 GR 8677927796770177 | # Login | Register

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 , where y is up. Which of the following is a Lagrangian for the particle?

Mechanics}Lagrangians

The kinetic energy, in general, is given by . The potential energy is just . The Lagrangian is given by .

Now, given the constraint , 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 . Square that to get , where one replaces the through the given relation .

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

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...