# Where does the $(\ell + x)^2\dot\theta^2$ term come from in the Lagrangian of a spring pendulum?

I am reading some notes about Lagrangian mechanics. I don't understand equation 6.9, which gives the Lagrangian for a spring pendulum (a massive particle on one end a spring).

$$T = \frac{1}{2}m\Bigl(\dot{x}^2 + (\ell + x)^2\dot{\theta}^2\Bigr)\tag{6.9}$$

I don't understand where the component $(\ell + x)^2\dot{\theta}^2$ is coming from. If we say the $x$-component is radial and $y$ is tangential, so we have according to this $\vec{v}^2 = v_{x}^2 + v_{y}^2$, then $y = (\ell + x)\sin\theta$ by small angle approximation we have $y = (\ell + x)\theta$, but then if we choose this coordinate system then $V(x,y)$ equation doesn't make sense specifically the potential from gravity! If someone could shed some light into this that would be nice.

• $y = (l+x)\sin\theta$ is the usual vertical $y$ coordinate. We are interested here in the tangential velocity, which has magnitude $(l+x)\dot{\theta}$: radius times angular velocity. – Javier Jul 18 '15 at 2:49
• yeah but this doesn't makes sense then because if we choose that y-component to be like that then how come the potential coming from gravity is $mg(l + x)cos(\theta)?$ – Dude Jul 18 '15 at 3:26
• Because I made a mistake. If you look at how $\theta$ is defined, the vertical coordinate has $\cos\theta$ and the horizontal component has $\sin\theta$. – Javier Jul 18 '15 at 13:20

Velocities in the kinetic part of Lagrangian

The variable $\;x\;$, that represents the displacement of the string from its position at rest, has been replaced by the variable $\;s\;$ in order not to be confused with the coordinate $\;x\;$ of a Cartesian system.

The velocity of the particle $\:\mathbf{v}\:$ is analysed as follows

$$\mathbf{v}=\mathbf{v}_{s}+\mathbf{v}_{\theta} \tag{01}$$

where $\:\mathbf{v}_{s}\:$ the component along the string line and $\:\mathbf{v}_{\theta}\:$ that normal to it.

Now, $$v_{s}=\dfrac{d\left(\ell+s\right)}{dt}=\underbrace{\dot\ell}_{=0}+\dot{s}=\dot{s} \tag{02}$$

$$v_{\theta}=\left(\ell+s\right)\omega =\left(\ell+s\right) \dfrac{d\theta}{dt}=\left(\ell+s\right)\dot{\theta} \tag{03}$$

$$v^{2}=v_{s}^2 + v_{\theta}^2=\dot{s}^2 + (\ell + s)^2\dot{\theta}^2 \tag{04}$$

I'm not sure why you're talking about an $x$ and $y$ component of velocity when you're working in a polar coordinate system. Maybe you're confusing $x(t)$ (the extension of the spring as a function of time) with the Cartesian coordinate $x$. These are very different things. To understand what the radial component of the velocity is, assume the pendulum isn't swinging ($\dot{\theta}=0$). In this case you know the speed of the mass is $$\frac{d}{dt}(\ell+x)=\dot{x}$$ since $\ell$ is a constant. This gives the radial component of the kinetic energy $$T_{r}=\frac{1}{2}m\dot{x}^2$$

Now assume the pendulum is swinging but there is no radial motion ($\dot{x}=0$). We know that the kinetic energy of an object in circular motion is $$T_{\theta}=\frac{1}{2}mv^2$$ where $v$ is the tangential speed of the object and $m$ is its mass. In this case, we know the tangential speed is $$v=\dot{\theta}r$$ where $r$ is the radius of the circle traced out by the path of the object. Since we're considering the case with $\dot{x}=0$, we have a well defined constant radius given by $r=\ell+x$. This gives the radial kinetic energy $$T_{\theta}=\frac{1}{2}m\dot{\theta}^2(\ell+x)^2$$ The important thing to note is this kinetic energy relation holds even with a changing radius. The Lagrangian is a function of time anyway (although not explicitly in this case), so this isn't a problem. You can then take the kinetic component of the Lagrangian to be the sum of these kinetic energy relations to give $$L=\frac{1}{2}m\Bigl(\dot{x}^2 + (\ell + x)^2\dot{\theta}^2\Bigr)-V$$

In general you can write the kinetic energy of a free particle as:

$$T = \frac{1}{2} m \,\vec{v}\cdot\vec{v}$$

which holds whatever coordinate system you choose (could a physical quantity such the trajectory of a particle depend on the coordinate system that you choose?).

We can rewrite this: $$T = \frac{1}{2} m \,\vec{v}\cdot\vec{v} = m/2 \sum_{\mu ,\nu} v_{\mu}v_{\nu}\left( \vec{e}_\mu \cdot \vec{e}_\nu\right)$$ where $\vec{e}_\mu$ is the $\mu^{th}$ basis vector. In particular, remembering that in polar coordinates $\vec{e}_r \cdot \vec{e}_r= 1$, $\vec{e}_r \cdot \vec{e}_\theta= 0$ and $\vec{e}_\theta \cdot \vec{e}_\theta = r^2$ you get: $$T= m/2 \left( v_r^2 +r^2 v_\theta^2\right)$$ which is your expression if you call the $r$ coordinate $x$ and the $\theta$ coordinate $y$ and you make a radial translation of the origin of your coordinate system of a quantity $l$ in order to center it properly.