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I know this may seem like a homework problem at first, but please bear with me...

In this problem, we have two masses sliding without friction in a horizontal and vertical track, connected by a rigid massless link. Choosing $\theta$ as the single generalized coordinate, we can derive a single equation of motion, e.g. via the Lagrangian method. enter image description here

The generalized force in context of Lagrangian dynamics is (often) defined as $$\begin{equation}\tag{1}Q_i = \sum_{j=1}^m \frac{\partial \mathbf{r}_j}{\partial q_i}\cdot \mathbf{F}_j\end{equation}$$ where $i$ is the index of the generalized coordinate, $m$ is the number of applied forces and $\mathbf{r}_j$ is the position vector to the $j$th force.

Question-1: Using equation (1), how is the applied force $F$ (or it's moment) included as a generalized force when our single coordinate is $q_1=\theta$, and what expression will the generalized force take ?


From a course at my university, the generalized force (for 2D systems) is defined $$\begin{equation}\tag{2}Q_i = \sum_{j=1}^m \frac{\partial \mathbf{r}_j}{\partial q_i}\cdot\mathbf{F}_j+\sum_{j=1}^p\frac{\partial \theta_j}{\partial q_i}M_j\end{equation}$$ where $m,p$ are nr of applied forces/torques

For this system, choosing $q_1=\theta$, we get $Q_1=M$.   The answer should be $Q_1 = FL\sin(\theta)$, which is kind of intuitive (but not quite)

Question-2: Please clarify the $Q=FL\sin(\theta)$  part. Would that also be the case if e.g. $m_2$ had no mass ?


Edit: I have derived the equations of motion to be $$ (m_1L^2 s_{\theta}^2 + m_2 L^2 c_{\theta}^2)\ddot{\theta} - L^2s_{\theta}c_{\theta}(m_1-m_2)\dot{\theta}^2 + m_2gLc_{\theta} = Q$$ where $c_\theta=\cos(\theta)$, $s_\theta=\sin(\theta)$, and with kinetic and potential energies $$E_k=\frac{1}{2}m_1L^2s_{\theta}^2\dot{\theta}^2 + \frac{1}{2}m_2L^2c_{\theta}^2\dot{\theta}^2$$ $$E_p = m_2gLs_{\theta}$$

Perhaps this is flawed, and the force $F$ should be part of the potential ?

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  • $\begingroup$ You have to start with the position vector r1 r2 $\endgroup$ – Eli Jul 25 at 4:52
  • $\begingroup$ $Q_1 = FL\sin(\theta)$ is a torque arm. Where is your Lagrangian? You can use any coordinates you want. $M$ is undefined. Since $q_{1}$ is an angle, then $Q_{1}q_{1}$ has to be torque. If the $q_{1}$ was a position, then $Q_{1}q_{1}$ would be a force. But your $Q_{1}$ does have dimensions of work. $\endgroup$ – Cinaed Simson Jul 25 at 6:35
  • $\begingroup$ $Q_1=FL\sin(\theta)$ is the moment from $F$. If the system was rigid with mass center at $m_2$, then this quantity makes sense to me. Yes, $M$ is undefined ! $\endgroup$ – Ronny Landsverk Jul 25 at 9:27
  • $\begingroup$ I don't want to answer my own question, but I guess you can do $$\mathbf{r}_1 = \left(x_0+L(1-\cos(\theta))\right)\hat{\mathbf{i}}$$ and then we get $Q_1=FL\sin(\theta)$ $\endgroup$ – Ronny Landsverk Aug 14 at 13:07

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