K$ and $U$ to get total energy, which in the absence of friction is constant (time invariant).

Derive that equation for $t$, apply the small angle approximation, rework a bit and obtain a very simple equation of motion, from which $\omega$ can be derived easily.

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