# Conservation of Energy on Pendulum

So I've been working on tons of these pendulum problems and they consistently get me stuck. My question this time is:

Consider a nonlinear pendulum. Show that the sum of the potential energy ( $mgy$ where $y$ is the vertical distance of the pendulum above its natural position ) and the kinetic energy ( $mv^{2}/2$, where $v$ is the speed of the mass ) is a constant. Hint: Suppose a mass m located at $\left(x,y\right)$ is acted upon by a force field, $\vec{F}$ ( i.e., $m\,{\rm d}^{2}\vec{x}/{\rm d}t^{2} = \vec{F}$ ). The kinetic energy is defined as $${1 \over 2}\,m\left[% \left({\rm d}x \over {\rm d}t\right)^{2} + \left({\rm d}y \over {\rm d}t\right)^{2} \right]$$ it again equals $mv^{2}/2$. The potential energy is defined as $$-\int_\vec{x_{1}}^{\vec{x}}\vec{F}\cdot{\rm d}\vec{s}$$ where $\vec{x_{1}}$ is any fixed position.

I'm not totally sure how the hint helps but I'm completely lost on this. I have some idea as I know that as the pendulum swings up it's potential energy will increase but it's kinetic energy will decrease thereby balancing the overall $E$ to be a constant energy value, thus energy is conserved. But I don't know how to show that. Any ideas ?.

• I find this exercise to be a little backwards. First and foremost, there is the principle of conservation of energy: $\frac{dE_{\mathrm{total}}}{dt}=0$. So unless there is reason to suspect a third form of energy present here, the result requires no calculation. Also, where do the equations defining kinetic energy and potential energy come from? They are a consequence of conservation of energy. So that's another reason it feel backwards to use those equations to deduce that energy is conserved. – alex.jordan Feb 9 '14 at 18:58

If this is a pendulum confined to a plane, with length $l$, local gravitational acceleration $g$, and polar coordinate $\theta$, then its equation of motion is $$\frac{d^2\theta}{dt^2}=-\frac{g}{l}\sin\theta.$$ The standard trick is to multiply both sides by $d\theta/dt$: $$\frac{d\theta}{dt}\frac{d^2\theta}{dt^2}=-\frac{g}{l}\frac{d\theta}{dt}\sin\theta.$$ Now note that the left hand side can be expressed as $$\frac{d\theta}{dt}\frac{d^2\theta}{dt^2}=\frac{1}{2}\frac{d}{dt}\left(\frac{d\theta}{dt}\right)^2$$ and the right hand side as $$-\frac{g}{l}\frac{d\theta}{dt}\sin\theta=-\frac{g}{l}\frac{d}{dt}\left(-\cos\theta\right).$$ Putting back these expressions in terms of derivatives, we have $$0=\frac{d}{dt}\left(\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2-\frac{g}{l}\cos\theta\right).$$ Integrating with respect to $t$ and multiplying through by $m$, the mass of the pendulum bob, and $l^2$, we get $$\frac{1}{2}ml^2\left(\frac{d\theta}{dt}\right)^2-mgl\cos\theta=\textrm{constant}.$$ Now you can recognize these pieces as the kinetic and potential energies. It's not an accident. If you search "first integral of motion", you'll find more details on the general phenomenon.
Strictly speaking, you are assuming that the tension of the string is always perpendicular to the velocity; if this wasn't true, the string may be doing some work. You have Newton equations: $$\mathbf a= m\mathbf g+\mathbf T.$$ Dot product by the velocity $\mathbf v$: $$\dfrac{\text d}{\text d t}\frac{mv^2}{2}= {\mathbf a}\cdot {\mathbf v}=m\mathbf g\cdot {\mathbf v }+\mathbf T \cdot \mathbf v=-mg\frac{\text dy}{\text dt},$$ since the second product vanishes and $g$ has only a (negative) $y$ component.
Let $s(t)=\left(x(t), y(t)\right)$ be a trajectory of a particle of mass $m$ in the given force field. Then according to $F=ma$ this trajectory satisfies $F(s)= m \,\ddot{s}$. The total energy as function of $t$ is $$\begin{eqnarray} E(t)&=&\tfrac{1}{2}m||\dot{s}(t)||^2-\int_{s(0)}^{s(t)}\langle F(s), ds\rangle\\ &=&\tfrac{1}{2}m||\dot{s}(t)||^2-\int_0^t\langle F(s(u)), \dot{s}(u)\rangle du. \end{eqnarray}$$
Differentiating this in the variable $t$ gives $$\dot{E}(t)= m\langle \ddot{s}(t),\dot{s}(t)\rangle-\langle F(s(t)),\dot{s}(t)\rangle=\langle m\,\ddot{s}(t)-F(s(t)),\dot{s}(t)\rangle=0$$
and so the total energy $E$ is constant in $t$.