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I am trying to better understand Lagrangian dynamics and am struggling to complete the following question:

A reel of thread of mass $m$ and radius $r$ is allowed to unwind under gravity, the upper end of the thread being fixed. Find the initial acceleration of the reel.

                                 Diagram of spool and thread

I believe there are three generalised co-ordinates here $(x,y,\theta)$, as shown in the diagram and the constraint that $x=0$ (as there is no acceleration in the $x$-direction). We therefore have kinetic energy:

$$T=\frac{1}{2}m\dot{y}^{2}+\frac{1}{2}m r^{2}\dot{\theta}^{2}y \cos(\theta)$$

And potential energy given by:

$$U=-mgy$$

We therefore can define the Lagrangian:

$$\mathcal{L}(y,\theta,t)=\frac{1}{2}m\dot{y}^{2}+\frac{1}{2}m r^{2}\dot{\theta}^{2}y \cos(\theta) + mgy$$

We therefore have the Euler-Lagrange equations:

\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) &= \frac{\partial \mathcal{L}}{\partial y} \\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) &= \frac{\partial \mathcal{L}}{\partial \theta} \end{align}

Expanding these we get:

$$m\ddot{y}=\frac{1}{2}mr^{2}\dot{\theta}^{2}\cos(\theta) + mg$$

And:

$$mr^{2}\ddot{\theta}y\cos(\theta)-mr^{2}\dot{\theta}^{2}\sin(\theta)+mr^{2}\dot{\theta}\dot{y}\cos(\theta)=0$$

Solving both of these simultaneously gives:

$$\ddot{\theta}=\frac{\dot{\theta}^{2}\tan(\theta)-\dot{y}\dot{\theta}}{y},\quad \ddot{y}=\frac{1}{2}\left(2g + r^{2}\dot{\theta}^{2}\cos(\theta)\right)$$

Which I am unable to solve to yield anything useful; and therefore am assuming I am on the wrong track.

I would be grateful for any pointers as to where I have misunderstood things.


Just to clarify what the issue is, given the expression above for $\ddot{y}$ and the boundary conditions $y_{0}=0$, $\dot{y}_{0}=0$, $\theta=0$ and $\dot{\theta}=0$ we get:

$$\ddot{y}_{0}=g$$

Which is what I expected, however the answer to the question states that $\ddot{y}_{0}=\frac{2g}{3}$.

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  • $\begingroup$ Isn't $d^2\theta/dt^2$ an acceleration? $\endgroup$ – Kyle Kanos Mar 4 '14 at 14:43
  • $\begingroup$ @KyleKanos Yes, but I thought the center of the object itself wasn't moving in the $x$-direction; only the orientation $\theta$ was varying? $\endgroup$ – Thomas Russell Mar 4 '14 at 14:45
  • $\begingroup$ @Shaktal: I'm not sure what you mean, how does $x$ come in here? $\endgroup$ – Kyle Kanos Mar 4 '14 at 14:52
  • $\begingroup$ @KyleKanos When you said isn't $\ddot{\theta}$ an acceleration, I assumed that was in response to my statement that there is no acceleration in the $x$-direction and therefore $x=0$ and hence why I left out $\frac{1}{2}mr^{2}\dot{\theta}^{2}\sin(\theta)$ in $T$? Is that not what you mean? $\endgroup$ – Thomas Russell Mar 4 '14 at 14:54
  • $\begingroup$ @Shaktal: Your question asks for initial acceleration and you've got two of them. They look okay (haven't double checked the work), but what have you tried with them? Saying "I am unable to solve to yield anything useful" is useless to those wanting to help. $\endgroup$ – Kyle Kanos Mar 4 '14 at 14:57
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I have managed to figure out what was going wrong. My mistake was not realising that the spool would have to be at the side of the string, which gives us the constraint: $\dot{y}=r\dot{\theta}$, we therefore have our Lagrangian:

$$\mathcal{L}=\frac{1}{2}m\dot{y}^{2}+\frac{1}{4}mr^{2}\dot{\theta}^{2}+mgy=\frac{3}{4}mr^{2}\dot{\theta}^{2}+mgr\theta$$

Therefore we have our Euler-Lagrange equation:

$$\frac{3}{2}mr^{2}\ddot{\theta}=mgr \implies \ddot{\theta}=\frac{2mgr}{3mr^{2}}=\frac{2g}{3r}$$

We therefore have:

$$\ddot{y}=r\ddot{\theta}=\frac{2}{3}g$$

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