I Equation of motion for a simple mechanical system

AI Thread Summary
The discussion focuses on the dynamics of a mechanical system consisting of a rod, a disk, and a collar, with specific velocities defined for each component based on the angle θ. The velocities of the collar and disk are derived from their respective positions and the generalized coordinate θ, while the translational and rotational velocities of the rod are also calculated. The total kinetic energy of the system is expressed as a sum of contributions from all components, including their translational and rotational motions. The main concern raised is whether the motion of the rod has been accurately accounted for in the analysis. Overall, the calculations and approach appear to be correct.
davidwinth
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I want to find an equation of motion for the following system through the use of the Lagrangian approach.
The system is shown below. It consists of a rod of length ##L## and mass ##m_b## connecting a disk of radius ##R## and mass ##m_d## to a collar of mass ##m_c## which is in turn free to slide without friction on a vertical and rigid pole. The disk rolls without slipping on the floor. The ends of the rod are attached such that they rotate without friction. I choose the angle ##\theta## between the horizontal and the rod as the generalized coordinate.The velocity of the center of mass of the collar is strictly vertical, and is given by: ##V{_c} = Lcos(\theta)\dot{\theta}##.

The velocity of the center of mass of the disk is strictly horizontal, and is given by ##V{_d}=-Lsin(\theta)\dot{\theta}##.

The translational velocity of the center of mass of the rod is obtained from its components as: ##V_{b} = \sqrt{\dot{x}_{b}^2 + \dot{y}_{b}^2} = \frac{L\dot{\theta}}{2}##.

The rotational velocity of the disk about its center of mass is given as: ##\omega_d = \frac{-Lsin(\theta)\dot{\theta}}{R}##.

The rotational velocity of the rod about its center of mass is given as: ##\omega_b = \dot{\theta}##.

Thus, the total kinetic energy for this system is the sum of all these contributions:

$$\frac{1}{2}m_{c}(Lcos(\theta)\dot{\theta})^2 + \frac{1}{2}m_{b}\left(\frac{L\dot{\theta}}{2}\right)^2 + \frac{1}{2}m_{d}\left(Lsin(\theta)\dot{\theta}\right)^2 + \frac{1}{2}I_{d}\left(\frac{Lsin(\theta)\dot{\theta}}{R}\right)^2 + \frac{1}{2}I_{b}(\dot{\theta})^2$$

Where the ##I_d## and ##I_b## are the moment of inertia about the centroid of the disk and rod, respectively.

My doubt is whether I have correctly accounted for the motion of the rod by decomposing it this way. Is there something I have neglected in considering the motion of the rod?
CollarSlip.png
 
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davidwinth said:
My doubt is whether I have correctly accounted for the motion of the rod by decomposing it this way. Is there something I have neglected in considering the motion of the rod?
So far, everything looks good to me.
 
Thank you!
 
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