GREPhysics.NET: GR9677 #68

GR9677 #68

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Mechanics $\Rightarrow$ }Lagrangians

The potential energy of the mass is obviously $U=mgs\cos\theta$ , and thus one eliminates choices (B) and (C). (To wit: $L=T-U$ , (C) has the wrong sign).

The translational part of the kinetic energy is easily just $1/2 m \dot{s}^2$ . The rotational part requires the calculation of the moment of inertia for a point particle, which is just $I=mr^2$ , where $r=s\sin\theta$ , in this case. Thus, the rotational kinetic energy is $1/2 m (s\sin\theta)^2 \omega^2$ . The only choice that has the right rotational kinetic energy term is choice (E).

Alternate Solutions

jmason86
2009-10-04 15:04:26

MOE:
As Yosun said: PE = mgscos $\theta$ . Eliminate (B) and (C)
Also as Yosun said: Translational KE = 1/2m $\dot{s}^2$ . Eliminate (D)
Only (A) and (E) remain.
Rotational KE = something... another term. (A) is only the Translational KE and the PE; it lacks a Rotational KE term. Eliminate (A).
(E) Remains.

Reply to this comment

Comments

ngendler
2015-10-23 05:06:16

Does no one else see a DOT over the m in choice (E)???

camcam
2019-09-12 16:14:14

It looks like a printing error, there are similar dots all over the page if you look closely. That one\'s just in an unfortunate spot, I guess.

Reply to this comment

nc
2014-10-16 15:48:23

Without actually calculating the Lagrangian, we know that kinetic energy term will depend on omega, so that eliminates A and D. We also know that L = T - U and that U = mgscos(theta). The only choice that makes sense, then, is E.

Reply to this comment

Sagan_fan
2013-05-23 15:08:36

The no-effort MOE: L = T - U

Here T has two components, rotational and translational, so we want an answer with 3 terms; Eliminate (A) and (D).

Since the potential should be subtracted, eliminate (C).

By inspection, U has an angle term, eliminating (B)

Reply to this comment

jmason86
2009-10-04 15:04:26

Reply to this comment

f4hy
2009-04-03 18:28:36

LIMITS!
I first eliminated C since the wrong sign and then B because no theta dependance.

From there I thought what would happen if $\theta =0$ . Well then it should be like a free falling particle so D is out. The spinning must give some kinetic energy so there goes A. E is the only one left.

Reply to this comment

r10101
2007-10-30 19:13:43

After eliminating (B) and (C) by the potential term, use variable dependency: T = T $(\omega)$ for sure, so (A) and (D) are both out as well.

p3ace
2008-04-11 10:25:59

This is a great observation, but if there were other choices with $\omega$ dependence in the middle term, one would have to solve more exactly.
I would use the expression for $\dot{r}}$ in spherical coordinates which should be memorized, and in cylindrical coord.'s as well.

p3ace
2008-04-11 10:44:23

I hit post prematurely. Sorry.rnAs I was saying, the expressions for velocity in the other coordinates are very handy for problems like this one because one can simply dot r vector with itself to get v^2 for kinetic energy. rnIn this case, The spherical symmetry lends itself perfectly to the problem because s is just r, and $\omega$ is $\phi$ '. Also, $\theta$ is fixed, so $\theta$ ' is zero. Then dot the velocity with itself and multiply by (1/2)m and Wah-Lah. You have it.rnI would type in the origianl expression for velocity in spherical coordinates but I'm not used to Latex and would rather be studying. I got it from Mechanics by Symon, and I've used it to solve numersous problems with cylindrical or spherical symmetry.

Reply to this comment

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone