Physics Problems & Solutions: Classical Mechanics

A particle of mass m on the Earth’s surface is confined to move on the parabolic curve y = ax², where y is up. Which of the following is a Lagrangian for the particle?

A. $L = \frac{1}{2} m\dot{y} ^2 \left (1+\frac{1}{4ay} \right )-mgy$

B. $L = \frac{1}{2} m\dot{y} ^2 \left (1-\frac{1}{4ay} \right )-mgy$

C. $L = \frac{1}{2} m\dot{x} ^2 \left (1+\frac{1}{4ay} \right )-mgx$

D. $L = \frac{1}{2} m\dot{x} ^2 \left (1+4a^2x^2 \right )+mgx$

E. $L = \frac{1}{2} m\dot{x} ^2 +\frac{1}{2} m\dot{y} ^2+mgy$

(GR9277 #44)

Solution:

Kinetic Energy, $T = \frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)$

Potential Energy, $U=mgy$

Lagrangian, $L=T-U=\frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)-mgy$

Given the curve y = ax²,

$\begin{aligned} \dot{y}&=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=2ax\dot{x}\\\rightarrow \dot{x}&=\frac{\dot{y}}{2ax} \\\dot{x}^2&=\frac{\dot{y}^2}{4a^2x^2}=\frac{\dot{y}^2}{4ay} \end{aligned}$

$\begin{aligned} L&=\frac{1}{2} m\left (\frac{\dot{y} ^2}{4ay}+\dot{y}^2 \right ) -mgy \\&=\frac{1}{2} m\dot{y} ^2\left (\frac{1}{4ay}+1 \right ) -mgy \end{aligned}$

Answer: A

Physics Problems & Solutions

Classical Mechanics - Lagrangian

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