Solution to 1992 Problem 93

Let l be the length of the string. Then
\begin{align*}x &= - l \cos \theta \\y &= - l \sin \theta\end{align*}
implies that
\begin{align*}\dot{x} &=  l \dot{\theta} \sin \theta \\\dot{y} &= - l  \dot{\theta} \cos \theta\end{align*}
which implies that
\begin{align*}\ddot{x} &=  l \dot{\theta}^2 \cos \theta + l \ddot{\theta} \sin \theta \\\ddot{y} &=  l  \dot{\theta}^...
\begin{align}a^2 &= \ddot{x}^2 + \ddot{y}^2 \label{eqn:3} \\&= l^2 \dot{\theta}^4 + l^2 \ddot{\theta}^2 \label{eqn:4}...
So, we need to find \dot{\theta} and \ddot{\theta} as functions of \theta. For \dot{\theta}, this can be done by equating the kinetic energy that the mass has when its angle is \theta to the change in the mass's potential energy between its initial position and its position when its angle is \theta:
\begin{align*}m g l \sin \theta = \frac{1}{2} m l^2 \dot{\theta}^2\end{align*}
\setcounter{equation}{2}\begin{align}\dot{\theta}^2 = \frac{2g}{l} \sin \theta \label{eqn:1}\end{align}
In order to find \ddot{\theta}, we use Lagrange's equation
\begin{align*}\frac{\partial L}{\partial \theta} = \frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}}\end{align*}
The kinetic energy of the particle is
\begin{align*}T = \frac{1}{2}m l^2 \dot{\theta}^2\end{align*}
The potential energy of the particle is
\begin{align*}V = - l \sin \theta m g\end{align*}
Therefore the Lagrangian of the particle is
\begin{align*}L = \frac{1}{2}m l^2 \dot{\theta}^2 +  l \sin \theta m g\end{align*}
Using Langrange's equation, we find that
\setcounter{equation}{3}\begin{align}\ddot{\theta} = \frac{g}{l} \cos \theta \label{eqn:2}\end{align}
Plugging equations (3) and (4) into equation (1)-(2) gives
\begin{align*}a^2 &= g^2 \left(4 \sin \theta^2 + \cos^2 \theta \right) \\&= g^2 \left(3 \sin \theta^2 +1 \right) \end...
\begin{align*}\boxed{a = g \sqrt{3 \sin \theta^2 +1 }} \\\end{align*}
Therefore, answer (E) is correct.

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