# Lagrangian approach to spinning thread reel

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.

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 $\ddot{y}_{0}=\frac{2g}{3}$.

• Isn't $d^2\theta/dt^2$ an acceleration? Commented Mar 4, 2014 at 14:43
• @KyleKanos Yes, but I thought the center of the object itself wasn't moving in the $x$-direction; only the orientation $\theta$ was varying? Commented Mar 4, 2014 at 14:45
• @Shaktal: I'm not sure what you mean, how does $x$ come in here? Commented Mar 4, 2014 at 14:52
• @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? Commented Mar 4, 2014 at 14:54
• @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. Commented Mar 4, 2014 at 14:57

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