Physics GRE - #7

A particle of mass $m$ on the Earth's surface is confined to move on the parabolic curve $y=ax^2$, where $y$ is up. Which of the following is a Lagrangian for the particle?

• $L=\frac{1}{2}m\dot{y}^2\left(1+\frac{1}{4ay}\right)-mgy$
• $L=\frac{1}{2}m\dot{y}^2\left(1-\frac{1}{4ay}\right)-mgy$
• $L=\frac{1}{2}m\dot{x}^2\left(1+\frac{1}{4ax}\right)-mgx$
• $L=\frac{1}{2}m\dot{x}^2\left(1+4a^2x^2\right)+mgx$
• $L=\frac{1}{2}m\dot{y}^2+\frac{1}{2}m\dot{x}^2+mgy$

Solution :

Quick solution:
We know right away it is either the first choice or the second choice as $L=T-U$ and $U=mgy$. We know it must be the first choice as $T=\frac{1}{2}m(\dot{y}^2+\dot{x}^2)$, so we need a positive term $\left(+\frac{1}{4ay}\right)$ inside the brackets multiplied by $\dot{y}^2$.

Full solution: \[\dot{x}=\frac{d}{dt}\sqrt{\frac{y}{a}}=\dot{y}\sqrt{\frac{1}{4ay}}\] \[T=\frac{1}{2}m(\dot{y}^2+\dot{x}^2)=\frac{1}{2}m\dot{y}^2\left(1+\frac{1}{4ay}\right)\] \[U=mgy\] \[L=T-U=\frac{1}{2}m\dot{y}^2\left(1+\frac{1}{4ay}\right)-mgy\]