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GR9277 #44
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{44}
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?

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

Mechanics$\Rightarrow$}Lagrangians

The kinetic energy, in general, is given by $T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$. The potential energy is just $V=mgy$. The Lagrangian is given by $L=T-V=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-mgy$.

Now, given the constraint $y=ax^2$, one can differentiate it and plug it into the Lagrangian above to reexpress the Lagrangian in terms of just y, for example.

Differentiating, one has $\frac{d}{dt}(y=ax^2) \Rightarrow \dot{y}=a2x\dot{x} \Rightarrow \dot{x}=\dot{y}/(2ax)$. Square that to get $\dot{x}^2=\frac{\dot{y}^2}{4a^2x^2}=\frac{\dot{y}^2}{4ay}$, where one replaces the $x^2$ through the given relation $y=ax^2$.

Plug that back into the Lagrangian above to get exactly choice (A).

Alternate Solutions
 rrfan2011-11-06 12:45:34 The Lagrangian is $L=T-V$, where $T=\frac{1}{2}(\dot{x}^2+\dot{y}^2)$ and $V=-mgy$. Only choices (A) and (B) contain $-mgy$. Of these two, only (A) is possible because $T$ must be greater than or equal to $\frac{1}{2}\dot{y}^2$.Reply to this comment
rrfan
2011-11-06 12:45:34
The Lagrangian is $L=T-V$, where $T=\frac{1}{2}(\dot{x}^2+\dot{y}^2)$ and $V=-mgy$. Only choices (A) and (B) contain $-mgy$. Of these two, only (A) is possible because $T$ must be greater than or equal to $\frac{1}{2}\dot{y}^2$.
 RusFortunat2015-10-22 16:32:57 $V=mgy$
Quark
2011-10-26 15:17:39
This is probably a silly question but, why would one be wrong in choosing choice (E)? I actually picked the right answer (A) on the test but glancing over it again, (E) is also correct in the most general form. Is it not explicit enough? lol
 rrfan2011-11-06 13:14:00 the +mgy is wrong.
 livieratos2011-11-08 14:41:41 also if i remember correctly, since the two coordinates of each other the Lagrangian has to be dependent only on one of them and on its time derivative, not both... but i could be wrong. guess i should read lagrange and hamilton again :P
$null 2009-10-28 16:39:52 Reposted to $typo$ section...(with sexy LaTeX) There is a typo in Answer A, it should be $\dot{y}^2$, not $\ddot{y}^2$ tan 2009-10-14 23:52:17 There is a typo in Answer A, it should be y dot, not y double dot. mdornfe1 2008-11-06 16:58:28 This can be done by noticing that potential energy has to be -mgy. A and B are only answers with this property. Second the kinetic energy must always be positive 1-(4ay)^-1 is not always positive. So choice A is the answer.  his dudeness2010-09-04 13:42:09 well done, brah  TeamGandalf2011-04-01 18:12:40 The Lagrangian is equal to T-V. Why doesn't the negative from the Potential cancel the negative in the equation? Why isn't it L = T - (-mgy)= T + mgy?  Quark2011-10-26 15:20:37 @TeamGandalf The lagrangian is L = T - U. You can't have a negative potential energy... U=mgy not (-mgy). Poop Loops 2008-10-12 00:45:40 So what happens when y = 0, as it inevitably will when the particle comes back down?  segfault2009-09-04 12:14:26 (I realize I'm replying to a 1 year old post--this is for the general public). When y->0, the $1+\frac{1}{4ay}$ term will blow up but $\dot{y}=2ax\dot{x}$ will be zero, so L won't blow up. Perhaps if L was written in terms of $x$ it would have a nicer form... etano 2007-06-16 14:12:37 There is a typo in Answer A, it should be y dot, not y double dot. Post A Comment!  Username: Password: Click here to register. This comment is best classified as a: (mouseover) Mouseover the respective type above for an explanation of each type. Bare Basic LaTeX Rosetta Stone LaTeX syntax supported through dollar sign wrappers$, ex., $\alpha^2_0$ produces $\alpha^2_0$.
type this... to get...
$\int_0^\infty$ $\int_0^\infty$
$\partial$ $\partial$
$\Rightarrow$ $\Rightarrow$
$\ddot{x},\dot{x}$ $\ddot{x},\dot{x}$
$\sqrt{z}$ $\sqrt{z}$
$\langle my \rangle$ $\langle my \rangle$
$\left( abacadabra \right)_{me}$ $\left( abacadabra \right)_{me}$
$\vec{E}$ $\vec{E}$
$\frac{a}{b}$ $\frac{a}{b}$

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