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Monday, 8 August 2011

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

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